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lv2ergyt | maths | probability | probability-distribution-of-a-random-variable | <p>If the mean of the following probability distribution of a radam variable $$\mathrm{X}$$ :</p>
<p><style type="text/css">
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<col style="width: 90px"/>
<col style="width: 91px"/>
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</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$\mathrm{X}$$</th>
<th class="tg-baqh">0</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">4</th>
<th class="tg-baqh">6</th>
<th class="tg-baqh">8</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$\mathrm{P(X)}$$</td>
<td class="tg-baqh">$$a$$</td>
<td class="tg-baqh">$$2a$$</td>
<td class="tg-baqh">$$a+b$$</td>
<td class="tg-baqh">$$2b$$</td>
<td class="tg-baqh">$$3b$$</td>
</tr>
</tbody>
</table></p>
<p>is $$\frac{46}{9}$$, then the variance of the distribution is</p> | [{"identifier": "A", "content": "$$\\frac{581}{81}$$\n"}, {"identifier": "B", "content": "$$\\frac{566}{81}$$\n"}, {"identifier": "C", "content": "$$\\frac{151}{27}$$\n"}, {"identifier": "D", "content": "$$\\frac{173}{27}$$"}] | ["B"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 422px">
<colgroup>
<col style="width: 69px">
<col style="width: 69px">
<col style="width: 71px">
<col style="width: 71px">
<col style="width: 71px">
<col style="width: 71px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$X$$</th>
<th class="tg-baqh">0</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">4</th>
<th class="tg-baqh">6</th>
<th class="tg-baqh">8</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$P(X)$$</td>
<td class="tg-baqh">$$a$$</td>
<td class="tg-baqh">$$2a$$</td>
<td class="tg-baqh">$$a+b$$</td>
<td class="tg-baqh">$$2b$$</td>
<td class="tg-baqh">$$3b$$</td>
</tr>
</tbody>
</table></p>
<p>$$\begin{aligned}
& \text { Mean }=\sum x_i P\left(x_i\right) \\
& \frac{46}{9}=4 a+4 a+4 b+12 b+24 b \\
& \frac{46}{9}=8 a+40 b \\
& \frac{23}{9}=4 a+20 b \\
& 36 a+180 b=23 \quad \text{.... (1)}
\end{aligned}$$</p>
<p>Sum of probability is 1</p>
<p>$$\Rightarrow 4 a+6 b=1 \quad \text{... (2)}$$</p>
<p>$$\begin{aligned}
& \text { Solving (1) and (2) } \\
& a=\frac{1}{12}, b=\frac{1}{9} \\
& \sigma^2=\sum x_i^2 P\left(x_i\right)-\left(\sum x_i P\left(x_i\right)\right)^2 \\
& =4 \times 2 a+16(a+b)+36(2 b)+64(3 b)-\left(\frac{46}{9}\right)^2 \\
& =8(a+2(a+b)+9 b+24 b)-\left(\frac{46}{9}\right)^2 \\
& =8(3 a+35 b)-\left(\frac{46}{9}\right)^2 \\
& =8\left(\frac{3}{12}+\frac{35}{9}\right)-\left(\frac{46}{9}\right)^2 \\
& =8\left(\frac{149}{36}\right)-\left(\frac{46}{9}\right)^2=\frac{566}{81} \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv5gs180 | maths | probability | probability-distribution-of-a-random-variable | <p>Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables $$X$$ and $$Y$$ respectively denote the number of blue and yellow balls. If $$\bar{X}$$ and $$\bar{Y}$$ are the means of $$X$$ and $$Y$$ respectively, then $$7 \bar{X}+4 \bar{Y}$$ is equal to ___________.</p> | [] | null | 17 | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
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<table class="tg" style="undefined;table-layout: fixed; width: 232px">
<colgroup>
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<col style="width: 47px">
<col style="width: 44px">
<col style="width: 46px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$X$$</th>
<th class="tg-baqh">3</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">1</th>
<th class="tg-baqh">0</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$Y$$</td>
<td class="tg-baqh">0</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">3</td>
</tr>
</tbody>
</table></p>
<p>$$\begin{aligned}
& \bar{X}=\sum X p(X) \\
& \bar{Y}=\sum Y p(Y) \\
& P(X=3)=P(Y=0)=\frac{{ }^5 C_3 \cdot C_0}{{ }^9 C_3}=\frac{{ }^5 C_2}{{ }^9 C_3}=\frac{5}{42} \\
& P(X=2)=P(Y=1)=\frac{{ }^5 C_2 \cdot C_1}{{ }^9 C_3}=\frac{10}{21} \\
& P(X=1)=P(Y=2)=\frac{{ }^5 C_1 \cdot C_2}{{ }^9 C_3}=\frac{5}{14} \\
& P(X=0)=P(Y=3)=\frac{{ }^5 C_0 \cdot C_3}{{ }^9 C_3}=\frac{4}{84}=\frac{1}{21} \\
& \bar{X}=3 \times \frac{5}{42}+2 \times \frac{10}{21}+\frac{5}{14}+0 \times \frac{1}{21}=\frac{15+40+15}{42}=\frac{70}{42} \\
& \bar{Y}=0 \times \frac{5}{42}+1 \times \frac{10}{21}+2 \times \frac{5}{14}+3 \times \frac{1}{21}=\frac{20+30+6}{42}=\frac{56}{42} \\
& 7 \bar{X}+4 \bar{Y}=17
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-morning-shift |
lv7v3quo | maths | probability | probability-distribution-of-a-random-variable | <p>From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $$X$$ denote the number of defective items in the sample. If the variance of $$X$$ is $$\sigma^2$$, then $$96 \sigma^2$$ is equal to __________.</p> | [] | null | 56 | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 592px">
<colgroup>
<col style="width: 88px">
<col style="width: 114px">
<col style="width: 115px">
<col style="width: 137px">
<col style="width: 138px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$x$$</th>
<th class="tg-baqh">0</th>
<th class="tg-baqh">1</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">3</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$P(x)$$</td>
<td class="tg-baqh">$$<br>\frac{{ }^7 C_5}{{ }^{10} C_5}=\frac{1}{12}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{C_4 \cdot{ }^3 C_1}{{ }^{10} C_5}=\frac{5}{12}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{{ }^7 C_3 \cdot{ }^3 C_2}{{ }^{10} C_5}=\frac{5}{12}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{{ }^7 C_2 \cdot{ }^3 C_3}{{ }^{10} C_5}=\frac{1}{12}<br>$$</td>
</tr>
<tr>
<td class="tg-baqh">$$xP(x)$$</td>
<td class="tg-baqh">0</td>
<td class="tg-baqh">$$\frac{5}{12}$$</td>
<td class="tg-baqh">$$\frac{10}{12}$$</td>
<td class="tg-baqh">$$\frac{3}{12}$$</td>
</tr>
</tbody>
</table></p>
<p>$$\begin{aligned}
& \mu=\sum x P(x)=0+\frac{5}{12}+\frac{10}{12}+\frac{3}{12}=\frac{3}{2} \\
& \sigma^2=\sum(x-\mu) P(x)=\sum\left(x-\frac{3}{2}\right)^2 P(x) \\
& =\frac{9}{4} \times \frac{1}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{9}{4} \times \frac{1}{12}=\frac{7}{12}
\end{aligned}$$</p>
<p>$$\Rightarrow \sigma^2 \cdot 96=8 \times 7=56$$</p> | integer | jee-main-2024-online-5th-april-morning-shift |
lvb2954k | maths | probability | probability-distribution-of-a-random-variable | <p>From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $$X$$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $$X$$ is $$\frac{m}{n}$$, where $$\operatorname{gcd}(m, n)=1$$, then $$n-m$$ is equal to _________.</p> | [] | null | 71 | <p>Given a lot of 12 items, 3 are defective.</p>
<p>Good items, $$12-3=9$$</p>
<p>Let $$X$$ denote the number of defective items.</p>
<p>So, value of $$X=0,1,2,3$$</p>
<p>A sample of $$S$$ items is drawn.</p>
<p>$$P(X=0)=G G G G G$$</p>
<p>(here $$G$$ is good item and $$d$$ is defective)</p>
<p>$$\begin{aligned}
& \frac{9}{12} \cdot \frac{8}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} \cdot \frac{5}{8}=\frac{21}{132}=\frac{7}{44} \\
& P(X=1)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 3}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{21}{44} \\
& P(X=2)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{14}{44} \\
& P(X=3)=5\left[\frac{3 \cdot 2 \cdot 1 \cdot 9 \cdot 8}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{2}{44} \\
& P(X=4)=0 \\
& P(X=5)=0
\end{aligned}$$</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 588px">
<colgroup>
<col style="width: 85px">
<col style="width: 90px">
<col style="width: 86px">
<col style="width: 77px">
<col style="width: 81px">
<col style="width: 85px">
<col style="width: 84px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$X$$</th>
<th class="tg-baqh">0</th>
<th class="tg-baqh">1</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">3</th>
<th class="tg-baqh">4</th>
<th class="tg-baqh">5</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$P(X)$$</td>
<td class="tg-baqh">$$<br>\frac{7}{44}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{21}{44}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{14}{44}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{2}{44}<br>$$</td>
<td class="tg-baqh">0</td>
<td class="tg-baqh">0</td>
</tr>
<tr>
<td class="tg-baqh">$$XP(X)$$</td>
<td class="tg-baqh">0</td>
<td class="tg-baqh">$$<br>\frac{21}{44}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{28}{44}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{6}{44}<br>$$</td>
<td class="tg-baqh">0</td>
<td class="tg-baqh">0</td>
</tr>
<tr>
<td class="tg-baqh">$$X^2P(X)$$</td>
<td class="tg-baqh">0</td>
<td class="tg-baqh">$$<br>\frac{21}{44}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{56}{44}<br>$$</td>
<td class="tg-baqh">$$<br>\frac{18}{44}<br>$$</td>
<td class="tg-baqh">0</td>
<td class="tg-baqh">0</td>
</tr>
</tbody>
</table></p>
<p>$$\begin{aligned}
& \sigma_x^2=\sum X^2 P(x)-\left(\sum x P(x)\right)^2 \\
& =\frac{95}{44}-\left(\frac{55}{44}\right)^2 \\
& =\frac{4180-3025}{1936}=\frac{1155}{1936}=\frac{105}{176}=\frac{m}{n} \\
& =n-m=71
\end{aligned}$$</p> | integer | jee-main-2024-online-6th-april-evening-shift |
NlmoqFBur2q2cMiy | maths | probability | total-probability-theorem | A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and
this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at
random from the bag, then the probability that this drawn ball is red, is : | [{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${3 \\over 10}$$"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$${1 \\over 5}$$"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264051/exam_images/ijwqqvgif3jlthy0b2tu.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Probability Question 169 English Explanation">
<br><br>If we follow path 1, then probability of getting 1st ball black $$ = {6 \over {10}}$$ and probability of getting 2nd ball red when there is 4 R and 8 B balls = $${4 \over {12}}$$.
<br><br>So, the probability of getting 1st ball black and 2nd ball red = $${6 \over {10}} \times {4 \over {12}}$$.
<br><br>If we follow path 2, then the probability of getting 1st ball red $$ = {4 \over {10}}$$ and probability of getting 2nd ball red when in the bag there is 6 red and 6 black balls = $${6 \over {12}}$$
<br><br>$$\therefore\,\,\,$$ Probability of getting 2nd ball as red
<br><br>$$ = {6 \over {10}} \times {4 \over {12}} + {4 \over {10}} \times {6 \over {12}}$$
<br><br>$$ = {1 \over 5} + {1 \over 5}$$
<br><br>$$ = {2 \over 5}$$ | mcq | jee-main-2018-offline |
lgnxp7gc | maths | probability | total-probability-theorem | A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is : | [{"identifier": "A", "content": "$\\frac{1}{4}$"}, {"identifier": "B", "content": "$\\frac{9}{50}$"}, {"identifier": "C", "content": "$\\frac{1}{5}$"}, {"identifier": "D", "content": "$\\frac{11}{50}$"}] | ["C"] | null | Let $X$ be the number rolled on the die, and let $W$ be the event that all balls drawn are white. We want to find the probability $P(W)$, which can be calculated using the law of total probability as follows :
<br/><br/>$$P(W) = \sum\limits_{x=1}^{6} P(W|X=x)P(X=x)$$
<br/><br/>The probability of rolling any number from 1 to 6 on the die is equal, so $P(X=x) = \frac{1}{6}$ for all $x \in \{1, 2, 3, 4, 5, 6\}$.
<br/><br/>Now let's calculate the conditional probabilities $P(W|X=x)$ for each possible value of $x$ :
<br/><br/>1. $P(W|X=1) = {{{}^6{C_1}} \over {{}^{10}{C_1}}}= \frac{6}{10} = \frac{3}{5}$, since there are 6 white balls out of a total of 10 balls.
<br/><br/>2. $P(W|X=2) = {{{}^6{C_2}} \over {{}^{10}{C_2}}} = \frac{15}{45} = \frac{1}{3}$, since there are 15 ways to choose 2 white balls out of 6, and 45 ways to choose 2 balls out of 10.
<br/><br/>3. $P(W|X=3) = {{{}^6{C_3}} \over {{}^{10}{C_3}}} = \frac{20}{120} = \frac{1}{6}$, since there are 20 ways to choose 3 white balls out of 6, and 120 ways to choose 3 balls out of 10.
<br/><br/>4. $P(W|X=4) = {{{}^6{C_4}} \over {{}^{10}{C_4}}} = \frac{15}{210} = \frac{1}{14}$, since there are 15 ways to choose 4 white balls out of 6, and 210 ways to choose 4 balls out of 10.
<br/><br/>5. $P(W|X=5) = {{{}^6{C_5}} \over {{}^{10}{C_5}}} = \frac{6}{252} = \frac{1}{42}$, since there are 6 ways to choose 5 white balls out of 6, and 252 ways to choose 5 balls out of 10.
<br/><br/>6. $P(W|X=6) = {{{}^6{C_6}} \over {{}^{10}{C_6}}} = \frac{1}{210}$, since there are 1 ways to choose 6 white balls out of 6, and 210 ways to choose 6 balls out of 10.
<br/><br/>Using the law of total probability, we have :
<br/><br/>$$P(W) = \frac{1}{6} \left(P(W|X=1) + P(W|X=2) + P(W|X=3) + P(W|X=4) + P(W|X=5) + P(W|X=6)\right)$$
<br/><br/>$$P(W) = \frac{1}{6} \left(\frac{3}{5} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14} + \frac{1}{42} + \frac{1}{210}\right)$$
<br/><br/>To simplify this expression, find a common denominator :
<br/><br/>$$P(W) = \frac{1}{6} \left(\frac{126}{210} + \frac{70}{210} + \frac{35}{210} + \frac{15}{210} + \frac{5}{210} + \frac{1}{210}\right)$$
<br/><br/>Add the fractions :
<br/><br/>$$P(W) = \frac{1}{6}\left(\frac{126+70+35+15+5+1}{210}\right)=\frac{42}{210}=\frac{1}{5}$$ | mcq | jee-main-2023-online-15th-april-morning-shift |
lsao86jd | maths | probability | total-probability-theorem | A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is : | [{"identifier": "A", "content": "$\\frac{2}{5}$"}, {"identifier": "B", "content": "$\\frac{2}{7}$"}, {"identifier": "C", "content": "$\\frac{1}{7}$"}, {"identifier": "D", "content": "$\\frac{1}{5}$"}] | ["B"] | null | $\begin{aligned} & \mathrm{P}(4 \mathrm{~W} 4 \mathrm{~B} / 2 \mathrm{~W} 2 \mathrm{~B})= \\\\ & \frac{P(4 W 4 B) \times P(2 W 2 B / 4 W 4 B)}{P(2 W 6 B) \times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \times P(2 W 2 B / 3 W 5 B)} \\ & +\ldots \ldots \ldots \ldots+P(6 W 2 B) \times P(2 W 2 B / 6 W 2 B)\end{aligned}$
<br/><br/>$\begin{aligned} & =\frac{\frac{1}{5} \times \frac{{ }^4 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}}{\frac{1}{5} \times \frac{{ }^2 \mathrm{C}_2 \times{ }^6 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\frac{1}{5} \times \frac{{ }^3 \mathrm{C}_2 \times{ }^5 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\ldots+\frac{1}{5} \times \frac{{ }^6 \mathrm{C}_2 \times{ }^2 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}} \\\\ & =\frac{2}{7}\end{aligned}$ | mcq | jee-main-2024-online-1st-february-morning-shift |
q6vgte5MQsjwFa1v | maths | probability | venn-diagram-and-set-theory | A problem in mathematics is given to three students $$A,B,C$$ and their respective probability of solving the problem is $${1 \over 2},{1 \over 3}$$ and $${1 \over 4}.$$ Probability that the problem is solved is : | [{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$ "}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["A"] | null | Given $$P\left( A \right) = {1 \over 2}$$, $$P\left( B \right) = {1 \over 3}$$, $$P\left( C \right) = {1 \over 4}$$
<br><br>So, $$P\left( {\overline A } \right) = {1 \over 2}$$ (Probablity that the problem can't be solve by A)
<br>$$P\left( {\overline B } \right) = {2 \over 3}$$ (Probablity that the problem can't be solve by B)
<br>and $$P\left( {\overline C } \right) = {3 \over 4}$$ (Probablity that the problem can't be solve by C)
<br><br>Now the probablity that the problem is solved by any one student of A, B and C = 1 - the probablity that the problem is solved by none of the students of A, B and C
<br><br>$$P\left( {A \cup B \cup C} \right)$$ = 1 - $$P\left( {\overline A } \right)P\left( {\overline B } \right)P\left( {\overline C } \right)$$
<br><br>$$ = 1 - {1 \over 2} \times {2 \over 3} \times {3 \over 4}$$
<br>$$=$$ $${3 \over 4}$$ | mcq | aieee-2002 |
Inc63WUwWyWr1GZE | maths | probability | venn-diagram-and-set-theory | $$A$$ and $$B$$ are events such that $$P\left( {A \cup B} \right) = 3/4$$,$$P\left( {A \cap B} \right) = 1/4,$$
<br/>$$P\left( {\overline A } \right) = 2/3$$ then $$P\left( {\overline A \cap B} \right)$$ is : | [{"identifier": "A", "content": "$$5/12$$"}, {"identifier": "B", "content": "$$3/8$$"}, {"identifier": "C", "content": "$$5/8$$ "}, {"identifier": "D", "content": "$$1/4$$ "}] | ["A"] | null | Given $$P\left( {A \cup B} \right) = 3/4$$,
<br>$$P\left( {A \cap B} \right) = 1/4,$$
<br>$$P\left( {\overline A } \right) = 2/3$$
<br><br>We know, $$P\left( A \right)$$ = 1 - $$P\left( {\overline A } \right)$$
<br>$$\therefore$$ $$P\left( A \right)$$ = 1 - $${2 \over 3}$$ = $${1 \over 3}$$
<br><br>We know $$P\left( {A \cup B} \right)$$ = $$P\left( A \right)$$ + $$P\left( B \right)$$ - $$P\left( {A \cap B} \right)$$
<br><br>$$ \Rightarrow $$$${3 \over 4}$$ = $${1 \over 3}$$ + $$P\left( B \right)$$ - $${1 \over 3}$$
<br><br>$$ \Rightarrow $$ 1 = $${1 \over 3}$$ + $$P\left( B \right)$$
<br><br>$$ \Rightarrow $$ $$P\left( B \right)$$ = $${2 \over 3}$$
<br><br>We know $$P\left( {\overline A \cap B} \right)$$ = $$P\left( B \right)$$ - $$P\left( {A \cap B} \right)$$
<br><br>So $$P\left( {\overline A \cap B} \right)$$ = $${2 \over 3}$$ - $${1 \over4}$$ = $${5 \over 12}$$ | mcq | aieee-2002 |
m09Dh9Fy3uo4ADUI | maths | probability | venn-diagram-and-set-theory | Events $$A, B, C$$ are mutually exclusive events such that $$P\left( A \right) = {{3x + 1} \over 3},$$ $$P\left( B \right) = {{1 - x} \over 4}$$ and $$P\left( C \right) = {{1 - 2x} \over 2}$$ The set of possible values of $$x$$ are in the interval. | [{"identifier": "A", "content": "$$\\left[ {0,1} \\right]$$ "}, {"identifier": "B", "content": "$$\\left[ {{1 \\over 3},{1 \\over 2}} \\right]$$ "}, {"identifier": "C", "content": "$$\\left[ {{1 \\over 3},{2 \\over 3}} \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {{1 \\<br>3},{13 \\over 3}} \\right]$$"}] | ["B"] | null | Given $$P\left( A \right) = {{3x + 1} \over 3},$$ $$P\left( B \right) = {{1 - x} \over 4}$$ and $$P\left( C \right) = {{1 - 2x} \over 2}$$
<br><br>We know for any event X, $$0 \le P\left( X \right) \le 1$$
<br><br>$$\therefore$$ $$0 \le {{3x + 1} \over 3} \le 1$$
<br>$$ \Rightarrow - 1 \le 3x \le 2$$
<br>$$ \Rightarrow - {1 \over 3} \le x \le {2 \over 3}$$
<br><br> $$0 \le {{1 - x} \over 4} \le 1$$
<br>$$ \Rightarrow - 3 \le x \le 1$$
<br><br> $$0 \le {{1 - 2x} \over 2} \le 1$$
<br>$$ \Rightarrow - 1 \le 2x \le 1$$
<br>$$ \Rightarrow - {1 \over 2} \le x \le {1 \over 2}$$
<br><br>In the question given that A, B and C are mutually exclusive
<br>So $$P\left( {A \cup B \cup C} \right)$$ = $$P\left( {A} \right)$$ + $$P\left( {B} \right)$$ + $$P\left( {C} \right)$$
<br> <br>$$ \Rightarrow P\left( {A \cup B \cup C} \right)$$ = $${{3x + 1} \over 3}$$ + $${{1 - x} \over 4}$$ + $${{1 - 2x} \over 2}$$
<br><br>$$\therefore$$ 0 $$ \le $$ $${{3x + 1} \over 3}$$ + $${{1 - x} \over 4}$$ + $${{1 - 2x} \over 2}$$ $$ \le $$ 1
<br><br>0 $$ \le $$ 13 - 3$$x$$ $$ \le $$ 12
<br><br>$$ \Rightarrow {1 \over 3} \le x \le {{13} \over 3}$$
<br><br>From all those relations, we get
<br><br>$$\max \left\{ { - {1 \over 3}, - 3, - {1 \over 2},{1 \over 3}} \right\}$$ $$ \le $$ $$x$$ $$ \le $$ $$\min \left\{ {{2 \over 3},1,{1 \over 2},{{13} \over 3}} \right\}$$
<br><br>So, $${1 \over 3} \le x \le {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $$ \Rightarrow x \in \left[ {{1 \over 3},{1 \over 2}} \right]$$ | mcq | aieee-2003 |
kJkLQp3LNsJbLPI9 | maths | probability | venn-diagram-and-set-theory | Let $$A$$ and $$B$$ two events such that $$P\left( {\overline {A \cup B} } \right) = {1 \over 6},$$ $$P\left( {A \cap B} \right) = {1 \over 4}$$ and $$P\left( {\overline A } \right) = {1 \over 4},$$ where $${\overline A }$$ stands for complement of event $$A$$. Then events $$A$$ and $$B$$ are : | [{"identifier": "A", "content": "equally likely and mutually exclusive"}, {"identifier": "B", "content": "equally likely but not independent "}, {"identifier": "C", "content": "independent but not equally likely"}, {"identifier": "D", "content": "mutually exclusive and independent"}] | ["C"] | null | <p>Given that,</p>
<p>$$P(\overline {A \cup B} ) = {1 \over 6}$$, $$P(A \cap B) = {1 \over 4}$$, $$P(\overline A ) = {1 \over 4}$$</p>
<p>$$\because$$ $$P(\overline {A \cup B} ) = {1 \over 6}$$</p>
<p>$$ \Rightarrow 1 - P(A \cup B) = {1 \over 6}$$</p>
<p>$$ \Rightarrow 1 - P(A) - P(B) + P(A \cap B) = {1 \over 6}$$</p>
<p>$$ \Rightarrow P(\overline A ) - P(B) + {1 \over 4} = {1 \over 6}$$</p>
<p>$$ \Rightarrow P(B) = {1 \over 4} + {1 \over 4} - {1 \over 6}$$</p>
<p>$$ \Rightarrow P(B) = {1 \over 3}$$ and $$P(A) = {3 \over 4}$$</p>
<p>Clearly, $$P(A \cap B) = P(A)P(B)$$,</p>
<p>so, the events A and B are independent events but not equally likely.</p> | mcq | aieee-2005 |
CvljnrLZQyp22oGR | maths | probability | venn-diagram-and-set-theory | For three events A, B and C, <br/><br/>P(Exactly one of A or B occurs) <br/>= P(Exactly one of B or C occurs) <br/>= P
(Exactly one of C or A occurs) = $${1 \over 4}$$
<br/>and P(All the three events occur simultaneously) = $${1 \over {16}}$$.
<br/><br/> Then the
probability that at least one of the events occurs, is : | [{"identifier": "A", "content": "$${7 \\over {16}}$$"}, {"identifier": "B", "content": "$${7 \\over {64}}$$"}, {"identifier": "C", "content": "$${3 \\over {16}}$$"}, {"identifier": "D", "content": "$${7 \\over {32}}$$"}] | ["A"] | null | Given, P (A $$ \cap $$ B $$ \cap $$ C) = $${1 \over {16}}$$
<br><br>P (exactly one of A or B occurs)
<br><br>= P(A) + P (B) – 2P (A $$ \cap $$ B) = $${1 \over 4}$$ .....(1)
<br><br>P (Exactly one of B or C occurs)
<br><br>= P(B) + P (C) – 2P (B $$ \cap $$ C) = $${1 \over 4}$$ .....(2)
<br><br>P (Exactly one of C or A occurs)
<br><br>= P(C) + P(A) – 2P (C $$ \cap $$ A) = $${1 \over 4}$$ .....(3)
<br><br>Adding (1), (2) and (3),we get
<br><br>2[ P(A) + P(B) + P (C) - P (A $$ \cap $$ B)
<br><br>- P (B $$ \cap $$ C) - P (C $$ \cap $$ A)] = $${3 \over 4}$$
<br><br>$$ \Rightarrow $$ P(A) + P(B) + P (C) - P (A $$ \cap $$ B)
<br><br>- P (B $$ \cap $$ C) - P (C $$ \cap $$ A) = $${3 \over 8}$$
<br><br>$$ \therefore $$ P(atleast one event occurs)
<br><br>= P (A $$ \cup $$ B $$ \cup $$ C)
<br><br>= P(A) + P(B) + P (C) - P (A $$ \cap $$ B)
<br><br>- P (B $$ \cap $$ C) - P (C $$ \cap $$ A) + P (A $$ \cap $$ B $$ \cap $$ C)
<br><br>= $${3 \over 8} + {1 \over {16}}$$ = $${7 \over {16}}$$ | mcq | jee-main-2017-offline |
dGx2Muc10197ojWmZcE9D | maths | probability | venn-diagram-and-set-theory | In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the students selected has opted neither for NCC
nor for NSS is : | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over 6}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${5 \\over 6}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264827/exam_images/atidu8b68l8usxxqwsp4.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Evening Slot Mathematics - Probability Question 146 English Explanation">
<br>A $$ \to $$ opted NCC
<br><br>B $$ \to $$ opted NSS
<br><br>$$ \therefore $$ P (nither A nor B) $$=$$ $${{10} \over {60}} = z{1 \over 6}$$ | mcq | jee-main-2019-online-12th-january-evening-slot |
QkHi8fNAnBXDglkK9K7k9k2k5hjz0pp | maths | probability | venn-diagram-and-set-theory | Let A and B be two events such that the
probability that exactly one of them occurs is $${2 \over 5}$$ and the probability that A or B occurs is $${1 \over 2}$$ ,
then the probability of both of them occur
together is : | [{"identifier": "A", "content": "0.20"}, {"identifier": "B", "content": "0.02"}, {"identifier": "C", "content": "0.01"}, {"identifier": "D", "content": "0.10"}] | ["D"] | null | Probability that exactly one of them occurs
<br><br>P(A) + P(B) – 2P (A $$ \cap $$ B) = $${2 \over 5}$$ .....(1)
<br><br>Probability
that A or B occurs is
<br><br>P(A) + P(B) – P(A $$ \cap $$ B) = $${1 \over 2}$$ ......(2)
<br><br>Doing (2) - (1)
<br><br>P(A $$ \cap $$ B) = $${1 \over 2} - {2 \over 5}$$ = 0.10 | mcq | jee-main-2020-online-8th-january-evening-slot |
eziF1WqbTH9NrxKj6Hjgy2xukg0cfpa9 | maths | probability | venn-diagram-and-set-theory | The probabilities of three events A, B and C are
given by <br/>P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.
<br/>If P(A$$ \cup $$B) = 0.8, P(A$$ \cap $$C) = 0.3, P(A$$ \cap $$B$$ \cap $$C) = 0.2,
P(B$$ \cap $$C) = $$\beta $$<br/> and P(A$$ \cup $$B$$ \cup $$C) = $$\alpha $$, where
0.85 $$ \le \alpha \le $$ 0.95, then $$\beta $$ lies in the interval : | [{"identifier": "A", "content": "[0.35, 0.36]\n"}, {"identifier": "B", "content": "[0.20, 0.25]"}, {"identifier": "C", "content": "[0.25, 0.35]"}, {"identifier": "D", "content": "[0.36, 0.40]"}] | ["C"] | null | P(A $$ \cup $$ B) = P(A) + P(B) – P(A $$ \cup $$ B)
<br><br>$$ \Rightarrow $$ 0.8 = 0.6 + 0.4 – P(A $$ \cap $$ B)
<br><br>$$ \Rightarrow $$ P(A $$ \cap $$ B) = 0.2
<br><br>P(A$$ \cup $$B$$ \cup $$C) = P(A) + P(B) + P(C) – P(A $$ \cap $$ B) – P(B $$ \cap $$ C) –P(C $$ \cap $$ A) + P(A $$ \cap $$ B $$ \cap $$ C)
<br><br>$$ \Rightarrow $$ $$\alpha $$ = 0.6 + 0.4 + 0.5 - 0.2 - $$\beta $$ - 0.3 + 0.2
<br><br>$$ \Rightarrow $$ $$\alpha $$ + $$\beta $$ = 1.2
<br><br>$$ \Rightarrow $$ $$\alpha $$ = 1.2 - $$\beta $$
<br><br>Given, 0.85 $$ \le \alpha \le $$ 0.95
<br><br>$$ \Rightarrow $$ 0.85 $$ \le $$ 1.2 - $$\beta $$ $$ \le $$ 0.95
<br><br>$$ \Rightarrow $$ 0.25 $$ \le \beta \le $$ 0.35 | mcq | jee-main-2020-online-6th-september-evening-slot |
1krrr6wf1 | maths | probability | venn-diagram-and-set-theory | Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 $$-$$ k), the probability that exactly one of B and C occurs is (1 $$-$$ 2k), the probability that exactly one of C and A occurs is (1 $$-$$ k) and the probability of all A, B and C occur simultaneously is k<sup>2</sup>, where 0 < k < 1. Then the probability that at least one of A, B and C occur is : | [{"identifier": "A", "content": "greater than $${1 \\over 8}$$ but less than $${1 \\over 4}$$"}, {"identifier": "B", "content": "greater than $${1 \\over 2}$$"}, {"identifier": "C", "content": "greater than $${1 \\over 4}$$ but less than $${1 \\over 2}$$"}, {"identifier": "D", "content": "exactly equal to $${1 \\over 2}$$"}] | ["B"] | null | $$P(\overline A \cap B) + P(A \cap \overline B ) = 1 - k$$<br><br>$$P(\overline A \cap C) + P(A \cap \overline C ) = 1 - 2k$$<br><br>$$P(\overline B \cap C) + P(B \cap \overline C ) = 1 - k$$<br><br>$$P(A \cap B \cap C) = {k^2}$$<br><br>$$P(A) + P(B) - 2P(A \cap B) = 1 - k$$ .....(i)<br><br>$$P(B) + P(C) - 2P(B \cap C) = 1 - k$$ ..... (ii)<br><br>$$P(C) + P(A) - 2P(A \cap C) = 1 - 2k$$ ..... (iii)<br><br>$$(i) + (ii) + (iii)$$<br><br>$$P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = {{ - 4k + 3} \over 2}$$<br><br>So,<br><br>$$P(A \cup B \cup C) = {{ - 4k + 3} \over 2} + {k^2}$$<br><br>$$P(A \cup B \cup C) = {{2{k^2} - 4k + 3} \over 2}$$<br><br>$$ = {{2{{(k - 1)}^2} + 1} \over 2}$$<br><br>$$P(A \cup B \cup C) > {1 \over 2}$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1l54tbhne | maths | probability | venn-diagram-and-set-theory | <p>The probability that a relation R from {x, y} to {x, y} is both symmetric and transitive, is equal to :</p> | [{"identifier": "A", "content": "$${5 \\over {16}}$$"}, {"identifier": "B", "content": "$${9 \\over {16}}$$"}, {"identifier": "C", "content": "$${11 \\over {16}}$$"}, {"identifier": "D", "content": "$${13 \\over {16}}$$"}] | ["A"] | null | Total number of relations $=2^{2^{2}}=2^{4}=16$
<br/><br/>
Relations that are symmetric as well as transitive are
<br/><br/>
$\phi,\{(x, x)\},\{(y, y)\},\{(x, x),(x, y),(y, y),(y, x)\},\{(x, x),(y, y)\}$
<br/><br/>
$\therefore \quad$ favourable cases $=5$
<br/><br/>
$\therefore \quad P_{r}=\frac{5}{16}$ | mcq | jee-main-2022-online-29th-june-evening-shift |
1l6givpji | maths | probability | venn-diagram-and-set-theory | <p>Let $$\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3}$$ be three mutually exclusive events such that $$\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{2+3 \mathrm{p}}{6}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{2-\mathrm{p}}{8}$$ and $$\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{1-\mathrm{p}}{2}$$. If the maximum and minimum values of $$\mathrm{p}$$ are $$\mathrm{p}_{1}$$ and $$\mathrm{p}_{2}$$, then $$\left(\mathrm{p}_{1}+\mathrm{p}_{2}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{2}{3}$$"}, {"identifier": "B", "content": "$$\\frac{5}{3}$$"}, {"identifier": "C", "content": "$$\\frac{5}{4}$$"}, {"identifier": "D", "content": "1"}] | ["B"] | null | <p>$$0 \le {{2 + 3P} \over 6} \le 1 \Rightarrow P \in \left[ { - {2 \over 3},{4 \over 3}} \right]$$</p>
<p>$$0 \le {{2 - P} \over 8} \le 1 \Rightarrow P \in [ - 6,2]$$</p>
<p>$$0 \le {{1 - P} \over 2} \le 1 \Rightarrow P \in [ - 1,1]$$</p>
<p>$$0 < P({E_1}) + P({E_2}) + P({E_3}) \le 1$$</p>
<p>$$0 < {{13} \over {12}} - {P \over 8} \le 1$$</p>
<p>$$P \in \left[ {{2 \over 3},{{26} \over 3}} \right]$$</p>
<p>Taking intersection of all</p>
<p>$$P \in \left[ {{2 \over 3},1} \right)$$</p>
<p>$${P_1} + {P_2} = {5 \over 3}$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6p2wf0h | maths | probability | venn-diagram-and-set-theory | <p>Let $$S=\{1,2,3, \ldots, 2022\}$$. Then the probability, that a randomly chosen number n from the set S such that $$\mathrm{HCF}\,(\mathrm{n}, 2022)=1$$, is :</p> | [{"identifier": "A", "content": "$$\\frac{128}{1011}$$"}, {"identifier": "B", "content": "$$\\frac{166}{1011}$$"}, {"identifier": "C", "content": "$$\\frac{127}{337}$$"}, {"identifier": "D", "content": "$$\\frac{112}{337}$$"}] | ["D"] | null | <p>S = {1, 2, 3, .......... 2022}</p>
<p>HCF (n, 2022) = 1</p>
<p>$$\Rightarrow$$ n and 2022 have no common factor</p>
<p>Total elements = 2022</p>
<p>2022 = 2 $$\times$$ 3 $$\times$$ 337</p>
<p>M : numbers divisible by 2.</p>
<p>{2, 4, 6, ........, 2022}$$\,\,\,\,$$ n(M) = 1011</p>
<p>N : numbers divisible by 3.</p>
<p>{3, 6, 9, ........, 2022}$$\,\,\,\,$$ n(N) = 674</p>
<p>L : numbers divisible by 6.</p>
<p>{6, 12, 18, ........, 2022}$$\,\,\,\,$$ n(L) = 337</p>
<p>n(M $$\cup$$ N) = n(M) + n(N) $$-$$ n(L)</p>
<p>= 1011 + 674 $$-$$ 337</p>
<p>= 1348</p>
<p>0 = Number divisible by 337 but not in M $$\cup$$ N</p>
<p>{337, 1685}</p>
<p>Number divisible by 2, 3 or 337</p>
<p>= 1348 + 2 = 1350</p>
<p>Required probability $$ = {{2022 - 1350} \over {2022}}$$</p>
<p>$$ = {{672} \over {2022}}$$</p>
<p>$$ = {{112} \over {337}}$$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1ldo6jbm8 | maths | probability | venn-diagram-and-set-theory | <p>Two dice are thrown independently. Let $$\mathrm{A}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is less than the number appeared on the $$2^{\text {nd }}$$ die, $$\mathrm{B}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is even and that on the second die is odd, and $$\mathrm{C}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is odd and that on the $$2^{\text {nd }}$$ is even. Then :</p> | [{"identifier": "A", "content": "A and B are mutually exclusive"}, {"identifier": "B", "content": "the number of favourable cases of the events A, B and C are 15, 6 and 6 respectively"}, {"identifier": "C", "content": "B and C are independent"}, {"identifier": "D", "content": "the number of favourable cases of the event $$(\\mathrm{A\\cup B)\\cap C}$$ is 6"}] | ["D"] | null | $\begin{aligned} & A=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\} \\\\ & n(A)=15 \\\\ & B=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\} \\\\ & n(B)=9 \\\\ & C=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\} \\\\ & n(C)=9\end{aligned}$
<br/><br/>$$
(4,5) \in A \text { and }(4,5) \in B
$$
<br/><br/>$\therefore A$ and $B$ are not exclusive events
<br/><br/>$$
\begin{aligned}
& n((A \cup B) \cap C)=n(A \cap C)+n(B \cap C)-n(A \cap B \cap C) \\\\
= & 3+3-0 \\\\
= & 6
\end{aligned}
$$
<br/><br/>Option (D) is correct.
<br/><br/>$$
\begin{aligned}
& n(B)=\frac{9}{36}, n(C)=\frac{9}{36}, n(B \cap C)=0 \\\\
& \Rightarrow n(B) \cdot n(C) \neq n(B \cap C) \\\\
& \therefore B \text { and } C \text { are not independent. }
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-evening-shift |
1lsgb3d4o | maths | probability | venn-diagram-and-set-theory | <p>A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics and Chemistry. It was found that all students passed in atleast one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, atmost 11 students passed in both Mathematics and Physics, atmost 15 students passed in both Physics and Chemistry, atmost 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is _________.</p> | [] | null | 10 | $\begin{aligned} & n(M)=20 \\\\ & n(P)=25 \\\\ & n(C)=16 \\\\ & n(M \cap P)=11 \\\\ & n(P \cap C)=15 \\\\ & n(M \cap C)=15\end{aligned}$
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnvub5/791e93e6-65e8-4ec4-acf1-6dac1a0eab27/74eb5a10-cde5-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnvub6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqnvub5/791e93e6-65e8-4ec4-acf1-6dac1a0eab27/74eb5a10-cde5-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnvub6.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Morning Shift Mathematics - Probability Question 15 English Explanation">
<br>$$
\begin{aligned}
& n(C \cup P \cup M) \leq n(U)=40 . \\\\
& n(C)+n(P)+n(M)-n(C \cap M)-n(P \cap M)-n(C \cap \\\\
& P)+n(C \cap P \cap M) \leq 40 \\\\
& 20+25+16-11-15-15+x \leq 40 \\\\
& x \leq 20
\end{aligned}
$$
<br><br>But $11-x \geq 0$ and $15-x \geq 0$
<br><br>$$
\Rightarrow x \geq 11
$$ | integer | jee-main-2024-online-30th-january-morning-shift |
PuWjyRwOQzS1fiJs | maths | properties-of-triangle | area-of-triangle | If in a $$\Delta ABC$$, the altitudes from the vertices $$A, B, C$$ on opposite sides are in H.P, then $$\sin A,\sin B,\sin C$$ are in : | [{"identifier": "A", "content": "G. P."}, {"identifier": "B", "content": "A. P."}, {"identifier": "C", "content": "A.P-G.P."}, {"identifier": "D", "content": "H. P"}] | ["B"] | null | $$\Delta = {1 \over 2}{p_1}a = {1 \over 2}{p_2}b = {1 \over 2}{p_3}b$$
<br><br>$${p_1},{p_2},{p_3},$$ are in $$H.P.$$
<br><br>$$ \Rightarrow {{2\Delta } \over a},{{2\Delta } \over b},{{2\Delta } \over c}$$ are in $$H.P.$$
<br><br>$$ \Rightarrow {1 \over a},{1 \over b},{1 \over c},$$ are in $$H.P.$$
<br><br>$$ \Rightarrow a,b,c$$ are in $$A.P.$$
<br><br>$$ \Rightarrow $$ $$K\sin A,K\sin B,K\sin C$$ are in $$A.P.$$
<br><br>$$ \Rightarrow $$ $$\sin A,\sin B,\sin C$$ are in $$A.P.$$ | mcq | aieee-2005 |
uF5vBTaoRp0w8iRS | maths | properties-of-triangle | area-of-triangle | In a $$\Delta PQR,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} $$ If $$3{\mkern 1mu} \sin {\mkern 1mu} P + 4{\mkern 1mu} \cos {\mkern 1mu} Q = 6$$ and $$4\sin Q + 3\cos P = 1,$$ then the angle R is equal to : | [{"identifier": "A", "content": "$${{5\\pi } \\over 6}$$ "}, {"identifier": "B", "content": "$${{\\pi } \\over 6}$$"}, {"identifier": "C", "content": "$${{\\pi } \\over 4}$$"}, {"identifier": "D", "content": "$${{3\\pi } \\over 4}$$"}] | ["B"] | null | Given $$3$$ $$\sin \,P + 4\cos Q = 6$$ $$\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$4\sin Q + 3\cos P = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Squaring and adding $$(i)$$ & $$(ii)$$ we get
<br><br>$$9\,{\sin ^2}P + 16{\cos ^2}Q + 24\sin P\cos Q$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 16\,{\sin ^2}Q + 9{\cos ^2}P + 24\sin Q\cos P$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 36 + 1 = 37$$
<br><br>$$ \Rightarrow 9\left( {{{\sin }^2}p + {{\cos }^2}P} \right) + 16\left( {{{\sin }^2}Q + {{\cos }^2}q} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 24\left( {\sin P\cos Q + \cos P\sin Q} \right) = 37$$
<br><br>$$ \Rightarrow 9 + 16 + 24\sin \left( {P + Q} \right) = 37$$
<br><br>[ As $${\sin ^2}\theta + {\cos ^2}\theta = 1$$ and
<br><br>$$\sin A\cos B + \cos A\sin B$$ $$ = \sin \left( {A + B} \right)$$ ]
<br><br>$$ \Rightarrow \sin \left( {P + Q} \right) = {1 \over 2}$$
<br><br>$$ \Rightarrow P + Q = {\pi \over 6}$$ or $${{5\pi } \over 6}$$
<br><br>$$ \Rightarrow R = {{5\pi } \over 6}$$ or $${\pi \over 6}$$
<br><br>(as $$P + Q + R = \pi $$ )
<br><br>If $$R = {{5\pi } \over 6}$$ then $$0 < P,Q < {\pi \over 6}$$
<br><br>$$ \Rightarrow \cos Q < 1$$ and $$\sin P < {1 \over 2}$$
<br><br>$$ \Rightarrow 3\sin P + 4\cos Q < {{11} \over 2}$$ which is not true.
<br><br>So $$R = {\pi \over 6}$$ | mcq | aieee-2012 |
fdmnaeLrvX4aMdxOwkjgy2xukf7gxjfo | maths | properties-of-triangle | area-of-triangle | A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If $$\angle BAC = {90^o}$$ and area$$\left( {\Delta ABC} \right) = 5\sqrt 5 $$ s units, then the abscissa of the vertex C is : | [{"identifier": "A", "content": "$$1 + 2\\sqrt 5 $$"}, {"identifier": "B", "content": "$$ 2\\sqrt 5 - 1$$"}, {"identifier": "C", "content": "$$1 + \\sqrt 5 $$"}, {"identifier": "D", "content": "$$2 + \\sqrt 5 $$"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264973/exam_images/fxiejzh5aeuzbgfxdpro.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267653/exam_images/matb2vixbiqzmc3xsv1z.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263515/exam_images/n3r9nhavsa80zjogtt4n.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264717/exam_images/ycmesahtltfopnsuq3ta.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 4th September Morning Slot Mathematics - Properties of Triangle Question 19 English Explanation"></picture>
<br>Distance between A and B<br><br>$$ = \sqrt {{{(2)}^2} + {{( - 1)}^2}} = \sqrt 5 $$<br><br>Area of triangle = $$5\sqrt 5 $$<br><br>$$ \Rightarrow {1 \over 2} \times \sqrt 5 \times x = 5\sqrt 5 $$<br><br>$$ \Rightarrow x = 10$$<br><br>Slope of AB, $${m_{AB}} = {{1 - 2} \over {3 - 1}} = - {1 \over 2}$$<br><br>AC is perpendicular to AB.<br><br>$$ \therefore $$ $${m_{AB}}\,.\,{m_{AC}} = - 1$$<br><br>$$ \Rightarrow {m_{AC}} = 2 = \tan \theta $$<br><br>So, $$\sin \theta = {2 \over {\sqrt 5 }}$$, $$\cos \theta = {1 \over {\sqrt 5 }}$$<br><br>$$ \therefore $$ $$a = {x_A} + r\cos \theta $$<br><br>$$ = 1 + 10 \times {1 \over {\sqrt 5 }}$$<br><br>$$ = 1 + 2\sqrt 5 $$ | mcq | jee-main-2020-online-4th-september-morning-slot |
JPpOTLK1LgUoYqsvMt1kluxiym4 | maths | properties-of-triangle | area-of-triangle | The triangle of maximum area that can be inscribed in a given circle of radius 'r' is : | [{"identifier": "A", "content": "An equilateral triangle having each of its side of length $$\\sqrt 3 $$r."}, {"identifier": "B", "content": "An equilateral triangle of height $${{2r} \\over 3}$$."}, {"identifier": "C", "content": "A right angle triangle having two of its sides of length 2r and r."}, {"identifier": "D", "content": "An isosceles triangle with base equal to 2r."}] | ["A"] | null | Area of triangle ABC<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266363/exam_images/t4j7msbaucclnkye1pro.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Properties of Triangle Question 17 English Explanation"><br><br>$$A = {1 \over 2} \times BC \times AM$$<br><br>$$ = {1 \over 2} \times 2\sqrt {{r^2} - {x^2}} \times (r + x)$$<br><br>$$A = (r + x)\sqrt {{r^2} - {x^2}} $$<br><br>$${{dA} \over {dx}} = \sqrt {{r^2} - {x^2}} - {x \over {\sqrt {{r^2} - {x^2}} }} \times (r + x) $$
<br><br>$$= {{{r^2} - {x^2} - rx - {x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{{r^2} - rx - 2{x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{ - (x + r)(2x - r)} \over {\sqrt {{r^2} - {x^2}} }}$$<br><br>$${{dA} \over {dx}} = 0 \Rightarrow x = {r \over 2}$$<br><br>Sign change of $${{dA} \over {dx}}$$ at $$x = {r \over 2}$$ <br><br>$$ \Rightarrow $$ A has maximum at $$x = {r \over 2}$$
<br><br>$$BC = 2\sqrt {{r^2} - {x^2}} = \sqrt 3 r$$, <br><br>$$AM = r + {1 \over 2}r$$ = $${3 \over 2}r$$<br><br>$$ \Rightarrow AB = AC = \sqrt 3 r$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
d8zoYrCZDGgB8OMG3I1kmhzdcqe | maths | properties-of-triangle | area-of-triangle | Let ABCD be a square of side of unit length. Let a circle C<sub>1</sub> centered at A with unit radius is drawn. Another circle C<sub>2</sub> which touches C<sub>1</sub> and the lines AD and AB are tangent to it, is also drawn. Let a tangent line from the point C to the circle C<sub>2</sub> meet the side AB at E. If the length of EB is $$\alpha$$ + $${\sqrt 3 }$$ $$\beta$$, where $$\alpha$$, $$\beta$$ are integers, then $$\alpha$$ + $$\beta$$ is equal to ____________. | [] | null | 1 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266007/exam_images/gqh1muo7zwcf1cjiz3fa.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Mathematics - Properties of Triangle Question 16 English Explanation"><br><br>(i) $$\sqrt 2 r + r = 1$$<br><br>$$r = {1 \over {\sqrt 2 + 1}}$$<br><br>$$r = \sqrt 2 - 1$$<br><br>(ii) $$C{C_2} = 2\sqrt 2 - 2 = 2\left( {\sqrt 2 - 1} \right)$$<br><br>From $$\Delta C{C_2}N = \sin \phi = {{\sqrt 2 - 1} \over {2\left( {\sqrt 2 - 1} \right)}}$$<br><br>$$\phi = 30^\circ $$<br><br>(iii) In $$\Delta$$ACE apply sine law<br><br>$${{AE} \over {\sin \phi }} = {{AC} \over {\sin 105^\circ }}$$<br><br>$$AE = {1 \over 2} \times {{\sqrt 2 } \over {\sqrt 3 + 1}}.2\sqrt 2 $$<br><br>$$AE = {2 \over {\sqrt 3 + 1}} = \sqrt 3 - 1$$<br><br>$$ \therefore $$ $$EB = 1 - \left( {\sqrt 3 - 1} \right)$$<br><br>= $$2 - \sqrt 3 $$<br><br>$$ \therefore $$ $$\alpha$$ = 2, $$\beta$$ = $$-$$1 $$ \Rightarrow $$ $$\alpha$$ + $$\beta$$ = 1 | integer | jee-main-2021-online-16th-march-morning-shift |
ZzEH5xuVZGeyr66uaK1kmiznmbw | maths | properties-of-triangle | area-of-triangle | In $$\Delta$$ABC, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of $$\Delta$$ABC is 30 cm<sup>2</sup> and R and r are respectively the radii of circumcircle and incircle of $$\Delta$$ABC, then the value of 2R + r (in cm) is equal to ___________. | [] | null | 15 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264566/exam_images/c0o3n3zizv5saixvrv1t.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Properties of Triangle Question 15 English Explanation 1"><br><br>Area = $${1 \over 2}(5)(12)\sin \theta = 30$$<br><br>$$\sin \theta = 1 \Rightarrow \theta = {\pi \over 2}$$<br><br>$$\Delta$$ is right angle $$\Delta$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264122/exam_images/hwpbih4rxxssh7b0owih.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Properties of Triangle Question 15 English Explanation 2"><br><br>$$r = (s - a)\tan {A \over 2}$$<br><br>$$r = (s - a)$$$$\tan {{90} \over 2}$$<br><br>$$r = (s - a)$$<br><br>$$2R + r = s$$, (As $$a = 2R$$)<br><br>$$2R + r = {{5 + 12 + 13} \over 2} = 15$$ | integer | jee-main-2021-online-16th-march-evening-shift |
1krzr2ib8 | maths | properties-of-triangle | area-of-triangle | If a rectangle is inscribed in an equilateral triangle of side length $$2\sqrt 2 $$ as shown in the figure, then the square of the largest area of such a rectangle is _____________.<br/><br/><img src="data:image/png;base64,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"/> | [] | null | 3 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266129/exam_images/ouqduiix1vmfx5nsutac.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263970/exam_images/qsqsvhbym8ueyy7jh4jv.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265279/exam_images/jk4yiktnfvwoeb8wtrgr.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267504/exam_images/tv5bb82om3ehrf6ij6ri.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263590/exam_images/tc0blbpawukgwb0ll2x0.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Evening Shift Mathematics - Properties of Triangle Question 13 English Explanation"></picture> <br><br>In $$\Delta$$DBF<br><br>$$\tan 60^\circ = {{2b} \over {2\sqrt 2 - l}} \Rightarrow b = {{\sqrt 3 \left( {2\sqrt 2 - l} \right)} \over 2}$$<br><br>A = Area of rectangle = l $$\times$$ b<br><br>$$A = l \times {{\sqrt 3 } \over 2}\left( {2\sqrt 2 - l} \right)$$<br><br>$${{dA} \over {dl}} = {{\sqrt 3 } \over 2}\left( {2\sqrt 2 - l} \right) - {{l.\sqrt 3 } \over 2} = 0$$<br><br>$$l = \sqrt 2 $$<br><br>$$A = l \times b = \sqrt 2 \times {{\sqrt 3 } \over 2}\left( {\sqrt 2 } \right) = \sqrt 3 $$<br><br>$$\Rightarrow$$ A<sup>2</sup> = 3 | integer | jee-main-2021-online-25th-july-evening-shift |
1l57o61t8 | maths | properties-of-triangle | area-of-triangle | <p>The lengths of the sides of a triangle are 10 + x<sup>2</sup>, 10 + x<sup>2</sup> and 20 $$-$$ 2x<sup>2</sup>. If for x = k, the area of the triangle is maximum, then 3k<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9umae/dcd93d6b-ea7f-4626-9b1a-432648434cd3/99642f60-0655-11ed-903e-c9687588b3f3/file-1l5q9umaf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5q9umae/dcd93d6b-ea7f-4626-9b1a-432648434cd3/99642f60-0655-11ed-903e-c9687588b3f3/file-1l5q9umaf.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Properties of Triangle Question 11 English Explanation"></p>
<p>$$CD = \sqrt {{{(10 + {x^2})}^2} - {{(10 - {x^2})}^2}} = 2\sqrt {10} |x|$$</p>
<p>Area $$ = {1 \over 2} \times CD \times AB = {1 \over 2} \times 2\sqrt {10} |x|(20 - 2{x^2})$$</p>
<p>$$A = \sqrt {10} |x|(10 - {x^2})$$</p>
<p>$${{dA} \over {dx}} = \sqrt {10} {{|x|} \over x}(10 - {x^2}) + \sqrt {10} |x|( - 2x) = 0$$</p>
<p>$$ \Rightarrow 10 - {x^2} = 2{x^2}$$</p>
<p>$$3{x^2} = 10$$</p>
<p>$$x = k$$</p>
<p>$$3{k^2} = 10$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
1ldr5mq8n | maths | properties-of-triangle | area-of-triangle | <p>A straight line cuts off the intercepts $$\mathrm{OA}=\mathrm{a}$$ and $$\mathrm{OB}=\mathrm{b}$$ on the positive directions of $$x$$-axis and $$y$$ axis respectively. If the perpendicular from origin $$O$$ to this line makes an angle of $$\frac{\pi}{6}$$ with positive direction of $$y$$-axis and the area of $$\triangle \mathrm{OAB}$$ is $$\frac{98}{3} \sqrt{3}$$, then $$\mathrm{a}^{2}-\mathrm{b}^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{392}{3}$$"}, {"identifier": "B", "content": "98"}, {"identifier": "C", "content": "196"}, {"identifier": "D", "content": "$$\\frac{196}{3}$$"}] | ["A"] | null | <p>$${1 \over 2}ab = {{98\sqrt 3 } \over 3}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leq0ppnt/4e6ea50d-9a35-4fc8-8434-5621ca292f68/d132e190-b85f-11ed-8195-4f3c56fa1eb5/file-1leq0ppnu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leq0ppnt/4e6ea50d-9a35-4fc8-8434-5621ca292f68/d132e190-b85f-11ed-8195-4f3c56fa1eb5/file-1leq0ppnu.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 30th January Morning Shift Mathematics - Properties of Triangle Question 8 English Explanation"></p>
<p>$$ \Rightarrow \sqrt 3 ab = 196$$ ..... (i)</p>
<p>$$OP = OB\cos 30^\circ = OA\cos 60^\circ $$</p>
<p>$$ \Rightarrow {{b\sqrt 3 } \over 2} = {a \over 2}$$</p>
<p>$$ \Rightarrow \sqrt 3 b = a$$ ..... (ii)</p>
<p>By (i) and (ii)</p>
<p>$${a^2} = 196$$</p>
<p>$$a = 14$$</p>
<p>$${b^2} = {{{a^2}} \over 3}$$</p>
<p>$${a^2} - {b^2} = {{2{a^2}} \over 3} = {{392} \over 3}$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1lgvqyw2i | maths | properties-of-triangle | area-of-triangle | <p>In the figure, $$\theta_{1}+\theta_{2}=\frac{\pi}{2}$$ and $$\sqrt{3}(\mathrm{BE})=4(\mathrm{AB})$$. If the area of $$\triangle \mathrm{CAB}$$ is $$2 \sqrt{3}-3$$ unit $${ }^{2}$$, when $$\frac{\theta_{2}}{\theta_{1}}$$ is the largest, then the perimeter (in unit) of $$\triangle \mathrm{CED}$$ is equal to _________.</p>
<p><img src="data:image/png;base64,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"/></p> | [] | null | 6 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnk6r0mj/2e1e2dc2-1bc0-46e6-95cc-c6b581c795c0/aac42bb0-6758-11ee-a06a-699a057b80c4/file-6y3zli1lnk6r0mk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnk6r0mj/2e1e2dc2-1bc0-46e6-95cc-c6b581c795c0/aac42bb0-6758-11ee-a06a-699a057b80c4/file-6y3zli1lnk6r0mk.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Properties of Triangle Question 5 English Explanation">
<br><br>We have, $\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(B E)=4 A B$
<br><br>Let $A B=x$ unit
<br><br>$$
\begin{aligned}
& A C=x \tan \theta_1 \\\\
& E D=x \tan \theta_2 \\\\
& B E=B D+D E
\end{aligned}
$$
<br><br>$$
\begin{array}{rlrl}
& \Rightarrow \frac{4}{\sqrt{3}} x =x\left(\tan \theta_1+\tan \theta_2\right) {[\because \sqrt{3} B E=4 A B]} \\\\
& \Rightarrow \frac{4}{\sqrt{3}}=\tan \theta_1+\tan \left(\frac{\pi}{2}-\theta_1\right) {\left[\because \theta_1+\theta_2=\frac{\pi}{2}\right]}
\end{array}
$$
<br><br>$$
\begin{aligned}
& \Rightarrow \tan \theta_1+\cot \theta_1=\frac{4}{\sqrt{3}}=\sqrt{3}+\frac{1}{\sqrt{3}} \\\\
& \Rightarrow \tan \theta_1=\sqrt{3} \text { or } \theta_1=\frac{\pi}{3} \text { and } \theta_2=\frac{\pi}{6} \\\\
& \operatorname{or} \theta_1=\frac{\pi}{6} \text { and } \theta_2=\frac{\pi}{3}
\end{aligned}
$$
<br><br>$\because \frac{\theta_2}{\theta_1}$ is largest
<br><br>$$
\therefore \theta_1=\frac{\pi}{6} \text { and } \theta_2=\frac{\pi}{3}
$$
<br><br>$$
\begin{aligned}
& \text { Area of } \triangle C A B=\frac{1}{2} \times x \times x \tan \theta_1 \\\\
& \Rightarrow \frac{x^2 \tan \theta_1}{2}=2 \sqrt{3}-3 \\\\
& \Rightarrow x^2=\frac{2(2 \sqrt{3}-3)}{\tan \frac{\pi}{6}}=12-6 \sqrt{3} \\\\\
& \Rightarrow x=3-\sqrt{3}
\end{aligned}
$$
<br><br>$$
\text { Also, } C E=\sqrt{x^2+x^2 \tan ^2 \frac{\pi}{3}}=(3-\sqrt{3}) \times 2=6-2 \sqrt{3}
$$
<br><br>Perimeter of $\triangle C E D$
<br><br>$$
\begin{aligned}
& =C D+D E+C E \\\\
& =(3-\sqrt{3})+(3-\sqrt{3}) \tan \frac{\pi}{3}+6-2 \sqrt{3} \\\\
& =3-\sqrt{3}+3 \sqrt{3}-3+6-2 \sqrt{3}=6
\end{aligned}
$$ | integer | jee-main-2023-online-10th-april-evening-shift |
jaoe38c1lseyc1v8 | maths | properties-of-triangle | circumcenter,-incenter-and-orthocenter | <p>Let $$\left(5, \frac{a}{4}\right)$$ be the circumcenter of a triangle with vertices $$\mathrm{A}(a,-2), \mathrm{B}(a, 6)$$ and $$C\left(\frac{a}{4},-2\right)$$. Let $$\alpha$$ denote the circumradius, $$\beta$$ denote the area and $$\gamma$$ denote the perimeter of the triangle. Then $$\alpha+\beta+\gamma$$ is</p> | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "62"}, {"identifier": "C", "content": "53"}, {"identifier": "D", "content": "30"}] | ["C"] | null | <p>$$\begin{aligned}
& A(a,-2), B(a, 6), C\left(\frac{a}{4},-2\right), O\left(5, \frac{a}{4}\right) \\
& A O=B O \\
& (a-5)^2+\left(\frac{a}{4}+2\right)^2=(a-5)^2+\left(\frac{a}{4}-6\right)^2 \\
& a=8 \\
& A B=8, A C=6, B C=10 \\
& \alpha=5, \beta=24, \gamma=24
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
luxwe7pg | maths | properties-of-triangle | circumcenter,-incenter-and-orthocenter | <p>Two vertices of a triangle $$\mathrm{ABC}$$ are $$\mathrm{A}(3,-1)$$ and $$\mathrm{B}(-2,3)$$, and its orthocentre is $$\mathrm{P}(1,1)$$. If the coordinates of the point $$\mathrm{C}$$ are $$(\alpha, \beta)$$ and the centre of the of the circle circumscribing the triangle $$\mathrm{PAB}$$ is $$(\mathrm{h}, \mathrm{k})$$, then the value of $$(\alpha+\beta)+2(\mathrm{~h}+\mathrm{k})$$ equals</p> | [{"identifier": "A", "content": "81"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "51"}, {"identifier": "D", "content": "5"}] | ["D"] | null | <p>$$\begin{aligned}
& m_{P A}=\frac{2}{-2}=-1 \\
& \therefore \quad m_{B C}=1
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1m11ba/909a9c7c-64f8-4f41-9021-c0d2d0a9d656/7865b860-0f4f-11ef-ad52-af909612c772/file-1lw1m11bb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1m11ba/909a9c7c-64f8-4f41-9021-c0d2d0a9d656/7865b860-0f4f-11ef-ad52-af909612c772/file-1lw1m11bb.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Evening Shift Mathematics - Properties of Triangle Question 3 English Explanation"></p>
<p>$$\begin{aligned}
& B C: y=x+5 \\
& m_{B P}=\frac{2}{-3}=\frac{-2}{3} \\
& \therefore m_{A C}=\frac{3}{2} \\
& A C: y=\frac{3}{2} x-\frac{11}{2} \quad \Rightarrow 2 y=3 x-11 \\
& \therefore \quad C:(21,26)
\end{aligned}$$</p>
<p>Let the circumcentre be $$(h, k)$$</p>
<p>$$\begin{aligned}
& (h-1)^2+(k-1)^2=(h+2)^2+(k-3)^2 \quad \text{... (i)}\\
& (h-1)^2+(k-1)^2=(h-3)^2+(k+1)^2 \quad \text{... (ii)}
\end{aligned}$$</p>
<p>Solving (i) and (ii)</p>
<p>$$\begin{aligned}
& h=\frac{-19}{2}, k=\frac{-23}{2} \\
& \alpha+\beta+2(h+k) \\
& =21+26-19-23 \\
& =2+3=5
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
lv2erz6q | maths | properties-of-triangle | cosine-rule | <p>Consider a triangle $$\mathrm{ABC}$$ having the vertices $$\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$$ and $$\mathrm{C}(\gamma, \delta)$$ and angles $$\angle A B C=\frac{\pi}{6}$$ and $$\angle B A C=\frac{2 \pi}{3}$$. If the points $$\mathrm{B}$$ and $$\mathrm{C}$$ lie on the line $$y=x+4$$, then $$\alpha^2+\gamma^2$$ is equal to _______.</p> | [] | null | 14 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhhd8gk/a86d447c-ca28-4c88-a2d0-2170f95b6e56/c85f5d40-1809-11ef-b156-f754785ad3ce/file-1lwhhd8gl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhhd8gk/a86d447c-ca28-4c88-a2d0-2170f95b6e56/c85f5d40-1809-11ef-b156-f754785ad3ce/file-1lwhhd8gl.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Properties of Triangle Question 2 English Explanation 1"></p>
<p>$$\begin{aligned}
& P=\frac{|2-1-4|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}} \\
& \sin \left(\frac{\pi}{6}\right)=\frac{3 / \sqrt{2}}{A B}=\frac{1}{2} \Rightarrow A B=\frac{6}{\sqrt{2}} \\
& \Rightarrow(\alpha-1)^2+(\alpha+4-2)^2=18
\end{aligned}$$</p>
<p>$$\Rightarrow 2 \alpha^2+2 \alpha-13=0 \rightarrow \alpha$$ and $$\gamma$$ satisfy same equation</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhhhojv/60e45a27-3bf3-4540-bde6-9ba4a53c1429/440a25b0-180a-11ef-b156-f754785ad3ce/file-1lwhhhojw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhhhojv/60e45a27-3bf3-4540-bde6-9ba4a53c1429/440a25b0-180a-11ef-b156-f754785ad3ce/file-1lwhhhojw.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Properties of Triangle Question 2 English Explanation 2"></p>
<p>$$\begin{aligned}
& \Rightarrow \alpha^2+\gamma^2=(\alpha+\gamma)^2-2 \alpha \gamma \\
& =(-1)^2-2\left(\frac{-13}{2}\right)=1+13=14
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-evening-shift |
lvb294yu | maths | properties-of-triangle | cosine-rule | <p>In a triangle $$\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N}$$ and $$\cos \mathrm{A}=\frac{2}{3}$$. If $$49 \cos (3 \mathrm{C})+42=\frac{\mathrm{m}}{\mathrm{n}}$$, where $$\operatorname{gcd}(m, n)=1$$, then $$m+n$$ is equal to _________.</p> | [] | null | 39 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwap0izi/fd6cee3f-e437-4a65-830c-d11ed7b1c7a4/3378e3e0-144e-11ef-860c-d121cbcdd1fc/file-1lwap0izj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwap0izi/fd6cee3f-e437-4a65-830c-d11ed7b1c7a4/3378e3e0-144e-11ef-860c-d121cbcdd1fc/file-1lwap0izj.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Properties of Triangle Question 1 English Explanation"></p>
<p>$$\begin{aligned}
& \cos A=\frac{2}{3}=\frac{\alpha^2+8^2-7^2}{2 \cdot \alpha \cdot 8} \\
& \Rightarrow\left(\alpha^2+15\right) 3=32 \alpha \\
& 3 \alpha^2-32 \alpha+45=0 \\
& \Rightarrow \alpha=\frac{5}{3}, 9 \\
& \because \alpha \in N \Rightarrow \alpha=9 \\
& \cos C=\frac{7^2+8^2-9^2}{2 \cdot 7 \cdot 8}=\frac{2}{7}
\end{aligned}$$</p>
<p>$$
\begin{aligned}
\cos 3 C & =4 \cos ^3 C-3 \cos C \\
= & \frac{4 \times 8}{7^3}-\frac{6}{7} \\
49 \cos 3 C & =\frac{32}{7}-42 \\
\Rightarrow \quad & 49 \cos 3 C+42=\frac{32}{7} \\
\Rightarrow \quad & m+n =39
\end{aligned}
$$</p> | integer | jee-main-2024-online-6th-april-evening-shift |
L8Rjt2UM5OY6mLNw | maths | properties-of-triangle | ex-circle | In a triangle with sides $$a, b, c,$$ $${r_1} > {r_2} > {r_3}$$ (which are the ex-radii) then : | [{"identifier": "A", "content": "$$a>b>c$$"}, {"identifier": "B", "content": "$$a < b < c$$ "}, {"identifier": "C", "content": "$$a > b$$ and $$b < c$$ "}, {"identifier": "D", "content": "$$a < b$$ and $$b > c$$ "}] | ["A"] | null | $${r_1} > {r_2} > {r_3}$$
<br><br>$$ \Rightarrow {\Delta \over {s - a}} > {\Delta \over {s - b}} > {\Delta \over {s - c}};$$
<br><br>$$ \Rightarrow s - a < s - b < s - c$$
<br><br>$$ \Rightarrow - a < - b < - c$$
<br><br>$$ \Rightarrow a > b > c$$ | mcq | aieee-2002 |
R4DzEtYltigsfO3b | maths | properties-of-triangle | half-angle-formulae | The sum of the radii of inscribed and circumscribed circles for an $$n$$ sided regular polygon of side $$a, $$ is : | [{"identifier": "A", "content": "$${a \\over 4}\\cot \\left( {{\\pi \\over {2n}}} \\right)$$ "}, {"identifier": "B", "content": "$$a\\cot \\left( {{\\pi \\over {n}}} \\right)$$"}, {"identifier": "C", "content": "$${a \\over 2}\\cot \\left( {{\\pi \\over {2n}}} \\right)$$"}, {"identifier": "D", "content": "$$a\\cot \\left( {{\\pi \\over {2n}}} \\right)$$ "}] | ["C"] | null | $$\tan \left( {{\pi \over n}} \right) = {a \over {2r}};\,\,\sin \left( {{\pi \over n}} \right) = {a \over {2R}}$$
<br><br>$$r + R = {a \over 2}\left[ {\cot {\pi \over n} + \cos ec{\pi \over n}} \right]$$
<br><br>
<br>$$ = {a \over 2}\left[ {{{\cos {\pi \over n} + 1} \over {\sin {\pi \over n}}}} \right]$$
<br><br>$$ = {a \over 2}\left[ {{{2{{\cos }^2}{\pi \over {2n}}} \over {2\sin {\pi \over {2n}}\cos {\pi \over {2n}}}}} \right]$$
<br><br>$$ = {a \over 2}\cot {\pi \over {2\pi }}$$<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwj5c728/3dbfb32f-aec8-426a-86ca-495696731c70/859754f0-503c-11ec-87d6-07ff3bfe6155/file-1kwj5c729.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="AIEEE 2003 Mathematics - Properties of Triangle Question 33 English Explanation"> | mcq | aieee-2003 |
9HMUIzA8ObB5e9Yg | maths | properties-of-triangle | half-angle-formulae | If in a $$\Delta ABC$$ $$a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2},$$ then the sides $$a, b$$ and $$c$$ : | [{"identifier": "A", "content": "satisfy $$a+b=c$$"}, {"identifier": "B", "content": "are in A.P"}, {"identifier": "C", "content": "are in G.P "}, {"identifier": "D", "content": "are in H.P"}] | ["B"] | null | If $$a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2}$$
<br><br>$$a\left[ {\cos C + 1} \right] + c\left[ {\cos A + 1} \right] = 3b$$
<br><br>$$\left( {a + c} \right) + \left( {a\cos C + c\cos \,B} \right) = 3b$$
<br><br>$$a + c + b = 3b$$ or $$a + c = 2b$$
<br><br>or $$a,b,c$$ are in $$A.P.$$ | mcq | aieee-2003 |
Y0Ghl4TgTp0GRL5s | maths | properties-of-triangle | half-angle-formulae | For a regular polygon, let $$r$$ and $$R$$ be the radii of the inscribed and the circumscribed circles. A $$false$$ statement among the following is : | [{"identifier": "A", "content": "There is a regular polygon with $${r \\over R} = {1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "There is a regular polygon with $${r \\over R} = {2 \\over 3}$$ "}, {"identifier": "C", "content": "There is a regular polygon with $${r \\over R} = {{\\sqrt 3 } \\over 2}$$ "}, {"identifier": "D", "content": "There is a regular polygon with $${r \\over R} = {1 \\over 2}$$ "}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265911/exam_images/oijmrfmejq2eg8qdfmzd.webp" loading="lazy" alt="AIEEE 2010 Mathematics - Properties of Triangle Question 27 English Explanation">
<br><br>If $$O$$ is center of polygon and
<br><br>$$AB$$ is one of the side, then by figure
<br><br>$$\cos {\pi \over n} = {r \over R}$$
<br><br>$$ \Rightarrow {r \over R} = {1 \over 2},{1 \over {\sqrt 2 }},{{\sqrt 3 } \over 2}\,\,for$$
<br><br>$$n = 3,4,6$$ respectively. | mcq | aieee-2010 |
lgnyq5za | maths | properties-of-triangle | half-angle-formulae | If the line $x=y=z$ intersects the line
<br/><br/>$x \sin A+y \sin B+z \sin C-18=0=x \sin 2 A+y \sin 2 B+z \sin 2 C-9$,
<br/><br/>where $A, B, C$ are the angles of a triangle $A B C$, then $80\left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)$
<br/><br/>is equal to ______________. | [] | null | 5 | $$
\begin{aligned}
&x= y=z=k(\text { let }) \\\\
&\therefore k(\sin A+\sin B+\sin C)=18 \\\\
&\Rightarrow k\left(4 \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}\right)=18 \\\\
& k(\sin 2 A+\sin 2 B+\sin 2 C)=9 \\\\
&\Rightarrow k(4 \sin A \cdot \sin B \cdot \sin C)=9 \ldots \text { (ii) } \\\\
& \text {(ii)} /\text {(i)} \\\\
& 8 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}=\frac{9}{18} \\\\
&\Rightarrow 80 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}=5
\end{aligned}
$$ | integer | jee-main-2023-online-15th-april-morning-shift |
6TY7IytUNsnzYJJy | maths | properties-of-triangle | mediun-and-angle-bisector | In a triangle $$ABC$$, medians $$AD$$ and $$BE$$ are drawn. If $$AD=4$$,
<br/>$$\angle DAB = {\pi \over 6}$$ and $$\angle ABE = {\pi \over 3}$$, then the area of the $$\angle \Delta ABC$$ is : | [{"identifier": "A", "content": "$${{64} \\over 3}$$ "}, {"identifier": "B", "content": "$${8 \\over 3}$$ "}, {"identifier": "C", "content": "$${{16} \\over 3}$$ "}, {"identifier": "D", "content": "$${{32} \\over {3\\sqrt 3 }}$$ "}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264957/exam_images/mahs6riio0a6znddsa2r.webp" loading="lazy" alt="AIEEE 2003 Mathematics - Properties of Triangle Question 32 English Explanation">
<br><br>$$AP = {2 \over 3}AD = {8 \over 3};\,\,PD = {4 \over 3};\,\,$$
<br><br>Let $$PB=x$$
<br><br>$$\tan {60^ \circ } = {{8/3} \over x}$$
<br><br>or $$x = {8 \over {3\sqrt 3 }}$$
<br><br>Area of $$\Delta ABD$$
<br><br>$$ = {1 \over 2} \times 4 \times {8 \over {3\sqrt 3 }} = {{16} \over {3\sqrt 3 }}$$
<br><br>$$\therefore$$ Area of $$\Delta ABC$$
<br><br>$$ = 2 \times {{16} \over {3\sqrt 3 }} = {{32} \over {3\sqrt 3 }}$$
<br><br>$$\left[ \, \right.$$ As median of a $$\Delta $$ divides it into two $$\Delta 's$$ of equal area. $$\left. \, \right]$$ | mcq | aieee-2003 |
aZw5wVknZ2tAwD4dSv3rsa0w2w9jxadfk62 | maths | properties-of-triangle | mediun-and-angle-bisector | A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (–1, 1) and (2, 3). Then the centroid of this triangle is : | [{"identifier": "A", "content": "$$\\left( {{1 \\over 3},2} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{1 \\over 3},{5 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {1,{7 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 3},1} \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265818/exam_images/u883z0iudbpn9rsvcpfy.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Properties of Triangle Question 20 English Explanation"><br>
Centroid = $$\left( {{{1 + 3 - 3} \over 3},{{2 + 0 + 4} \over 3}} \right) = \left( {{1 \over 3},2} \right)$$ | mcq | jee-main-2019-online-12th-april-evening-slot |
2Pi5cD2rl2m48jUoBJ1klrkm8wt | maths | properties-of-triangle | mediun-and-angle-bisector | Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be $$\left( {{{10} \over 3},{7 \over 3}} \right)$$. If $$\alpha$$, $$\beta$$ are the roots of the equation $$a{x^2} + bx + 1 = 0$$, then the value of $${\alpha ^2} + {\beta ^2} - \alpha \beta $$ is : | [{"identifier": "A", "content": "$${{69} \\over {256}}$$"}, {"identifier": "B", "content": "$${{71} \\over {256}}$$"}, {"identifier": "C", "content": "$$ - {{71} \\over {256}}$$"}, {"identifier": "D", "content": "$$ - {{69} \\over {256}}$$"}] | ["C"] | null | 2b = a + c<br><br>$${{2a + 2} \over 3} = {{10} \over 3}$$ and $${{2b + c} \over 3} = {7 \over 3}$$<br><br>a = 4, <br><br>$$\left\{ \matrix{
2b + c = 7 \hfill \cr
2b - c = 4 \hfill \cr} \right\}$$, solving<br><br>$$b = {{11} \over 4}$$<br><br>$$c = {3 \over 2}$$<br><br>$$ \therefore $$ Quadratic Equation is $$4{x^2} + {{11} \over 4}x + 1 = 0$$<br><br>$$ \therefore $$ The value of
<br><br>$${\alpha ^2} + {\beta ^2} - \alpha \beta $$
<br><br>= $${\alpha ^2} + {\beta ^2} + 2\alpha \beta - 3\alpha \beta $$
<br><br>= $${(\alpha + \beta )^2} - 3\alpha \beta $$
<br><br>$$= {{121} \over {256}} - {3 \over 4} = - {{71} \over {256}}$$ | mcq | jee-main-2021-online-24th-february-evening-slot |
oNsrvtAu54sUcSHeU7ufP | maths | properties-of-triangle | sine-rule | In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x<sup>2</sup> – c<sup>2</sup> = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is : | [{"identifier": "A", "content": "$${y \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${c \\over 3}$$"}, {"identifier": "C", "content": "$${c \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$y"}] | ["C"] | null | Given a + b = x and ab = y
<br><br>If x<sup>2</sup> $$-$$ c<sup>2</sup> = y $$ \Rightarrow $$ (a + b)<sup>2</sup> $$-$$ c<sup>2</sup> = ab
<br><br>$$ \Rightarrow $$ a<sup>2</sup> + b<sup>2</sup> $$-$$ c<sup>2</sup> = $$-$$ ab
<br><br>$$ \Rightarrow $$ $${{{a^2} + {b^2} - {c^2}} \over {2ab}} = - {1 \over 2}$$
<br><br>$$ \Rightarrow \cos C = - {1 \over 2}$$
<br><br>$$ \Rightarrow \angle C = {{2\pi } \over 3}$$
<br><br>$$R = {c \over {2\sin C}} = {c \over {\sqrt 3 }}$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
T5cYEycLTxY97vpIuZ3rsa0w2w9jx22m6nv | maths | properties-of-triangle | sine-rule | The angles A, B and C of a triangle ABC are in A.P. and a : b = 1 : $$\sqrt 3 $$. If c = 4 cm, then the area (in sq. cm)
of this triangle is : | [{"identifier": "A", "content": "2$$\\sqrt 3 $$"}, {"identifier": "B", "content": "4$$\\sqrt 3 $$"}, {"identifier": "C", "content": "$${4 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$"}] | ["A"] | null | From the question 2B = A + C & A + B + C = $$\pi $$<br><br>
$$ \Rightarrow $$ 3B = $$\pi $$<br><br>
$$ \Rightarrow $$ B = $${\pi \over 3}$$<br><br>
$$ \therefore A + C = {{2\pi } \over 3}\sigma $$<br><br>
$${a \over b} = {1 \over {\sqrt 3 }}$$<br><br>
$${{2R\sin A} \over {2R\sin B}} = {1 \over {\sqrt 3 }}$$<br><br>
sin A = $${1 \over 2}$$<br><br>
$$ \therefore $$ A = 30<sup>o</sup><br><br>
$$ \therefore $$ a = 2, b = 2$$\sqrt 3$$, c = 4<br><br>
$$\Delta = {1 \over 2} \times 2\sqrt 3 \times 2 = 2\sqrt 3 $$
| mcq | jee-main-2019-online-10th-april-evening-slot |
1krpur0db | maths | properties-of-triangle | sine-rule | If in a triangle ABC, AB = 5 units, $$\angle B = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$ and radius of circumcircle of $$\Delta$$ABC is 5 units, then the area (in sq. units) of $$\Delta$$ABC is : | [{"identifier": "A", "content": "$$10 + 6\\sqrt 2 $$"}, {"identifier": "B", "content": "$$8 + 2\\sqrt 2 $$"}, {"identifier": "C", "content": "$$6 + 8\\sqrt 3 $$"}, {"identifier": "D", "content": "$$4 + 2\\sqrt 3 $$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l14vgj3y/f007ae64-29a9-4ba7-a06b-961129209c7b/fea546d0-ab5f-11ec-ba67-db8ddd9d738c/file-1l14vgj3z.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l14vgj3y/f007ae64-29a9-4ba7-a06b-961129209c7b/fea546d0-ab5f-11ec-ba67-db8ddd9d738c/file-1l14vgj3z.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 20th July Morning Shift Mathematics - Properties of Triangle Question 14 English Explanation"><br>As, $$\cos B = {3 \over 5} \Rightarrow B = 53^\circ $$<br><br>As, $$R = 5 \Rightarrow {c \over {\sin c}} = 2R$$<br><br>$$ \Rightarrow {5 \over {10}} = \sin c \Rightarrow C = 30^\circ $$<br><br>Now, $${b \over {\sin B}} = 2R \Rightarrow b = 2(5)\left( {{4 \over 5}} \right) = 8$$<br><br>Now, by cosine formula<br><br>$$\cos B = {{{a^2} + {c^2} - {b^2}} \over {2ac}}$$<br><br>$$ \Rightarrow {3 \over 5} = {{{a^2} + 25 - 64} \over {2(5)a}}$$<br><br>$$ \Rightarrow {a^2} - 6a - 3g = 0$$<br><br>$$\therefore$$ $$a = {{6 \pm \sqrt {192} } \over 2} = {{6 \pm 8\sqrt 3 } \over 2}$$<br><br>$$ \Rightarrow 3 + 4\sqrt 3 $$ (Reject $$a = 3 - 4\sqrt 3 $$)<br><br>Now, $$\Delta = {{abc} \over {4R}} = {{(3 + 4\sqrt 3 )(8)(5)} \over {4(5)}} = 2(3 + 4\sqrt 3 )$$<br><br>$$ \Rightarrow \Delta = (6 + 8\sqrt 3 )$$<br><br>$$\Rightarrow$$ Option (3) is correct. | mcq | jee-main-2021-online-20th-july-morning-shift |
1ktemdhtu | maths | properties-of-triangle | sine-rule | Let $${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$$, where A, B, C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then : | [{"identifier": "A", "content": "b<sup>2</sup> $$-$$ a<sup>2</sup> = a<sup>2</sup> + c<sup>2</sup>"}, {"identifier": "B", "content": "b<sup>2</sup>, c<sup>2</sup>, a<sup>2</sup> are in A.P."}, {"identifier": "C", "content": "c<sup>2</sup>, a<sup>2</sup>, b<sup>2</sup> are in A.P."}, {"identifier": "D", "content": "a<sup>2</sup>, b<sup>2</sup>, c<sup>2</sup> are in A.P."}] | ["B"] | null | $${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$$<br><br>As A, B, C are angles of triangle.<br><br>A + B + C = $$\pi$$<br><br>A = $$\pi$$ $$-$$ (B + C) ...... (1)<br><br>Similarly sinB = sin(A + C) ..... (2)<br><br>From (1) and (2)<br><br>$${{\sin (B + C)} \over {\sin (A + C)}} = {{\sin (A - C)} \over {\sin (C - B)}}$$<br><br>$$\sin (C + B).\sin (C - B) = \sin (A - C)\sin (A + C)$$<br><br>$${\sin ^2}C - {\sin ^2}B = {\sin ^2}A - {\sin ^2}C$$<br><br>$$\because$$ $$\{ \sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y\} $$<br><br>$$2{\sin ^2}C = {\sin ^2}A + {\sin ^2}B$$<br><br>By sine rule<br><br>$$2{c^2} = {a^2} + {b^2}$$<br><br>$$\Rightarrow$$ b<sup>2</sup>, c<sup>2</sup> and a<sup>2</sup> are in A.P. | mcq | jee-main-2021-online-27th-august-morning-shift |
sz85xbGPliuKueJlPT7k9k2k5irbfkf | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The number of real roots of the equation,
<br/>e<sup>4x</sup> + e<sup>3x</sup> – 4e<sup>2x</sup> + e<sup>x</sup> + 1 = 0 is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["A"] | null | e<sup>4x</sup> + e<sup>3x</sup> – 4e<sup>2x</sup> + e<sup>x</sup> + 1 = 0
<br><br>Dividing by e<sup>2x</sup>, we get
<br><br>e<sup>2x</sup> + e<sup>x</sup> - 4 + $${1 \over {{e^x}}}$$ + $${1 \over {{e^{2x}}}}$$ = 0
<br><br>$$ \Rightarrow $$ $$\left( {{e^{2x}} + {1 \over {{e^{2x}}}}} \right) + \left( {{e^x} + {1 \over {{e^x}}}} \right)$$ - 4 = 0
<br><br>$$ \Rightarrow $$ $${\left( {{e^x} + {1 \over {{e^x}}}} \right)^2} - 2$$ + $$\left( {{e^x} + {1 \over {{e^x}}}} \right)$$ - 4 = 0
<br><br>Let $${{e^x} + {1 \over {{e^x}}} = z}$$
<br><br>(z<sup>2</sup>
– 2) + (z) – 4 = 0
<br><br>$$ \Rightarrow $$ z<sup>2</sup>
+ z – 6 = 0
<br><br>$$ \Rightarrow $$ z = –3, 2
<br><br>$$ \therefore $$ $${{e^x} + {1 \over {{e^x}}} = 2}$$
<br><br>$$ \Rightarrow $$ (e<sup>x</sup> – 1)<sup>2</sup> = 0 $$ \Rightarrow $$ x = 0.
<br><br>$$ \therefore $$ Number of real roots = 1 | mcq | jee-main-2020-online-9th-january-morning-slot |
2aAWMYAQBGnzJe5QFG1kluhn0w9 | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The sum of 162<sup>th</sup> power of the roots of the equation x<sup>3</sup> $$-$$ 2x<sup>2</sup> + 2x $$-$$ 1 = 0 is ________. | [] | null | 3 | x<sup>3</sup> $$-$$ 2x<sup>2</sup> + 2x $$-$$ 1 = 0<br><br>x = 1 satisfying the equation<br><br>$$ \therefore $$ x $$-$$ 1 is factor of <br><br>x<sup>3</sup> $$-$$ 2x<sup>2</sup> + 2x $$-$$ 1<br><br>= (x $$-$$ 1) (x<sup>2</sup> $$-$$ x + 1) = 0<br><br>x = 1, $${{1 + i\sqrt 3 } \over 2},{{1 - i\sqrt 3 } \over 2}$$<br><br>x = 1, $$-$$ $$\omega$$<sup>2</sup>, $$-$$$$\omega$$<br><br>Sum of 162<sup>th</sup> power of roots<br><br>= (1)<sup>162</sup> + ($$-$$$$\omega$$<sup>2</sup>)<sup>162</sup> + ($$-$$$$\omega$$)<sup>162</sup><br><br>= 1 + ($$\omega$$)<sup>324</sup> + ($$\omega$$)<sup>162</sup><br><br>= 1 + 1 + 1 = 3 | integer | jee-main-2021-online-26th-february-morning-slot |
1krw18wj6 | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The number of real roots of the equation $${e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0$$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $${e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0$$<br><br>$$ \Rightarrow {\left( {{e^{3x}} - 1} \right)^2} - {e^x}\left( {{e^{3x}} - 1} \right) = 12{e^{2x}}$$<br><br>$${\left( {{e^{3x}} - 1} \right)^2}\left( {{e^x} - {e^{ - x}} - {e^{ - 2x}}} \right) = 12$$<br><br>$$ \Rightarrow \underbrace {{e^x} - {e^{ - x}} - {e^{ - 2x}}}_{increa{\mathop{\rm sing}\nolimits} \,(let\,f(x))} = {{12} \over {\underbrace {{e^{3x}} - 1}_{decrea{\mathop{\rm sing}\nolimits} \,(let\,g(x))}}}$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263594/exam_images/bip7efus8gpxzmiurj3w.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - Quadratic Equation and Inequalities Question 79 English Explanation"><br>$$\Rightarrow$$ No. of real roots = 2 | mcq | jee-main-2021-online-25th-july-morning-shift |
1krygctud | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The number of real roots of the equation e<sup>4x</sup> $$-$$ e<sup>3x</sup> $$-$$ 4e<sup>2x</sup> $$-$$ e<sup>x</sup> + 1 = 0 is equal to ______________. | [] | null | 2 | t<sup>4</sup> $$-$$ t<sup>3</sup> $$-$$ 4t<sup>2</sup> $$-$$ t + 1 = 0, e<sup>x</sup> = t > 0<br><br>$$ \Rightarrow {t^2} - t - 4 - {1 \over t} + {1 \over {{t^2}}} = 0$$<br><br>$$ \Rightarrow {\alpha ^2} - \alpha - 6 = 0,\alpha = t + {1 \over t} \ge 2$$<br><br>$$ \Rightarrow \alpha = 3, - 2$$ (reject)<br><br>$$ \Rightarrow t + {1 \over t} = 3$$<br><br>$$\Rightarrow$$ The number of real roots = 2 | integer | jee-main-2021-online-27th-july-evening-shift |
1ks09h31j | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | Let $$\alpha$$, $$\beta$$ be two roots of the <br/><br/>equation x<sup>2</sup> + (20)<sup>1/4</sup>x + (5)<sup>1/2</sup> = 0. Then $$\alpha$$<sup>8</sup> + $$\beta$$<sup>8</sup> is equal to | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "100"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "160"}] | ["C"] | null | x<sup>2</sup> + (20)<sup>1/4</sup>x + (5)<sup>1/2</sup> = 0
<br/><br/>$$ \Rightarrow $$ x<sup>2</sup> + $$\sqrt 5$$ = - (20)<sup>1/4</sup>x
<br/><br/>Squaring both sides, we get
<br/><br/>$${\left( {{x^2} + \sqrt 5 } \right)^2} = \sqrt {20} {x^2}$$<br><br>$$ \Rightarrow $$ x<sup>4</sup> = $$-$$5 $$\Rightarrow$$ x<sup>8</sup> = 25<br><br>$$ \Rightarrow $$ $$\alpha$$<sup>8</sup> + $$\beta$$<sup>8</sup> = 50 | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktk93082 | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | The sum of the roots of the equation <br/><br/>$$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$$, is : | [{"identifier": "A", "content": "log<sub>2</sub> 14"}, {"identifier": "B", "content": "log<sub>2</sub> 11"}, {"identifier": "C", "content": "log<sub>2</sub> 12"}, {"identifier": "D", "content": "log<sub>2</sub> 13"}] | ["B"] | null | $$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$$<br><br>$${\log _2}({2^{x + 1}}) - {\log _2}{(3 + {2^x})^2} + {\log _2}(10 - {2^{ - x}}) = 0$$<br><br>$$lo{g_2}\left( {{{{2^{x + 1}}.(10 - {2^{ - x}})} \over {{{(3 + {2^x})}^2}}}} \right) = 0$$<br><br>$${{2({{10.2}^{ - x}} - 1)} \over {{{(3 + {2^x})}^2}}} = 1$$<br><br>$$ \Rightarrow {20.2^x} - 2 = 9 + {2^{2x}} + {6.2^x}$$<br><br>$$\therefore$$ $${({2^x})^2} - 14({2^x}) + 11 = 0$$<br><br>Roots are 2<sup>x<sub>1</sub></sup> & 2<sup>x<sub>2</sub></sup><br><br>$$\therefore$$ 2<sup>x<sub>1</sub></sup> . 2<sup>x<sub>2</sub></sup> = 11<br><br>x<sub>1</sub> + x<sub>2</sub> = log<sub>2</sub>(11) | mcq | jee-main-2021-online-31st-august-evening-shift |
1ktob9klc | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | Let f(x) be a polynomial of degree 3 such that<br/> $$f(k) = - {2 \over k}$$ for k = 2, 3, 4, 5. Then the value of 52 $$-$$ 10f(10) is equal to : | [] | null | 26 | $$k\,f(k) + 2 = \lambda (x - 2)(x - 3)(x - 4)(x - 5)$$ .... (1)<br><br>put x = 0<br><br>we get $$\lambda = {1 \over {60}}$$<br><br>Now, put $$\lambda$$ in equation (1)<br><br>$$ \Rightarrow kf(k) + 2 = {1 \over {60}}(x - 2)(x - 3)(x - 4)(x - 5)$$<br><br>Put x = 10<br><br>$$ \Rightarrow 10f(10) + 2 = {1 \over {60}}(8)(7)(6)(5)$$<br><br>$$ \Rightarrow 52 - 10f(10) = 52 - 26 = 26$$ | integer | jee-main-2021-online-1st-september-evening-shift |
1l54aj2cv | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>Let $$\alpha$$ be a root of the equation 1 + x<sup>2</sup> + x<sup>4</sup> = 0. Then, the value of $$\alpha$$<sup>1011</sup> + $$\alpha$$<sup>2022</sup> $$-$$ $$\alpha$$<sup>3033</sup> is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$\\alpha$$"}, {"identifier": "C", "content": "1 + $$\\alpha$$"}, {"identifier": "D", "content": "1 + 2$$\\alpha$$"}] | ["A"] | null | <p>Given, $$\alpha$$ is a root of the equation 1 + x<sup>2</sup> + x<sup>4</sup> = 0</p>
<p>$$\therefore$$ $$\alpha$$ will satisfy the equation.</p>
<p>$$\therefore$$ 1 + $$\alpha$$<sup>2</sup> + $$\alpha$$<sup>4</sup> = 0</p>
<p>$${\alpha ^2} = {{ - 1 \pm \sqrt {1 - 4} } \over 2}$$</p>
<p>$$ = {{ - 1 \pm \sqrt 3 i} \over 2}$$</p>
<p>$$\therefore$$ $${\alpha ^2} = \omega \,ar\,{\omega ^2}$$</p>
<p>Now,</p>
<p>$${\alpha ^{1011}} + {\alpha ^{2022}} - {\alpha ^{3033}}$$</p>
<p>$$ = \alpha \,.\,{({\alpha ^2})^{505}} + {({\alpha ^2})^{1011}} - \alpha \,.\,{({\alpha ^2})^{1516}}$$</p>
<p>$$ = \alpha {(\omega )^{505}} + {(\omega )^{1011}} - \alpha \,.\,{(\omega )^{1516}}$$</p>
<p>$$ = \alpha \,.\,{({\omega ^3})^{168}}\,.\,\omega + {({\omega ^3})^{337}} - \alpha \,.\,{({\omega ^3})^{505}}\,.\,\omega $$</p>
<p>$$ = \alpha \,\omega + 1 - \alpha \,\omega $$</p>
<p>$$ = 1$$</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l57oztwx | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>If the sum of all the roots of the equation <br/><br/>$${e^{2x}} - 11{e^x} - 45{e^{ - x}} + {{81} \over 2} = 0$$ is $${\log _e}p$$, then p is equal to ____________.</p> | [] | null | 45 | Given that
<br/><br/>$$e^{2 x}-11 e^x-45 e^{-x}+\frac{81}{2}=0 $$
<br/><br/>$$\Rightarrow 2 e^{3 x}-22 e^{2 x}-90+81 e^x=0 $$
<br/><br/>$$\Rightarrow 2\left(e^x\right)^3-22\left(e^x\right)^2+81 e^x-90=0$$
<br/><br/>Let $ e^x=y$
<br/><br/>$$
\Rightarrow 2 y^3-22 y^2+81 y-90=0
$$
<br/><br/>Product of roots $\left(y_1, y_2, y_3\right)$
<br/><br/>$$
y_1 \cdot y_2 \cdot y_3=\frac{-(-90)}{2}=45
$$
<br/><br/>Let $x_1, x_2$, and $x_3$ be roots of given equation
<br/><br/>$$\Rightarrow e^{x_1} \cdot e^{x_2} \cdot e^{x_3} = 45 $$
<br/><br/>$$\Rightarrow e^{x_1+x_2+x_3} =45$$
<br/><br/>$$\Rightarrow x_1+x_2+x_3 =\log _e 45=\log _e p $$
<br/><br/>$$\Rightarrow p = 45$$ | integer | jee-main-2022-online-27th-june-morning-shift |
1l6duvpmp | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>If $$\alpha, \beta, \gamma, \delta$$ are the roots of the equation $$x^{4}+x^{3}+x^{2}+x+1=0$$, then $$\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$4"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}] | ["B"] | null | <p>When, $${x^5} = 1$$</p>
<p>then $${x^5} - 1 = 0$$</p>
<p>$$ \Rightarrow (x - 1)({x^4} + {x^3} + {x^2} + x + 1) = 0$$</p>
<p>Given, $${x^4} + {x^3} + {x^2} + x + 1 = 0$$ has roots $$\alpha$$, $$\beta$$, $$\gamma$$ and 8.</p>
<p>$$\therefore$$ Roots of $${x^5} - 1 = 0$$ are 1, $$\alpha$$, $$\beta$$, $$\gamma$$ and 8.</p>
<p>We know, Sum of p<sup>th</sup> power of n<sup>th</sup> roots of unity = 0. (If p is not multiple of n) or n (If p is multiple of n)</p>
<p>$$\therefore$$ Here, Sum of p<sup>th</sup> power of n<sup>th</sup> roots of unity</p>
<p>$$ = {1^p} + {\alpha ^p} + {\beta ^p} + {\gamma ^p} + {8^p} = \left\{ {\matrix{
0 & ; & {\mathrm{If\,p\,is\,not\,multiple\,of\,5}} \cr
5 & ; & {\mathrm{If\,p\,is\,multiple\,of\,5}} \cr
} } \right.$$</p>
<p>Here, $$p = 2021$$, which is not multiple of 5.</p>
<p>$$\therefore$$ $${1^{2021}} + {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {8^{2021}} = 0$$</p>
<p>$$ \Rightarrow {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {8^{2021}} = - 1$$</p>
<p></p> | mcq | jee-main-2022-online-25th-july-morning-shift |
jaoe38c1lscn5p78 | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>If $$\alpha, \beta$$ are the roots of the equation, $$x^2-x-1=0$$ and $$S_n=2023 \alpha^n+2024 \beta^n$$, then :</p> | [{"identifier": "A", "content": "$$2 S_{12}=S_{11}+S_{10}$$\n"}, {"identifier": "B", "content": "$$S_{12}=S_{11}+S_{10}$$\n"}, {"identifier": "C", "content": "$$S_{11}=S_{10}+S_{12}$$\n"}, {"identifier": "D", "content": "$$2 S_{11}=S_{12}+S_{10}$$"}] | ["B"] | null | <p>$$\begin{aligned}
& x^2-x-1=0 \\
& S_n=2023 \alpha^n+2024 \beta^n \\
& S_{n-1}+S_{n-2}=2023 \alpha^{n-1}+2024 \beta^{n-1}+2023 \alpha^{n-2}+2024 \beta^{n-2} \\
& =2023 \alpha^{n-2}[1+\alpha]+2024 \beta^{n-2}[1+\beta] \\
& =2023 \alpha^{n-2}\left[\alpha^2\right]+2024 \beta^{n-2}\left[\beta^2\right] \\
& =2023 \alpha^n+2024 \beta^n \\
& S_{n-1}+S_{n-2}=S_n \\
& P_{u t} n=12 \\
& S_{11}+S_{10}=S_{12}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsfl1j5e | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>Let the set $$C=\left\{(x, y) \mid x^2-2^y=2023, x, y \in \mathbb{N}\right\}$$. Then $$\sum_\limits{(x, y) \in C}(x+y)$$ is equal to _________.</p> | [] | null | 46 | <p>First, let's consider the equation $$x^2 - 2^y = 2023$$ where $$x$$ and $$y$$ are natural numbers. Our goal is to find all the pairs $$(x, y)$$ that satisfy this equation and then sum the values of $$x+y$$ for each pair in set $$C$$.
<p>Since $$2023$$ is an odd number, and $$x^2$$, the square of any natural number, is even when $$x$$ is even and odd when $$x$$ is odd, we can determine that for the left-hand side of the equation to be odd (thus equal to $$2023$$), $$x$$ must be odd since the right-hand side of the equation ($$2^y$$) is always even as it represents a power of two.</p>
<p>Also, $$2023$$ can be factored into prime factors to further analyze the possible solutions:</p>
<p>$$2023 = 7 \times 17 \times 17$$</p>
<p>Thus, allowing us to rewrite the equation as:</p>
<p>$$x^2 - 2^y = 7 \times 17^2$$</p>
<p>The next step is to check for potential values of $$x$$ that would fit the equation, keeping in mind that $$x$$ must be odd. We can try to express $$x^2$$ as $$7 \times 17^2$$ plus a power of $$2$$, recognizing that we are looking for the decomposition of the form:</p>
<p>$$x^2 = 7 \times 17^2 + 2^y$$</p>
<p>By examining the powers of $$2$$ and keeping in mind that they grow very quickly, we can reason that $$y$$ cannot be very large because $$x^2$$ must not exceed $$2023$$ by a large margin.</p>
<p>Let's start by trying the lowest values for $$y$$ since that would make $$2^y$$ small and $$x$$ has a better chance of being a natural number:</p>
<ol>
<li>For $$y=1$$:</li>
</ol>
<p>$$x^2 = 2023 + 2^1 = 2023 + 2 = 2025$$</p>
<p>Surprisingly, we find a perfect square since $$45^2 = 2025$$. Therefore, $$(x, y) = (45, 1)$$ is one solution.</p>
<ol>
<li>For $$y=2$$ or higher:</li>
</ol>
<p>$$2^y$$ becomes at least $$4$$ and increases exponentially, so $$x^2$$ must be at least $$2027$$ or higher in such cases. There's no natural number between $$45$$ and $$46$$, and $$46^2$$ far exceeds the target (2116), making it impossible for $$x^2$$ to be less than $$2116$$ for any larger $$y$$.</p>
<p>Hence, it appears there is only one possible solution: $$(x, y) = (45, 1)$$.</p>
<p>Therefore, the sum $$\sum_\limits{(x, y) \in C}(x+y)$$ for this set will consist of only this one pair:</p>
<p>$$\sum_\limits{(x, y) \in C}(x+y) = 45 + 1 = 46$$</p>
<p>So the answer is $$46$$. </p></p> | integer | jee-main-2024-online-29th-january-evening-shift |
lvc57nwy | maths | quadratic-equation-and-inequalities | algebraic-equations-of-higher-degree | <p>Let $$x_1, x_2, x_3, x_4$$ be the solution of the equation $$4 x^4+8 x^3-17 x^2-12 x+9=0$$ and $$\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$$. Then the value of $$m$$ is _________.</p> | [] | null | 221 | <p>$$\begin{aligned}
& 4 x^4+8 x^3-17 x^2-12 x+9=0 \\
& (x+1)\left(4 x^3+4 x^2-21 x+9\right)=0 \\
& (x+1)(x+3)\left(4 x^2-8 x+3\right)=0 \\
& (x+1)(x+3)\left(4 x^2-6 x-2 x+3\right)=0 \\
& (x+1)(x+3)(2 x(2 x-3)-1(2 x-3))=0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd5jqj1/376c6606-3ddf-4373-89b3-5a3bdffa64e6/6d099bd0-15a8-11ef-99cc-63deb36dcc4a/file-1lwd5jqj2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd5jqj1/376c6606-3ddf-4373-89b3-5a3bdffa64e6/6d099bd0-15a8-11ef-99cc-63deb36dcc4a/file-1lwd5jqj2.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Morning Shift Mathematics - Quadratic Equation and Inequalities Question 1 English Explanation"></p>
<p>$$\begin{aligned}
& (4+1)(4+9)\left(4+\frac{1}{4}\right)\left(4+\frac{9}{4}\right)=\frac{125}{16} m \\
& 5 \times 13 \times\left(\frac{17}{4}\right) \times\left(\frac{25}{4}\right)=\frac{125}{16} m \\
& \frac{125}{16} \times[13 \times 17]=\frac{125}{16} m \\
& m=13 \times 17 \\
& m=221
\end{aligned}$$</p> | integer | jee-main-2024-online-6th-april-morning-shift |
i4NFcZin92dKvntr | maths | quadratic-equation-and-inequalities | common-roots | If one root of the equation $${x^2} + px + 12 = 0$$ is 4, while the equation $${x^2} + px + q = 0$$ has equal roots,
<br/>then the value of $$'q'$$ is | [{"identifier": "A", "content": "4 "}, {"identifier": "B", "content": "12 "}, {"identifier": "C", "content": "3 "}, {"identifier": "D", "content": "$${{49} \\over 4}$$ "}] | ["D"] | null | $$4$$ is a root of $${x^2} + px + 12 = 0$$
<br><br>$$ \Rightarrow 16 + 4p + 12 = 0$$
<br><br>$$ \Rightarrow p = - 7$$
<br><br>Now, the equation $${x^2} + px + q = 0$$
<br><br>has equal roots.
<br><br>$$\therefore$$ $${p^2} - 4q = 0$$ $$ \Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$ | mcq | aieee-2004 |
Y1ywUGSBBkETN6q8 | maths | quadratic-equation-and-inequalities | common-roots | The quadratic equations $${x^2} - 6x + a = 0$$ and $${x^2} - cx + 6 = 0$$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is | [{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "4 "}, {"identifier": "C", "content": "3 "}, {"identifier": "D", "content": "2"}] | ["D"] | null | Let the roots of equation $${x^2} - 6x + a = 0$$ be $$\alpha $$
<br><br>and $$4$$ $$\beta $$ and that of the equation
<br><br>$${x^2} - cx + 6 = 0$$ be $$\alpha $$ and $$3\beta .$$ Then
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a$$
<br><br>and $$\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 3\beta = c;\,\,\,\,\,\,\,3\alpha \beta = 6$$
<br><br>$$ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 8$$
<br><br>$$\therefore$$ The equation becomes
<br><br>$${x^2} - 6x + 8 = 0$$
<br><br>$$ \Rightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0$$
<br><br>$$ \Rightarrow $$ roots are $$2$$ and $$4$$
<br><br>$$ \Rightarrow \alpha = 2,\beta = 1$$
<br><br>$$\therefore$$ Common root is $$2.$$ | mcq | aieee-2008 |
hpSOJ5blAQVflwER | maths | quadratic-equation-and-inequalities | common-roots | If the equations $${x^2} + 2x + 3 = 0$$ and $$a{x^2} + bx + c = 0,$$ $$a,\,b,\,c\, \in \,R,$$ have a common root, then $$a\,:b\,:c\,$$ is | [{"identifier": "A", "content": "$$1:2:3$$ "}, {"identifier": "B", "content": "$$3:2:1$$"}, {"identifier": "C", "content": "$$1:3:2$$"}, {"identifier": "D", "content": "$$3:1:2$$"}] | ["A"] | null | Given equations are
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 2x + 3 = 0\,\,\,\,\,...\left( i \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,a{x^2} + bx + c = 0\,\,\,...\left( {ii} \right)$$
<br><br>Roots of equation $$(i)$$ are imaginary roots.
<br><br>According to the question $$(ii)$$ will also have both roots same as $$(i).$$
<br><br>Thus $${a \over 1} = {b \over 2} = {c \over 3} = \lambda \left( {say} \right)$$
<br><br>$$ \Rightarrow a = \lambda ,b = 2\lambda ,c = 3\lambda $$
<br><br>Hence, required ratio is $$1:2:3$$ | mcq | jee-main-2013-offline |
pHnNgwGeazDuA1PHdH1JN | maths | quadratic-equation-and-inequalities | common-roots | If the equations x<sup>2</sup> + bx−1 = 0 and x<sup>2</sup> + x + b = 0 have a common root different from −1, then $$\left| b \right|$$ is equal to : | [{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\sqrt 3 $$"}] | ["D"] | null | Given,
<br><br>x<sup>2</sup> + bx $$-$$ 1 = 0 . . . . .(1)
<br><br>and x<sup>2</sup> + x + b = 0 . . . . . (2)
<br><br>Performing (1) $$-$$ (2) we get,
<br><br>bx $$-$$ 1 $$-$$ x $$-$$ b = 0
<br><br>$$ \Rightarrow $$ x(b $$-$$ 1) = b + 1
<br><br>$$ \Rightarrow $$ x = $${{b + 1} \over {b - 1}}$$
<br><br>putting value of x in equation (2),
<br><br>$${\left( {{{b + 1} \over {b - 1}}} \right)^2} + \left( {{{b + 1} \over {b - 1}}} \right) + b = 0$$
<br><br>$$ \Rightarrow $$ (b + 1)<sup>2</sup> + (b + 1) (b $$-$$ 1) + b (b $$-$$ 1)<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ b<sup>2</sup> + 2b + 1 + b<sup>2</sup> $$-$$ 1 + b (b<sup>2</sup> $$-$$ 2b + 1) = 0
<br><br>$$ \Rightarrow $$ 2b<sup>3</sup> + 2b + b<sup>3</sup> $$-$$ 2b<sup>2</sup> + b = 0
<br><br>$$ \Rightarrow $$ b<sup>3</sup> + 3b = 0
<br><br>$$ \Rightarrow $$ b(b<sup>2</sup> + 3) = 0
<br><br>b<sup>2</sup> = $$-$$ 3, b = 0
<br><br>$$ \therefore $$ b = $$ \pm \sqrt 3 i$$
<br><br>$$ \Rightarrow $$ $$\left| b \right|$$ = $$\sqrt 3 $$ | mcq | jee-main-2016-online-9th-april-morning-slot |
2ZbFQGhj0vnf1RUxo53rsa0w2w9jxaoprqa | maths | quadratic-equation-and-inequalities | common-roots | If $$\alpha $$, $$\beta $$ and $$\gamma $$ are three consecutive terms of a non-constant G.P. such that the equations $$\alpha $$x
<sup>2</sup>
+ 2$$\beta $$x + $$\gamma $$ = 0 and
x<sup>2</sup>
+ x – 1 = 0 have a common root, then $$\alpha $$($$\beta $$ + $$\gamma $$) is equal to : | [{"identifier": "A", "content": "$$\\alpha $$$$\\gamma $$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$\\beta $$$$\\gamma $$"}, {"identifier": "D", "content": "$$\\alpha $$$$\\beta $$"}] | ["C"] | null | Let the common ratio of G.P is <b>r</b><br><br>
Then the equation of $$\alpha {x^2} + 2\beta x + \gamma = 0$$<br><br>
$$ \Rightarrow $$ $$\alpha {x^2} + 2\alpha rx + \alpha {r^2} = 0$$<br><br>
$$ \Rightarrow $$ x<sup>2</sup> + 2rx + r<sup>2</sup> = 0 ........<b>(i)</b><br>
Equation (i) and x<sup>2</sup> + x - 1 = 0 ......... <b>(ii)</b> has a common root.<br><br>
Now (i) - (ii) $$ \Rightarrow $$ (2r-1)x + (r<sup>2</sup>+1) = 0<br><br>
$$ \Rightarrow $$ x = $${{ - \left( {{r^2} + 1} \right)} \over {2r - 1}}$$ ....... <b>(iii)</b><br><br>
Now putting (iii) in equation (ii) <br><br>
$$ \Rightarrow $$ (r<sup>2</sup>+1)<sup>2</sup> - (r<sup>2</sup>+1) (2r - 1) - (2r - 1)<sup>2</sup> = 0<br><br>
$$ \Rightarrow $$ r<sup>4</sup> - 2r<sup>3</sup> - r<sup>2</sup> + 2r + 1 = 0 ..........<b> (iv)</b><br><br>
dividing equation (iv) by r<sup>2</sup><br><br>
$$ \Rightarrow $$ r<sup>2</sup> - 2r - 1 + $${2 \over r} + {1 \over {{r^2}}} = 0$$<br><br>
$$ \Rightarrow $$ $${\left( {r - {1 \over r}} \right)^2} - 2\left( {r - {1 \over r}} \right) + 1 = 0$$<br><br>
$$ \Rightarrow $$ $${\left( {r - {1 \over r} - r} \right)^2} = 0$$<br><br>
$$ \Rightarrow $$ $${{{r^2} - 1} \over r} = 1$$ $$ \Rightarrow $$ r<sup>2</sup> = 1 + r<br><br>
Now $$\alpha $$($$\beta $$ + $$\gamma $$) $$ \Rightarrow $$ $$\alpha $$($$\alpha $$r + $$\alpha $$r<sup>2</sup>)<br><br> $$ \Rightarrow $$ $$\alpha ^2 r (1 + r)$$ [$$ \because $$ r<sup>2</sup> = 1 + r]<br><br>
$$ \Rightarrow $$ $$\alpha ^2 r . r^2$$ $$ \Rightarrow $$ $$( \alpha r )\,( \alpha r^2 )$$ $$ \Rightarrow $$ $$\beta $$$$\gamma $$
| mcq | jee-main-2019-online-12th-april-evening-slot |
zmYqIAqG6Q4t0K97bk7k9k2k5k7hec6 | maths | quadratic-equation-and-inequalities | common-roots | Let a, b $$ \in $$ R, a $$ \ne $$ 0 be such that the equation,
ax<sup>2</sup> – 2bx + 5 = 0 has a repeated root $$\alpha $$, which
is also a root of the equation, x<sup>2</sup> – 2bx – 10 = 0.
If $$\beta $$ is the other root of this equation, then
$$\alpha $$<sup>2</sup> + $$\beta $$<sup>2</sup> is equal to : | [{"identifier": "A", "content": "28"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "26"}, {"identifier": "D", "content": "25"}] | ["D"] | null | Roots of equation ax<sup>2</sup>
– 2bx + 5 = 0 are $$\alpha $$, $$\alpha $$.
<br><br>$$ \therefore $$ $$\alpha $$ + $$\alpha $$ = $${{2b} \over a}$$
<br><br>$$ \Rightarrow $$ 2$$\alpha $$ = $${{2b} \over a}$$
<br><br>$$ \Rightarrow $$ $$\alpha $$ = $${{b} \over a}$$ ....(1)
<br><br>and $$\alpha $$<sup>2</sup> = $${5 \over a}$$ ....(2)
<br><br>From (1) and (2), we get
<br><br>$${{{b^2}} \over {{a^2}}} = {5 \over a}$$
<br><br>$$ \Rightarrow $$ b<sup>2</sup> = 5a
<br><br>$$\alpha $$, $$\beta $$ are the roots of equation x<sup>2</sup>
– 2bx – 10 = 0
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = 2b
<br><br>and $$\alpha $$$$\beta $$ = -10
<br><br>As $$\alpha $$ is a root of the equation x<sup>2</sup>
– 2bx – 10 = 0.
<br><br>$$ \therefore $$ $$\alpha $$<sup>2</sup> - 2b$$\alpha $$ - 10 = 0
<br><br>$$ \Rightarrow $$ $${{{b^2}} \over {{a^2}}} - {{2{b^2}} \over a} - 10 = 0$$
<br><br>$$ \Rightarrow $$ $${{5a} \over {{a^2}}} - {{10a} \over a} - 10 = 0$$
<br><br>$$ \Rightarrow $$ $${5 \over a} - 10 - 10 = 0$$
<br><br>$$ \Rightarrow $$ $$a$$ = $${1 \over 4}$$
<br><br>$$ \Rightarrow $$ $${\alpha ^2}$$ = 20
<br><br>$$\alpha $$$$\beta $$ = -10
<br><br>$$ \Rightarrow $$$${\beta ^2}$$ = 5
<br><br>$$ \therefore $$ $${\alpha ^2} + {\beta ^2}$$ = 20 + 5 = 25 | mcq | jee-main-2020-online-9th-january-evening-slot |
1ktd4mp2o | maths | quadratic-equation-and-inequalities | common-roots | Let $$\lambda$$ $$\ne$$ 0 be in R. If $$\alpha$$ and $$\beta$$ are the roots of the equation x<sup>2</sup> $$-$$ x + 2$$\lambda$$ = 0, and $$\alpha$$ and $$\gamma$$ are the roots of equation 3x<sup>2</sup> $$-$$ 10x + 27$$\lambda$$ = 0, then $${{\beta \gamma } \over \lambda }$$ is equal to ____________. | [] | null | 18 | 3$$\alpha$$<sup>2</sup> $$-$$ 10$$\alpha$$ + 27$$\lambda$$ = 0 ..... (1)<br><br>$$\alpha$$<sup>2</sup> $$-$$ $$\alpha$$ + 2$$\lambda$$ = 0 ...... (2)<br><br>(1) $$-$$ 3(2) gives<br><br>$$-$$7$$\alpha$$ + 21$$\lambda$$ = 0 $$\Rightarrow$$ $$\alpha$$ = 3$$\lambda$$<br><br>Put $$\alpha$$ = 3$$\lambda$$ in equation (1) we get<br><br>9$$\lambda$$<sup>2</sup> $$-$$ 3$$\lambda$$ + 2$$\lambda$$ $$-$$ 0<br><br>9$$\lambda$$<sup>2</sup> = $$\lambda$$ $$\Rightarrow$$ $$\lambda$$ = $${1 \over 9}$$ as $$\lambda$$ $$\ne$$ 0<br><br>Now, $$\alpha$$ = 3$$\lambda$$ $$\Rightarrow$$ $$\lambda$$ = $${1 \over 3}$$ <br><br>$$\alpha$$ + $$\beta$$ = 1 $$\Rightarrow$$ $$\beta$$ = 2/3<br><br>$$\alpha$$ + $$\gamma$$ = $${10 \over 3}$$ $$\Rightarrow$$ $$\gamma$$ = 3<br><br>$${{\beta \gamma } \over \lambda } = {{{2 \over 3} \times 3} \over {{1 \over 9}}} = 18$$ | integer | jee-main-2021-online-26th-august-evening-shift |
1l59jpipu | maths | quadratic-equation-and-inequalities | common-roots | <p>Let a, b $$\in$$ R be such that the equation $$a{x^2} - 2bx + 15 = 0$$ has a repeated root $$\alpha$$. If $$\alpha$$ and $$\beta$$ are the roots of the equation $${x^2} - 2bx + 21 = 0$$, then $${\alpha ^2} + {\beta ^2}$$ is equal to :</p> | [{"identifier": "A", "content": "37"}, {"identifier": "B", "content": "58"}, {"identifier": "C", "content": "68"}, {"identifier": "D", "content": "92"}] | ["B"] | null | <p>$$a{x^2} - 2bx + 15 = 0$$ has repeated root so $${b^2} = 15a$$ and $$\alpha = {{15} \over b}$$</p>
<p>$$\because$$ $$\alpha$$ is a root of $${x^2} - 2bx + 21 = 0$$</p>
<p>So $${{225} \over {{b^2}}} = 9 \Rightarrow {b^2} = 25$$</p>
<p>Now $${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta = 4{b^2} - 42 = 100 - 42 = 58$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l6gj9sdd | maths | quadratic-equation-and-inequalities | common-roots | <p>If for some $$\mathrm{p}, \mathrm{q}, \mathrm{r} \in \mathbf{R}$$, not all have same sign, one of the roots of the equation $$\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right) x^{2}-2 \mathrm{q}(\mathrm{p}+\mathrm{r}) x+\mathrm{q}^{2}+\mathrm{r}^{2}=0$$ is also a root of the equation $$x^{2}+2 x-8=0$$, then $$\frac{\mathrm{q}^{2}+\mathrm{r}^{2}}{\mathrm{p}^{2}}$$ is equal to ____________,</p> | [] | null | 272 | <p>Let roots of</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbuiql/3cbb58ae-c17d-4fc8-9ae2-606226e918d5/7db780d0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbuiqm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbuiql/3cbb58ae-c17d-4fc8-9ae2-606226e918d5/7db780d0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbuiqm.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Morning Shift Mathematics - Quadratic Equation and Inequalities Question 49 English Explanation"> </p>
<p>$$\therefore$$ $$\alpha$$ + $$\beta$$ > 0 and $$\alpha$$$$\beta$$ > 0</p>
<p>Also, it has a common root with $${x^2} + 2x - 8 = 0$$</p>
<p>$$\therefore$$ The common root between above two equations is 4.</p>
<p>$$ \Rightarrow 16({p^2} + {q^2}) - 8q(p + r) + {q^2} + {r^2} = 0$$</p>
<p>$$ \Rightarrow (16{p^2} - 8pq + {q^2}) + (16{q^2} - 8qr + {r^2}) = 0$$</p>
<p>$$ \Rightarrow {(4p - q)^2} + {(4q - r)^2} = 0$$</p>
<p>$$ \Rightarrow q = 4p$$ and $$r = 16p$$</p>
<p>$$\therefore$$ $${{{q^2} + {r^2}} \over {{p^2}}} = {{16{p^2} + 256{p^2}} \over {{p^2}}} = 272$$</p> | integer | jee-main-2022-online-26th-july-morning-shift |
ldr04kaw | maths | quadratic-equation-and-inequalities | common-roots | If the value of real number $a>0$ for which $x^2-5 a x+1=0$ and $x^2-a x-5=0$
<br/><br/>have a common real root is $\frac{3}{\sqrt{2 \beta}}$ then $\beta$ is equal to ___________. | [] | null | 13 | <p>$${x^2} - 5\alpha x + 1 = 0$$ ..... (1)</p>
<p>$${x^2} - \alpha x - 5 = 0$$ ...... (2)</p>
<p>have a common root.</p>
<p>Subtracting (1) with (2) we'll get $$x = {6 \over {4\alpha }}$$</p>
<p>Substituting in (1)</p>
<p>$${{36} \over {16{\alpha ^2}}} - {{30} \over 4} + 1 = 0$$</p>
<p>$$ \Rightarrow {\alpha ^2} = {9 \over {26}}$$</p>
<p>$$\alpha = {3 \over {\sqrt {2 \times 13} }}$$</p>
<p>$$\therefore$$ $$\beta = 13$$</p> | integer | jee-main-2023-online-30th-january-evening-shift |
Jkj398nu3J6NteP2 | maths | quadratic-equation-and-inequalities | graph-and-sign-of-quadratic-expression | If both the roots of the quadratic equation $${x^2} - 2kx + {k^2} + k - 5 = 0$$ are less than 5, then $$k$$ lies in the interval | [{"identifier": "A", "content": "$$\\left( {5,6} \\right]$$ "}, {"identifier": "B", "content": "$$\\left( {6,\\,\\infty } \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { - \\infty ,\\,4} \\right)$$ "}, {"identifier": "D", "content": "$$\\left[ {4,\\,5} \\right]$$ "}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267782/exam_images/zwhawncrjdcscooeyd20.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Quadratic Equation and Inequalities Question 154 English Explanation">
<br><br>both roots are less than $$5,$$
<br><br>then $$(i)$$ Discriminant $$ \ge 0$$
<br><br>$$\left( {ii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p\left( 5 \right) > 0$$
<br><br>$$\left( {iii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2} < 5$$
<br><br>Hence $$\left( i \right)\,\,\,\,\,\,4{k^2} - 4\left( {{k^2} + k - 5} \right) \ge 0$$
<br><br>$$4{k^2} - 4{k^2} - 4k + 20 \ge 0$$
<br><br>$$4k \le 20 \Rightarrow k \le 5$$
<br><br>$$\left( {ii} \right)\,\,\,\,\,f\left( 5 \right) > 0;25 - 10k + {k^2} + k - 5 > 0$$
<br><br>or $${k^2} - 9k + 20 > 0$$
<br><br>or $$k\left( {k - 4} \right) - 5\left( {k - 4} \right) > 0$$
<br><br>or $$\left( {k - 5} \right)\left( {k - 4} \right) > 0$$
<br><br>$$ \Rightarrow k \in \left( { - \infty ,4} \right) \cup \left( { - \infty ,5} \right)$$
<br><br>$$\left( {iii} \right)\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2}$$
<br><br>$$ = - {b \over {2a}} = {{2k} \over 2} < 5$$
<br><br>The intersection of $$(i)$$, $$(ii)$$ & $$(iii)$$ gives
<br><br>$$k \in \left( { - \infty ,4} \right).$$ | mcq | aieee-2005 |
krncAZPJvsT4aI3N | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | If $$a \in R$$ and the equation $$ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$$ (where [$$x$$] denotes the greater integer $$ \le x$$) has no integral solution, then all possible values of a lie in the interval : | [{"identifier": "A", "content": "$$\\left( { - 2, - 1} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - \\infty , - 2} \\right) \\cup \\left( {2,\\infty } \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { - 1,0} \\right) \\cup \\left( {0,1} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {1,2} \\right)$$ "}] | ["C"] | null | Given, $$ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$$
<br>As we know, $$\left[ x \right] + \left\{ x \right\} = x$$
<br>where $$\left[ x \right]$$ is integral part and $$\left\{ x \right\}$$ is fractional part.
<br>$$\therefore$$$$\left\{ x \right\} = x - \left[ x \right]$$
<br>Now put $$\left\{ x \right\}$$ inplace of $$x - \left[ x \right]$$ in the equation.
<br>The new equation is $$ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} + {a^2} = 0$$
<p>[<b>Note :</b> Question says this equation has no integral solution, it means $$\left\{ x \right\} \ne $$ 0. So, $$x$$ is not a integer.]</p>
$$\therefore$$ $$\left\{ x \right\}$$ = $${{ - 2 \pm \sqrt {4 - 4 \times \left( { - 3} \right){a^2}} } \over { - 6}}$$
<br> = $${{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}$$
<br>As $$\left\{ x \right\}$$ is fractional part so it is lies between 0 to 1($$0 \le \left\{ x \right\} < 1$$).
<br><br>By considering positive sign, we get
<br><br>$$0 \le {{ - 2 + \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$
<br><br>$$ \Rightarrow $$$$0 \ge - 2 + \sqrt {4 + 12{a^2}} > - 6$$
<br><br>$$ \Rightarrow $$$$2 \ge + \sqrt {4 + 12{a^2}} > - 4$$
<br><br>$$\because$$$$ + \sqrt {4 + 12{a^2}} $$ is always positive which is greater than any negative no. So can ignore the inequality $$ + \sqrt {4 + 12{a^2}} > - 4$$
<br><br>Consider this inequality,
<br>$$2 \ge + \sqrt {4 + 12{a^2}} $$
<br>$$ \Rightarrow $$ $$4 \ge 4 + 12{a^2}$$
<br>$$ \Rightarrow $$ $$12{a^2} \le 0$$
<br>$$ \Rightarrow $$ $${a^2} \le 0$$
<br>$$ \Rightarrow $$ $${a^2} = 0$$
<br>$$ \Rightarrow $$ $${a} = 0$$
<br><br>If $$a$$ = 0 then $$ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} = 0$$ so $$\left\{ x \right\}$$ becomes 0 but question says $$\left\{ x \right\}$$ $$ \ne $$ 0.
<br>So $$a$$ can't be 0.
<br><br>Now by considering negative sign, we get
<br><br><br>$$0 \le {{ - 2 - \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$
<br><br>$$ \Rightarrow $$$$0 \ge - 2 - \sqrt {4 + 12{a^2}} > - 6$$
<br><br>$$ \Rightarrow $$$$2 \ge - \sqrt {4 + 12{a^2}} > - 4$$
<br><br>As 2 is always greater than $${ - \sqrt {4 + 12{a^2}} }$$. Ignore this inequality.
<br><br>Now consider this inequality,
<br><br>$$ - \sqrt {4 + 12{a^2}} > - 4$$
<br><br>$$ \Rightarrow $$ $$\sqrt {4 + 12{a^2}} < 4$$
<br><br>$$ \Rightarrow $$ $$4 + 12{a^2} < 16$$
<br><br>$$ \Rightarrow $$ $$12{a^2} < 12$$
<br><br>$$ \Rightarrow $$ $${a^2} < 1$$
<br><br>$$ \Rightarrow $$ $$\left( {{a^2} - 1} \right) < 0$$
<br><br>$$ \Rightarrow $$ $$\left( {a + 1} \right)\left( {a - 1} \right) < 0$$
<br><br>$$ \Rightarrow $$ $$ - 1 < a < 1$$
<br><br>But earlier we found that $$a$$ $$ \ne $$ 0.
<br><br>So, the range of $$a$$ is = $$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$ | mcq | jee-main-2014-offline |
t5TJfuWEpRBw5JFGnLjgy2xukf7fkxwm | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | Let [t] denote the greatest integer $$ \le $$ t. Then the equation in x,
<br/>[x]<sup>2</sup> + 2[x+2] - 7 = 0 has : | [{"identifier": "A", "content": "no integral solution."}, {"identifier": "B", "content": "exactly two solutions."}, {"identifier": "C", "content": "exactly four integral solutions."}, {"identifier": "D", "content": "infinitely many solutions.\n"}] | ["D"] | null | $${[x]^2} + 2[x + 2] - 7 = 0$$
<br><br>$$ \Rightarrow $$ $${[x]^2} + 2[x] + 4 - 7 = 0$$
<br><br>Using the property [x + n] = [x] + n ; n $$ \in $$ I
<br><br>$$ \Rightarrow $$ $${[x]^2} + 2[x] - 3 = 0$$<br><br>let [x] = y<br><br>$${y^2} + 3y - y - 3 = 0$$<br><br>$$ \Rightarrow $$ $$(y - 1)(y + 3) = 0$$<br><br>$$[x] = 1\,or\,[x] = - 3$$<br><br>$$ \therefore $$ $$x \in \left[ {1,2} \right)\,\& \, \in \left[ { - 3, - 2} \right)$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
1kru42urc | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | Let [x] denote the greatest integer less than or equal to x. Then, the values of x$$\in$$R satisfying the equation $${[{e^x}]^2} + [{e^x} + 1] - 3 = 0$$ lie in the interval : | [{"identifier": "A", "content": "$$\\left[ {0,{1 \\over e}} \\right)$$"}, {"identifier": "B", "content": "[log<sub>e</sub>2, log<sub>e</sub>3)"}, {"identifier": "C", "content": "[1, e)"}, {"identifier": "D", "content": "[0, log<sub>e</sub>2)"}] | ["D"] | null | $${[{e^x}]^2} + [{e^x} + 1] - 3 = 0$$<br><br>$$ \Rightarrow {[{e^x}]^2} + [{e^x}] + 1 - 3 = 0$$<br><br>Let $$[{e^x}] = t$$<br><br>$$ \Rightarrow {t^2} + t - 2 = 0$$<br><br>$$ \Rightarrow t = - 2,1$$<br><br>$$[{e^x}] = - 2$$ (Not possible)<br><br>or $$[{e^x}] = 1$$ $$\therefore$$ $$1 \le {e^x} < 2$$<br><br>$$ \Rightarrow \ln (1) \le x < \ln (2)$$<br><br>$$ \Rightarrow 0 \le x < \ln (2)$$<br><br>$$ \Rightarrow x \in [0,\ln 2)$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krzmnrse | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | If [x] be the greatest integer less than or equal to x, <br/><br/>then $$\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]} $$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$2"}, {"identifier": "D", "content": "2"}] | ["B"] | null | $$\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]} $$<br><br>= [4] + [-4.5] + [5] + [-5.5] + [6] +..... + [-49.5] + [50]
<br><br>= 4 - 5 + 5 - 6 + 6 ......-50 + 50
<br><br> = 4 | mcq | jee-main-2021-online-25th-july-evening-shift |
1ldybbre4 | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | <p>The equation $${x^2} - 4x + [x] + 3 = x[x]$$, where $$[x]$$ denotes the greatest integer function, has :</p> | [{"identifier": "A", "content": "exactly two solutions in ($$-\\infty,\\infty$$)"}, {"identifier": "B", "content": "no solution"}, {"identifier": "C", "content": "a unique solution in ($$-\\infty,\\infty$$)"}, {"identifier": "D", "content": "a unique solution in ($$-\\infty,1$$)"}] | ["C"] | null | <p>$${x^2} - 4x + [x] + 3 = x[x]$$</p>
<p>$$ \Rightarrow {x^2} - 4x + [x] + 3 - x[x] = 0$$</p>
<p>$$ \Rightarrow (x - 1)(x - 3) - [x](x - 1) = 0$$</p>
<p>$$ \Rightarrow (x - 1)(x - [x] - 3) = 0$$</p>
<p>$$\therefore$$ $$x = 1$$</p>
<p>or</p>
<p>$$x - [x] - 3 = 0$$</p>
<p>$$ \Rightarrow \{ x\} - 3 = 0$$ [As $$\{ x\} = x - [x]$$]</p>
<p>$$ \Rightarrow \{ x\} = 3$$</p>
<p>But we know, $$0 < \{ x\} < 1$$</p>
<p>$$\therefore$$ $$\{ x\} \ne 3$$</p>
<p>$$\therefore$$ $$x$$ has only one solution in ($$-\infty,\infty$$) which is $$x = 1$$.</p> | mcq | jee-main-2023-online-24th-january-morning-shift |
1lgoy6y4d | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | <p>Let $$[\alpha]$$ denote the greatest integer $$\leq \alpha$$. Then $$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[\sqrt{120}]$$ is equal to __________</p> | [] | null | 825 | $$
\begin{aligned}
& {[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots .+[120]} \\\\
& E=1+1+1+2+2+2+2+2+3+3+3+3+3 \\\\
& +3+3+4+4+\ldots \\\\
& E=3 \times 1+5 \times 2+7 \times 3+\ldots .+19 \times 9+10 \times 21 \\\\
& =\sum_{r=1}^{10}(2 r+1) r=2\left[\frac{10 \times 11 \times 21}{6}\right]+\frac{10 \times 11}{2} \\\\
& =770+55 \\\\
& =825 \\\\
&
\end{aligned}
$$ | integer | jee-main-2023-online-13th-april-evening-shift |
1lh1zzike | maths | quadratic-equation-and-inequalities | greatest-integer-and-fractional-part-functions | <p>Let $$A = \{ x \in R:[x + 3] + [x + 4] \le 3\} ,$$
<br/><br/>$$B = \left\{ {x \in R:{3^x}{{\left( {\sum\limits_{r = 1}^\infty {{3 \over {{{10}^r}}}} } \right)}^{x - 3}} < {3^{ - 3x}}} \right\},$$ where [t] denotes greatest integer function. Then,</p> | [{"identifier": "A", "content": "$$B \\subset C,A \\ne B$$"}, {"identifier": "B", "content": "$$A \\subset B,A \\ne B$$"}, {"identifier": "C", "content": "$$A = B$$"}, {"identifier": "D", "content": "$$A \\cap B = \\phi $$"}] | ["C"] | null | We have,
<br/><br/>$$
\begin{aligned}
& A=\{x \in R:[x+3]+[x+4] \leq 3\} \\\\
& \text { Here, }[x+3]+[x+4] \leq 3 \\\\
& \Rightarrow [x]+3+[x]+4 \leq 3 \\\\
& (\because[x+n]=[x]+n, n \in I) \\\\
& \Rightarrow 2[x]+4 \leq 0 \Rightarrow[x] \leq-2 \\\\
& \Rightarrow x \in(-\infty,-1) \\\\
& A \equiv(-\infty,-1) ...........(i)
\end{aligned}
$$
<br/><br/>Also,
<br/><br/>$$
B=\left\{x \in R: 3^x\left(\sum\limits_{r=1}^{\infty} \frac{3}{10^r}\right)^{x-3}<3^{-3 x}\right\}
$$
<br/><br/>Here,
<br/><br/>$$
\begin{aligned}
& 3^x\left(\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+\ldots\right)^{x-3}<3^{-3 x} \\\\
\Rightarrow & 3^x\left(\frac{\frac{3}{10}}{1-\frac{1}{10}}\right)^{x-3}<3^{-3 x} \\\\
\Rightarrow & 3^x\left(\frac{1}{3}\right)^{x-3}<3^{-3 x}
\end{aligned}
$$
<br/><br/>$$
\begin{array}{ll}
\Rightarrow & 3^{x-x+3}<3^{-3 x} \Rightarrow 3^3<3^{-3 x} \\\\
\Rightarrow & -3 x>3 \Rightarrow x<-1 \\\\
\Rightarrow & x \in(-\infty,-1) \\\\
\Rightarrow & B \equiv(-\infty,-1) ...........(ii)
\end{array}
$$
<br/><br/>From equations (i) and (ii), we get
<br/><br/>$$
A=B
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
GvxG3TohU0ETSsR9 | maths | quadratic-equation-and-inequalities | inequalities | If $$a,\,b,\,c$$ are distinct $$ + ve$$ real numbers and $${a^2} + {b^2} + {c^2} = 1$$ then $$ab + bc + ca$$ is | [{"identifier": "A", "content": "less than 1 "}, {"identifier": "B", "content": "equal to 1 "}, {"identifier": "C", "content": "greater than 1 "}, {"identifier": "D", "content": "any real no."}] | ["A"] | null | As $$\,\,\,\,\,{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} > 0$$
<br><br>$$ \Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0$$
<br><br>$$ \Rightarrow 2 > 2\left( {ab + bc + ca} \right)$$
<br><br>$$ \Rightarrow ab + bc + ca < 1$$ | mcq | aieee-2002 |
1hqxM4JgTHL47dSP | maths | quadratic-equation-and-inequalities | inequalities | <b>STATEMENT - 1 :</b> For every natural number $$n \ge 2,$$
$$${1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + ........ + {1 \over {\sqrt n }} > \sqrt n .$$$
<p><b>STATEMENT - 2 :</b> For every natural number $$n \ge 2,$$,
$$$\sqrt {n\left( {n + 1} \right)} < n + 1.$$$</p>
| [{"identifier": "A", "content": "Statement - 1 is false, Statement - 2 is true "}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for statement - 1"}, {"identifier": "C", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1 "}, {"identifier": "D", "content": "Statement - 1 is true, Statement - 2 is false "}] | ["B"] | null | Statements $$2$$ is $$\sqrt {n\left( {n + 1} \right)} < n + 1,n \ge 2$$
<br><br>$$ \Rightarrow \sqrt n < \sqrt {n + 1} ,n \ge 2$$ which is true
<br><br>$$ \Rightarrow \sqrt 2 < \sqrt 3 < \sqrt 4 < \sqrt 5 < - - - - - - \sqrt n $$
<br><br>Now $$\sqrt 2 < \sqrt n \Rightarrow {1 \over {\sqrt 2 }} > {1 \over {\sqrt n }}$$
<br><br>$$\sqrt 3 < \sqrt n \Rightarrow {1 \over {\sqrt 3 }} > {1 \over {\sqrt n }};$$
<br><br>$$\sqrt n \le \sqrt n \Rightarrow {1 \over {\sqrt n }} \ge {1 \over {\sqrt n }}$$
<br><br>Also $${1 \over {\sqrt 1 }} > {1 \over {\sqrt n }}$$
<br><br>$$\therefore$$ Adding all, we get
<br><br>$${1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + ....... + {1 \over n} > {n \over {\sqrt n }} = \sqrt n $$
<br><br>Hence both the statements are correct and statement $$2$$ is a correct explanation of statement $$-1.$$ | mcq | aieee-2008 |
4ho1wNjg6e6O80jXth3rsa0w2w9jwxdf2pn | maths | quadratic-equation-and-inequalities | inequalities | All the pairs (x, y) that satisfy the inequality
<br/>$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$
<br/>also satisfy the equation | [{"identifier": "A", "content": "sin x = |sin y|"}, {"identifier": "B", "content": "sin x = 2sin y"}, {"identifier": "C", "content": "2 sin x = sin y"}, {"identifier": "D", "content": "2 |sin x | = 3 sin y"}] | ["A"] | null | $${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }} \le {2^{2{{\sin }^2}y}}$$<br><br>
$$ \Rightarrow $$ $$\sqrt {{{\sin }^2}x - 2\sin x + 5} \le 2{\sin ^2}y$$<br><br>
$$ \Rightarrow \sqrt {{{\left( {\sin x - 1} \right)}^2} + 4} \le 2{\sin ^2}y$$<br><br>
it is true when sinx = 1, |siny| = 1<br><br>
so sinx = |siny|
| mcq | jee-main-2019-online-10th-april-morning-slot |
VdoWy5AixI4ojhT7Ph1klrgdqnh | maths | quadratic-equation-and-inequalities | inequalities | Let p and q be two positive numbers such that p + q = 2 and p<sup>4</sup>+q<sup>4</sup> = 272. Then p and q are
roots of the equation : | [{"identifier": "A", "content": "x<sup>2</sup> \u2013 2x + 8 = 0"}, {"identifier": "B", "content": "x<sup>2</sup> - 2x + 136=0"}, {"identifier": "C", "content": "x<sup>2</sup> \u2013 2x + 16 = 0"}, {"identifier": "D", "content": "x<sup>2</sup> \u2013 2x + 2 = 0"}] | ["C"] | null | $${p^2} + {q^2} = {(p + q)^2} - 2pq$$<br><br>$$ = 4 - 2pq$$<br><br>Now, $${\left( {{p^2} + {q^2}} \right)^2} = {p^4} + {q^4} + 2{p^2}{q^2}$$<br><br>$$ \Rightarrow {\left( {4 - 2pq} \right)^2} = 272 + 2{p^2}{q^2}$$<br><br>$$ \Rightarrow 16 + 4{p^2}{q^2} - 16pq = 272 + 2{p^2}{q^2}$$<br><br>$$ \Rightarrow 2{p^2}{q^2} - 16pq - 256 = 0$$<br><br>$$ \Rightarrow {p^2}{q^2} - 8pq - 128 = 0$$<br><br>$$ \Rightarrow (pq - 16)(pq + 8) = 0$$<br><br>$$ \Rightarrow pq = 16, - 8$$
<br/><br/>Here, pq = - 8 is not possible as p and q are positive.
<br/><br/>$$ \therefore $$ pq = 16
<br/><br/>Now, the equation whose roots are p and q is
<br><br>$${x^2} - 2x + 16 = 0$$ | mcq | jee-main-2021-online-24th-february-morning-slot |
ABe8PX8aWeMsgSLkwa1kls48put | maths | quadratic-equation-and-inequalities | inequalities | The integer 'k', for which the inequality x<sup>2</sup> $$-$$ 2(3k $$-$$ 1)x + 8k<sup>2</sup> $$-$$ 7 > 0 is valid for every x in R, is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "0"}] | ["C"] | null | $${x^2} - 2(3k - 1)x + 8{k^2} - 7 > 0$$<br><br>Now, D < 0<br><br>$$ \Rightarrow 4{(3k - 1)^2} - 4 \times 1 \times (8{k^2} - 7) < 0$$<br><br>$$ \Rightarrow 9{k^2} - 6k + 1 - 8{k^2} + 7 < 0$$<br><br>$$ \Rightarrow {k^2} - 6k + 8 < 0$$<br><br>$$ \Rightarrow (k - 4)(k - 2) < 0$$<br><br>2 < k < 4
<br><br>then k = 3 | mcq | jee-main-2021-online-25th-february-morning-slot |
1krzrgqrk | maths | quadratic-equation-and-inequalities | inequalities | If a + b + c = 1, ab + bc + ca = 2 and abc = 3, then the value of a<sup>4</sup> + b<sup>4</sup> + c<sup>4</sup> is equal to ______________. | [] | null | 13 | (a + b + c)<sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ a<sup>2</sup>
+ b<sup>2</sup>
+ c<sup>2</sup>
+ 2(ab + bc + ca) = 1
<br><br>$$ \Rightarrow $$ a<sup>2</sup> + b<sup>2</sup>
+ c<sup>2</sup> = – 3 ….(i)
<br><br>$$ \Rightarrow $$ ab + bc + ca = 2 ….(ii)
<br><br>Squaring of equation (ii),
<br><br>$$ \Rightarrow $$ a<sup>2</sup>b<sup>2</sup>
+ b<sup>2</sup>c<sup>2</sup>
+ c<sup>2</sup>a<sup>2</sup>
+ 2(ab<sup>2</sup>c + bc<sup>2</sup>a + ca<sup>2</sup>b) = 4
<br><br>$$ \Rightarrow $$ a<sup>2</sup>b<sup>2</sup>
+ b<sup>2</sup>c<sup>2</sup>
+ c<sup>2</sup>a<sup>2</sup>
+ 2abc(a + b + c) = 4
<br><br>$$ \Rightarrow $$ a<sup>2</sup>b<sup>2</sup>
+ b<sup>2</sup>c<sup>2</sup>
+ c<sup>2</sup>a<sup>2</sup>
+ 6 = 4
<br><br>$$ \Rightarrow $$ a<sup>2</sup>b<sup>2</sup>
+ b<sup>2</sup>c<sup>2</sup>
+ c<sup>2</sup>a<sup>2</sup> = – 2 ….(iii)
<br><br>Squaring of equation (i),
<br><br>$$ \Rightarrow $$ a<sup>4</sup>
+ b<sup>4</sup>
+ c<sup>4</sup>
+ 2(a<sup>2</sup>b<sup>2</sup>
+ b<sup>2</sup>c<sup>2</sup>
+ c<sup>2</sup>a<sup>2</sup>) = 9
<br><br>$$ \Rightarrow $$ a<sup>4</sup>
+ b<sup>4</sup>
+ c<sup>4</sup> – 4 = 9
<br><br>$$ \Rightarrow $$ a<sup>4</sup>
+ b<sup>4</sup>
+ c<sup>4</sup>
= 13 | integer | jee-main-2021-online-25th-july-evening-shift |
1l5vz0zg9 | maths | quadratic-equation-and-inequalities | inequalities | <p>Let $${S_1} = \left\{ {x \in R - \{ 1,2\} :{{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0} \right\}$$ and $${S_2} = \left\{ {x \in R:{3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0} \right\}$$. Then, $${S_1} \cup {S_2}$$ is equal to :</p> | [{"identifier": "A", "content": "$$( - \\infty , - 2] \\cup (1,2)$$"}, {"identifier": "B", "content": "$$( - \\infty , - 2] \\cup [1,2]$$"}, {"identifier": "C", "content": "$$( - 2,1] \\cup [2,\\infty )$$"}, {"identifier": "D", "content": "$$( - \\infty ,2]$$"}] | ["B"] | null | <p>Given,</p>
<p>$${{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0$$</p>
<p>$${x^2} + 3x + 5$$ is a quadratic equation</p>
<p>$$a = 1 > 0$$ and $$D = {( - 3)^2} - 4\,.\,1\,.\,5 = - 11 < 0$$</p>
<p>$$\therefore$$ $${x^2} + 3x + 5 > 0$$ (always)</p>
<p>So, we can ignore this quadratic term</p>
<p>$${{(x + 2)} \over { - 2 + 3x - {x^2}}} \ge 0$$</p>
<p>$$ \Rightarrow {{x + 2} \over { - ({x^2} - 3x + 2)}} \ge 0$$</p>
<p>$$ \Rightarrow {{x + 2} \over {{x^2} - 3x + 2}} \le 0$$</p>
<p>$$ \Rightarrow {{x + 2} \over {{x^2} - 2x - x + 2}} \le 0$$</p>
<p>$$ \Rightarrow {{x + 2} \over {(x - 1)(x - 2)}} \le 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5xuvzep/d1d9533f-57ee-41d4-a9fe-c531cfba2e5b/5b97b000-0a81-11ed-808f-7334e319a570/file-1l5xuvzeq.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5xuvzep/d1d9533f-57ee-41d4-a9fe-c531cfba2e5b/5b97b000-0a81-11ed-808f-7334e319a570/file-1l5xuvzeq.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Quadratic Equation and Inequalities Question 52 English Explanation 1"></p>
<p>$$\therefore$$ $$x \in ( - \alpha , - 2] \cup (1,2)$$</p>
<p>$$\therefore$$ $${S_1} = ( - \alpha , - 2] \cup (1,2)$$</p>
<p>Now,</p>
<p>$${3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0$$</p>
<p>$$ \Rightarrow {({3^x})^2} - 3\,.\,{3^x} - {3^2}\,.\,{3^x} + 27 \le 0$$</p>
<p>Let $${3^x} = t$$</p>
<p>$$ \Rightarrow {t^2} - 3\,.\,t - {3^2}\,.\,t + 27 \le 0$$</p>
<p>$$ \Rightarrow t(t - 3) - 9(t - 3) \le 0$$</p>
<p>$$ \Rightarrow (t - 3)(t - 9) \le 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5xuwndt/2f07a224-3889-4add-a96c-03b381708852/6e1d0d10-0a81-11ed-808f-7334e319a570/file-1l5xuwndu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5xuwndt/2f07a224-3889-4add-a96c-03b381708852/6e1d0d10-0a81-11ed-808f-7334e319a570/file-1l5xuwndu.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Quadratic Equation and Inequalities Question 52 English Explanation 2"></p>
<p>$$\therefore$$ $$3 \le t \le 9$$</p>
<p>$$ \Rightarrow {3^1} \le {3^x} \le {3^2}$$</p>
<p>$$ \Rightarrow 1 \le x \le 2$$</p>
<p>$$\therefore$$ $$x \in [1,2]$$</p>
<p>$$\therefore$$ $${S_2} = [1,2]$$</p>
<p>$$\therefore$$ $${S_1} \cup {S_2} = ( - \alpha ,2] \cup (1,2) \cup [1,2]$$</p>
<p>$$ = ( - \alpha ,2] \cup [1,2]$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
jaoe38c1lse4uha6 | maths | quadratic-equation-and-inequalities | inequalities | <p>Let $$\mathrm{S}$$ be the set of positive integral values of $$a$$ for which $$\frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32} < 0, \forall x \in \mathbb{R}$$. Then, the number of elements in $$\mathrm{S}$$ is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$\\infty$$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $x^2-8 x+32>0 \forall x \in R$ as discriminant of this quadratic is $64-4 \times 32<0$
<br/><br/>$$
\Rightarrow a x^2+2(a+1) x+9 a+4<0 \forall x \in R
$$
<br/><br/>$\Rightarrow$ Only possible when $a<0$ and $D<0$
<br/><br/>$\Rightarrow$ Since $S$ is set of positive
values of $a \Rightarrow S$ is a null set
<br/><br/>$$
\Rightarrow n(S)=0
$$ | mcq | jee-main-2024-online-31st-january-morning-shift |
pQ8kOFxyh1NO0kIK | maths | quadratic-equation-and-inequalities | location-of-roots | All the values of $$m$$ for which both roots of the equation $${x^2} - 2mx + {m^2} - 1 = 0$$ are greater than $$ - 2$$ but less then 4, lie in the interval | [{"identifier": "A", "content": "$$ - 2 < m < 0$$ "}, {"identifier": "B", "content": "$$m > 3$$ "}, {"identifier": "C", "content": "$$ - 1 < m < 3$$ "}, {"identifier": "D", "content": "$$1 < m < 4$$ "}] | ["C"] | null | Equation $${x^2} - 2mx + {m^2} - 1 = 0$$
<br><br>$${\left( {x - m} \right)^2} - 1 = 0$$
<br><br>or $$\left( {x - m + 1} \right)\left( {x - m - 1} \right) = 0$$
<br><br>$$x = m - 1,m + 1$$
<br><br>$$m - 1 > - 2$$ and $$m + 1 < 4$$
<br><br>$$ \Rightarrow m > - 1$$ and $$m<3$$
<br><br>or $$\,\,\, - 1 < m < 3$$
| mcq | aieee-2006 |
pX6k4PzHt3u5DKh2xhfDB | maths | quadratic-equation-and-inequalities | location-of-roots | If both the roots of the quadratic equation x<sup>2</sup> $$-$$ mx + 4 = 0 are real and distinct and they lie in the interval [1, 5], then m lies in the interval : | [{"identifier": "A", "content": "($$-$$5, $$-$$4)"}, {"identifier": "B", "content": "(4, 5)"}, {"identifier": "C", "content": "(5, 6)"}, {"identifier": "D", "content": "(3, 4)"}] | ["B"] | null | x<sup>2</sup> $$-$$mx + 4 = 0
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264994/exam_images/bkaah5pafcac92p7jsty.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Mathematics - Quadratic Equation and Inequalities Question 124 English Explanation">
<b>Case-I :</b>
<br><br>D > 0
<br><br>m<sup>2</sup> $$-$$ 16 > 0
<br><br>$$ \Rightarrow $$ m $$ \in $$ ($$-$$ $$\infty $$, $$-$$ 4) $$ \cup $$ (4, $$\infty $$)
<br><br><b>Case-II :</b>
<br><br>$$ \Rightarrow \,\,1 < {{ - b} \over {2a}} < 5$$
<br><br>$$ \Rightarrow \,\,1 < {m \over 2} < 5 \Rightarrow \,m \in \left( {2,10} \right)$$
<br><br><b>Case-III :</b>
<br><br>f(1) > 0 and f(5) > 0
<br><br>1 $$-$$ m + 4 > 0 and 25 $$-$$ 5m + 4 > 0
<br><br>m < 5 and m < $${{29} \over 5}$$
<br><br><b>Case-IV :</b>
<br><br>Let one root is x = 1
<br><br>1 $$-$$ m + 4 = 0
<br><br>m = 5
<br><br>Now equation x<sup>2</sup> $$-$$ 5x + 4 = 0
<br><br>(x $$-$$ 1) (x $$-$$ 4) = 0
<br><br>x = 1 i.e. m = 5 is also included
<br><br>hence m $$ \in $$ (4, 5]
<br><br>So given option is (4, 5)
| mcq | jee-main-2019-online-9th-january-evening-slot |
mSNXXAHDAMqkcYrI59bPZ | maths | quadratic-equation-and-inequalities | location-of-roots | Consider the quadratic equation (c – 5)x<sup>2</sup> – 2cx + (c – 4) = 0, c $$ \ne $$ 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is - | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "11"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267172/exam_images/rziw7accnofwe87kl8ml.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Mathematics - Quadratic Equation and Inequalities Question 122 English Explanation">
<br><br>Let f(x) = (c $$-$$ 5)x<sup>2</sup> $$-$$ 2cx + c $$-$$ 4
<br><br>$$ \therefore $$ f(0)f(2) < 0 . . . . .(1)
<br><br>& f(2)f(3) < 0 . . . . .(2)
<br><br>from (1) and (2)
<br><br>(c $$-$$ 4)(c $$-$$ 24) < 0
<br><br>& (c $$-$$ 24)(4c $$-$$ 49) < 0
<br><br>$$ \Rightarrow $$ $${{49} \over 4}$$ < c < 24
<br><br>$$ \therefore $$ s = {113, 14, 15, . . . . . 23}
<br><br>Number of elements in set S = 11 | mcq | jee-main-2019-online-10th-january-morning-slot |
NjnfwqqKfFebUKmJODjgy2xukf44aor4 | maths | quadratic-equation-and-inequalities | location-of-roots | The set of all real values of $$\lambda $$ for which the
quadratic equations, <br/>($$\lambda $$<sup>2</sup>
+ 1)x<sup>2</sup>
– 4$$\lambda $$x + 2 = 0
always have exactly one root in the interval
(0, 1) is : | [{"identifier": "A", "content": "(\u20133, \u20131)"}, {"identifier": "B", "content": "(2, 4]"}, {"identifier": "C", "content": "(0, 2)"}, {"identifier": "D", "content": "(1, 3]"}] | ["D"] | null | Given quadratic equation,<br><br>$$({\lambda ^2} + 1){x^2} - 4\lambda x + 2 = 0$$<br><br>Here coefficient of x<sup>2</sup> is ($$\lambda $$<sup>2</sup> + 1) which is always positive. So quadratic equation is upward parabola.<br><br><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264642/exam_images/ubgriloi3sciyx6c5v8t.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266993/exam_images/nimn4csiiz1sk1zxtbdc.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267548/exam_images/p5a65zhnpiheonjz4tbp.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265324/exam_images/tjhjf9qkm4vxoi6n2nwk.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265329/exam_images/qnfpvne2zajqnb56tkig.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267176/exam_images/rbxyileicz9fwclti9fe.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267110/exam_images/ewtlfvgz5wvf04zgzuwf.webp"><source media="(max-width: 1580px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267012/exam_images/mjxsciy8b1tuw0pgypnx.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266183/exam_images/zoc6r8adwxvl4esfzzpb.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Evening Slot Mathematics - Quadratic Equation and Inequalities Question 98 English Explanation"></picture><br><br>So, either f(0) < 0 and f(1) > 1<br><br>or f(0) > 0 and f(1) < 0<br><br>$$ \therefore $$ In both those cases,<br><br>f(0) f(1) $$ \le $$ 0<br><br>$$ \Rightarrow $$ 2($$\lambda $$<sup>2</sup> $$-$$ 4$$\lambda $$ + 3) $$ \le $$ 0<br><br>$$ \Rightarrow $$ $$\lambda $$$$ \in $$ [1, 3]<br><br>At $$\lambda $$ = 1 :<br><br>Quadratic equation becomes<br><br>2x<sup>2</sup> $$-$$ 4x + 2 = 0<br><br>$$ \Rightarrow $$ (x $$-$$ 1)<sup>2</sup> = 0<br><br>$$ \Rightarrow $$ x = 1, 1<br><br>As both roots can't lie between (0, 1)<br><br>So, $$\lambda $$ = 1 can't be possible.<br><br>At $$\lambda $$ = 3 :<br><br>10x<sup>2</sup> $$-$$ 12x + 2 = 0<br><br>$$ \Rightarrow $$ 5x<sup>2</sup> $$-$$ 6x + 1 = 0<br><br>$$ \Rightarrow $$ (5x $$-$$ 1) (x $$-$$ 1) = 0<br><br>$$ \Rightarrow $$ x = 1, $${1 \over 5}$$<br><br>In the interval (0, 1) exactly one root $${1 \over 5}$$ present.<br><br>$$ \therefore $$ $$\lambda $$ $$ \in $$ (1, 3] | mcq | jee-main-2020-online-3rd-september-evening-slot |
1ldo68yjn | maths | quadratic-equation-and-inequalities | location-of-roots | <p>The number of integral values of k, for which one root of the equation $$2x^2-8x+k=0$$ lies in the interval (1, 2) and its other root lies in the interval (2, 3), is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5hv330/3bf12c29-312e-4d43-a5a3-c05d3e6a864c/6c2ba9c0-ad16-11ed-8a8c-4d67f5492755/file-1le5hv331.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5hv330/3bf12c29-312e-4d43-a5a3-c05d3e6a864c/6c2ba9c0-ad16-11ed-8a8c-4d67f5492755/file-1le5hv331.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Evening Shift Mathematics - Quadratic Equation and Inequalities Question 39 English Explanation">
<br><br>$$
\begin{aligned}
& f(1)>0 \Rightarrow k>6 \\\\
& f(2)<0 \Rightarrow k<8 \\\\
& f(3)>0 \Rightarrow k>6 \\\\
& k \in(6,8)
\end{aligned}
$$
<br><br>Only 1 integral value of $k$ is 7. | mcq | jee-main-2023-online-1st-february-evening-shift |
J4uBIzApDyw2Lggl | maths | quadratic-equation-and-inequalities | modulus-function | Product of real roots of equation $${t^2}{x^2} + \left| x \right| + 9 = 0$$ | [{"identifier": "A", "content": "is always positive "}, {"identifier": "B", "content": "is always negative "}, {"identifier": "C", "content": "does not exist"}, {"identifier": "D", "content": "none of these "}] | ["A"] | null | Product of real roots $$ = {9 \over {{t^2}}} > 0,\forall \,t\, \in R$$
<br><br>$$\therefore$$ Product of real roots is always positive. | mcq | aieee-2002 |
YCu5If6M2eZmxcua | maths | quadratic-equation-and-inequalities | modulus-function | The number of real solutions of the equation $${x^2} - 3\left| x \right| + 2 = 0$$ is | [{"identifier": "A", "content": "3 "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "4 "}, {"identifier": "D", "content": "1 "}] | ["C"] | null | $${x^2} - 3\left| x \right| + 2 = 0$$
<br><br>$$ \Rightarrow {\left| x \right|^2} - 3\left| x \right| + 2 = 0$$
<br><br>$$\left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 1} \right) = 0$$
<br><br>$$\left| x \right| = 1,2$$ or $$x = \pm 1, \pm 2$$
<br><br>$$\therefore$$ No. of solution $$=4$$ | mcq | aieee-2003 |
8zfNJ8kRVBDzRMei | maths | quadratic-equation-and-inequalities | modulus-function | Let S = { $$x$$ $$ \in $$ R : $$x$$ $$ \ge $$ 0 and
<br/><br/>$$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$}. Then S | [{"identifier": "A", "content": "contains exactly four elements"}, {"identifier": "B", "content": "is an empty set"}, {"identifier": "C", "content": "contains exactly one element"}, {"identifier": "D", "content": "contains exactly two elements"}] | ["D"] | null | Given,
<br><br>$$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$
<br><br><b><u>Case 1</u> :</b>
<br><br>When $$\sqrt x - 3 \ge 0,$$ then equation becomes
<br><br>$$2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$
<br><br>$$ \Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0$$
<br><br>$$ \Rightarrow \,\,\,\,x\, - 4\sqrt x = 0$$
<br><br>$$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0$$
<br><br>$$\therefore\,\,\,$$ $$\sqrt x = 0,4$$
<br><br>but as $$\sqrt x - 3 \ge 0$$ or $$\sqrt x \ge 3$$ then $$\sqrt x \ne 0$$
<br><br>$$\therefore\,\,\,$$ $$\sqrt x = 4$$ value is possible.
<br><br><b><u>Case 2</u> :</b>
<br><br>When $$\sqrt x - 3 < 0$$ or $$\sqrt x < 3.$$ There equation becomes
<br><br>$$ - 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$
<br><br>$$ \Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0$$
<br><br>$$ \Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0$$
<br><br>$$ \Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0$$
<br><br>$$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0$$
<br><br>$$ \Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0$$
<br><br>$$\therefore\,\,\,$$ $$\sqrt x = 2,6$$
<br><br>as $$\sqrt x < 3$$ so $$\sqrt x \ne 6$$
<br><br>$$\therefore\,\,\,$$ $$\sqrt x = 2$$ is possible.
<br><br>So, total possible value of $$\sqrt x = 2,4$$
<br><br>or for x possible values are 4, 16.
<br><br>$$\therefore\,\,\,$$ Set S contains exactly two elements 4 and 16. | mcq | jee-main-2018-offline |
GucnZdzvR3MFEr855XuFJ | maths | quadratic-equation-and-inequalities | modulus-function | The sum of the solutions of the equation <br/>
$$\left| {\sqrt x - 2} \right| + \sqrt x \left( {\sqrt x - 4} \right) + 2 = 0$$<br/>
(x > 0) is equal to: | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "10"}] | ["D"] | null | <b>Case 1 :</b> When $$\sqrt x \ge 2$$
<br><br>then $$\left| {\sqrt x - 2} \right| = \sqrt x - 2$$
<br><br>$$ \therefore $$ The given equation becomes,
<br><br>$$\left( {\sqrt x - 2} \right)$$ + $$\sqrt x \left( {\sqrt x - 4} \right) + 2$$ = 0
<br><br>$$ \Rightarrow $$ $$\left( {\sqrt x - 2} \right)$$ + $$x - 4\sqrt x $$ + 2 = 0
<br><br>$$ \Rightarrow $$ $$x - 3\sqrt x $$ = 0
<br><br>$$ \Rightarrow $$ $$\sqrt x \left( {\sqrt x - 3} \right)$$ = 0
<br><br>$$ \therefore $$ $$\sqrt x $$ = 0 or 3
<br><br>$$\sqrt x $$ = 0 is not possible as $$\sqrt x \ge 2$$.
<br><br>So, $$\sqrt x $$ = 3
<br><br>or $$x$$ = 9
<br><br><b>Case 2 :</b> When $$\sqrt x < 2$$
<br><br>then $$\left| {\sqrt x - 2} \right| = $$$$ - \left( {\sqrt x - 2} \right)$$ = $$2 - \sqrt x $$
<br><br>$$ \therefore $$ The given equation becomes,
<br><br>$$\left( {2 - \sqrt x } \right)$$ + $$\sqrt x \left( {\sqrt x - 4} \right) + 2$$ = 0
<br><br>$$ \Rightarrow $$ $${2 - \sqrt x }$$ + $$x - 4\sqrt x $$ + 2 = 0
<br><br>$$ \Rightarrow $$ $$x - 5\sqrt x + 4$$ = 0
<br><br>$$ \Rightarrow $$ $$x - 4\sqrt x - \sqrt x + 4$$ = 0
<br><br>$$ \Rightarrow $$ $$\sqrt x \left( {\sqrt x - 4} \right)$$$$-\left( {\sqrt x - 4} \right)$$ = 0
<br><br>$$ \Rightarrow $$ $$\left( {\sqrt x - 4} \right)$$$$\left( {\sqrt x - 1} \right)$$ = 0
<br><br>$$ \therefore $$ $$\sqrt x $$ = 4 or 1
<br><br>$$\sqrt x $$ = 4 is not possible as $$\sqrt x < 2$$.
<br><br>$$ \therefore $$ $$\sqrt x $$ = 1
<br><br>or $$x$$ = 1
<br><br>So, Sum of all solutions = 9 + 1 = 10 | mcq | jee-main-2019-online-8th-april-morning-slot |
ZaQAnwZJq0uaay0nM43rsa0w2w9jx1ywhvd | maths | quadratic-equation-and-inequalities | modulus-function | The number of real roots of the equation
<br/> 5 + |2<sup>x</sup>
– 1| = 2<sup>x</sup>
(2<sup>x</sup>
– 2) is | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["B"] | null | When 2<sup>x</sup> $$ \ge $$ 1 <br><br>
5 + 2<sup>x</sup> –1 = 2<sup>x</sup> (2<sup>x</sup> – 2) <br><br>
Let 2x = t <br><br>
$$ \Rightarrow $$5 + t – 1 = t (t – 2) <br><br>
$$ \Rightarrow $$ t = 4, – 1(rejected) <br><br>
$$ \Rightarrow $$ 2x = 4 <br><br>
$$ \Rightarrow $$ x = 2 <br><br>
Now when 2<sup>x</sup> < 1 <br><br>
5 + 1 – 2<sup>x</sup> = 2<sup>x</sup> (2<sup>x</sup> – 2)<br><br>
Let 2<sup>x</sup> = t <br><br>
$$ \Rightarrow $$ 5 + 1 – t = t (t – 2) <br><br>
$$ \Rightarrow $$ 0 = t<sup>2</sup> – t – 6 <br><br>
$$ \Rightarrow $$ 0 = (t – 3) (t – 2) <br><br>
$$ \Rightarrow $$ t = 3, – 2 <br><br>
2<sup>x</sup> = 3, 2<sup>x</sup> = – 2 (rejected)<br><br>
$$ \therefore $$ Only one real roots.
| mcq | jee-main-2019-online-10th-april-evening-slot |
tLULgaIgrnpDTGKEEF7k9k2k5hk1kck | maths | quadratic-equation-and-inequalities | modulus-function | Let S be the set of all real roots of the equation,<br/>
3<sup>x</sup>(3<sup>x</sup> – 1) + 2 = |3<sup>x</sup> – 1| + |3<sup>x</sup> – 2|. Then S : | [{"identifier": "A", "content": "contains exactly two elements."}, {"identifier": "B", "content": "is an empty set."}, {"identifier": "C", "content": "is a singleton."}, {"identifier": "D", "content": "contains at least four elements."}] | ["C"] | null | Let 3<sup>x</sup> = t ; t $$>$$ 0
<br><br>t(t – 1) + 2 = |t – 1| + |t – 2|
<br><br>t<sup>2</sup> – t + 2 = |t – 1| + |t – 2|
<br><br><b>Case-I</b> : t $$<$$ 1
<br><br>t<sup>2</sup> – t + 2 = 1 – t + 2 – t
<br><br>$$ \Rightarrow $$ t<sup>2</sup> + 2 = 3 – t
<br><br>$$ \Rightarrow $$ t<sup>2</sup> + t – 1 = 0
<br><br>$$ \Rightarrow $$ t = $${{ - 1 \pm \sqrt 5 } \over 2}$$
<br><br>$$ \Rightarrow $$ t = $${{\sqrt 5 - 1} \over 2}$$ [ As t $$>$$ 0]
<br><br><b>Case-II</b> : 1 $$ \le $$ t $$<$$ 2
<br><br>$$ \Rightarrow $$ t<sup>2</sup> – t + 2 = t – 1 + 2 – t
<br><br>$$ \Rightarrow $$ t<sup>2</sup> – t + 1 = 0
<br><br>D $$<$$ 0 so no real solution.
<br><br><b>Case-III</b> : t $$ \ge $$ 2
<br><br>$$ \Rightarrow $$ t<sup>2</sup> – t + 2 = t – 1 + t – 2
<br><br>$$ \Rightarrow $$ t<sup>2</sup> – 3t - 5 = 0
<br><br>$$ \Rightarrow $$ D $$<$$ 0 so no real solution. | mcq | jee-main-2020-online-8th-january-evening-slot |
RcJaa2kkStUK6mm6FBjgy2xukfg6wcxg | maths | quadratic-equation-and-inequalities | modulus-function | The product of the roots of the <br/>equation 9x<sup>2</sup> - 18|x| + 5 = 0 is : | [{"identifier": "A", "content": "$${{5} \\over {9}}$$"}, {"identifier": "B", "content": "$${{5} \\over {27}}$$"}, {"identifier": "C", "content": "$${{25} \\over {81}}$$"}, {"identifier": "D", "content": "$${{25} \\over {9}}$$"}] | ["C"] | null | $$9{x^2} - 18\left| x \right| + 5 = 0$$<br><br>$$ \Rightarrow $$ $$9{x^2} - 15\left| x \right| - 3\left| x \right| + 5 = 0$$ ($$ \because $$ x<sup>2</sup> = $${\left| x \right|^2}$$)<br><br>$$ \Rightarrow $$ $$3\left| x \right|(3\left| x \right| - 5) - (3\left| x \right| - 5) = 0$$<br><br>$$ \Rightarrow $$$$\left| x \right| = {1 \over 3},\,{5 \over 3}$$<br><br>$$ \Rightarrow $$ $$x = \pm {1 \over 3}, \pm \,{5 \over 3}$$<br><br>$$ \therefore $$ Product of roots <br><br>= $$\left( \frac{1}{3} \right) \left( -\frac{1}{3} \right) \left( \frac{5}{3} \right) \left( -\frac{5}{3} \right) $$
= $${{25} \over {81}}$$ | mcq | jee-main-2020-online-5th-september-morning-slot |
YmhKJLqLfmHUC0KJZD1klrmwr0o | maths | quadratic-equation-and-inequalities | modulus-function | The number of the real roots of the equation $${(x + 1)^2} + |x - 5| = {{27} \over 4}$$ is ________. | [] | null | 2 | When $$x > 5$$<br><br>$${(x + 1)^2} + (x - 5) = {{27} \over 4}$$<br><br>$$ \Rightarrow {x^2} + 3x - 4 = {{27} \over 4}$$<br><br>$$ \Rightarrow {x^2} + 3x - {{43} \over 4} = 0$$<br><br>$$ \Rightarrow 4{x^2} + 12x - 43 = 0$$<br><br>$$x = {{ - 12 \pm \sqrt {144 + 688} } \over 8}$$<br><br>$$x = {{ - 12 \pm \sqrt {832} } \over 8} = {{ - 12 \pm 28.8} \over 8}$$<br><br>$$ = {{ - 3 + 7.2} \over 2}$$<br><br>$$ = {{ - 3 + 7.2} \over 2},{{ - 3 - 7.2} \over 2}$$
<br><br>= 2.1, -5.1 [ both are rejected as x should be > 5 ]
<br><br>(Therefore no solution)<br><br>For $$x \le 5$$<br><br>$${(x + 1)^2} - (x - 5) = {{27} \over 4}$$<br><br>$${x^2} + x + 6 - {{27} \over 4} = 0$$<br><br>$$4{x^2} + 4x - 3 = 0$$<br><br>$$x = {{ - 4 \pm \sqrt {16 + 48} } \over 8}$$<br><br>$$x = {{ - 4 \pm 8} \over 8} \Rightarrow x = - {{12} \over 8},{4 \over 8}$$<br><br>$$ \therefore $$ So, the equation have two real roots. | integer | jee-main-2021-online-24th-february-evening-slot |
1krzn03g8 | maths | quadratic-equation-and-inequalities | modulus-function | The number of real solutions of the equation, x<sup>2</sup> $$-$$ |x| $$-$$ 12 = 0 is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}] | ["A"] | null | |x|<sup>2</sup> $$-$$ |x| $$-$$ 12 = 0<br><br>$$ \Rightarrow $$ (|x| + 3)(|x| $$-$$ 4) = 0<br><br>$$ \Rightarrow $$ |x| = 4
<br><br>$$\Rightarrow$$ x = $$\pm$$4 | mcq | jee-main-2021-online-25th-july-evening-shift |
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