question_id
stringlengths 8
35
| subject
stringclasses 1
value | chapter
stringclasses 32
values | topic
stringclasses 178
values | question
stringlengths 26
9.64k
| options
stringlengths 2
1.63k
| correct_option
stringclasses 5
values | answer
stringclasses 293
values | explanation
stringlengths 13
9.38k
| question_type
stringclasses 3
values | paper_id
stringclasses 149
values |
|---|---|---|---|---|---|---|---|---|---|---|
ssSsGJs4IgV0TRS9sbJkO | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | If the system of linear equations<br/><br/>
x – 2y + kz = 1<br/>
2x + y + z = 2<br/>
3x – y – kz = 3<br/><br/>
has a solution (x,y,z), z $$ \ne $$ 0, then (x,y) lies on
the straight line whose equation is : | [{"identifier": "A", "content": "4x \u2013 3y \u2013 4 = 0"}, {"identifier": "B", "content": "3x \u2013 4y \u2013 1 = 0"}, {"identifier": "C", "content": "4x \u2013 3y \u2013 1 = 0"}, {"identifier": "D", "content": "3x \u2013 4y \u2013 4 = 0"}] | ["A"] | null | x – 2y + kz = 1 ......(1)<br><br>
2x + y + z = 2 .........(2)<br><br>
3x – y – kz = 3 ........(3)
<br><br>for locus of (x, y)
add equation (1) + (3)
<br><br>4x – 3y = 4
<br><br>$$ \Rightarrow $$ 4x – 3y - 4 = 0 | mcq | jee-main-2019-online-8th-april-evening-slot |
OoONgLpzJPuR08YqU4ZzC | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | If a straight line passing through the point P(–3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is : | [{"identifier": "A", "content": "x \u2013 y + 7 = 0"}, {"identifier": "B", "content": "4x \u2013 3y + 24 = 0"}, {"identifier": "C", "content": "4x + 3y = 0"}, {"identifier": "D", "content": "3x \u2013 4y + 25 = 0"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263333/exam_images/sgkffsp4mfuv6idliumw.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Evening Slot Mathematics - Straight Lines and Pair of Straight Lines Question 96 English Explanation">
<br>Let the line be $${x \over a} + {y \over b} = 1$$
<br><br>($$-$$ 3, 4) = $$\left( {{a \over 2},{b \over 2}} \right)$$
<br><br>a = $$-$$6, b = 8
<br><br>equation of line is 4x $$-$$ 3y + 24 = 0 | mcq | jee-main-2019-online-12th-january-evening-slot |
DpbIaKmUJaLDxujcxegRK | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the
equation of the diagonal AD is : | [{"identifier": "A", "content": "5x + 3y \u2013 11 = 0"}, {"identifier": "B", "content": "5x \u2013 3y + 1 = 0"}, {"identifier": "C", "content": "3x \u2013 5y + 7 = 0"}, {"identifier": "D", "content": "3x + 5y \u2013 13 = 0"}] | ["B"] | null | co-ordinates of point D are (4, 7)
<br><br>$$ \Rightarrow $$ line AD is 5x $$-$$ 3y + 1 = 0 | mcq | jee-main-2019-online-11th-january-evening-slot |
LMDd8M7tN0EYa27fmeMWM | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | Two sides of a parallelogram are along the lines, x + y = 3 & x – y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is : | [{"identifier": "A", "content": "(2, 1)"}, {"identifier": "B", "content": "(2, 6)"}, {"identifier": "C", "content": "(3, 5)"}, {"identifier": "D", "content": "(3, 6)"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263689/exam_images/shi6jmtimy8lhgilqtoi.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Straight Lines and Pair of Straight Lines Question 100 English Explanation">
<br>Solving
<br><br>$$\matrix{
{x + y = 3} \cr
{x - y = - 3} \cr
} \,\, > \,\,A\left( {0,3} \right)$$
<br><br>and $${{{x_1} + 0} \over 2} = 2;\,\,{x_i} = 4$$
<br><br>similarly y<sub>1</sub> = 5
<br><br>C $$ \Rightarrow $$ (4, 5)
<br><br>Now equation of BC is x $$-$$ y = $$-$$ 1
<br><br>and equation of CD is x + y = 9
<br><br>Solving x + y = 9 and x $$-$$ y = $$-$$ 3
<br><br>Point D is (3, 6) | mcq | jee-main-2019-online-10th-january-evening-slot |
kLBkRZEqREDdFmqosAann | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | A point on the straight line, 3x + 5y = 15 which is equidistant from the coordinate axes will lie only
in : | [{"identifier": "A", "content": "1<sup>st </sup> and 2<sup>nd</sup> qudratants"}, {"identifier": "B", "content": "4<sup>th</sup> qudratant"}, {"identifier": "C", "content": "1<sup>st </sup> and 2<sup>nd</sup> and 4<sup>th</sup> qudratants"}, {"identifier": "D", "content": "1<sup>st </sup> qudratant"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264028/exam_images/c5iqjp4havmvitlpvih5.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264572/exam_images/kiwz4enqututlxqicwwd.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266516/exam_images/xlp8x7fisukawcfaja1f.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263884/exam_images/aumkosbopsf2ogx1v0ai.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 95 English Explanation"></picture>
<br>Let the point is P(x, y).
<br><br>According to the question, the point P(x, y) is equidistance from both x and y axis.
<br><br>$$ \therefore $$ |x| = |y|
<br><br>$$ \Rightarrow $$ x = $$ \pm $$ y
<br><br>So the point P lies on the either x = y or x = - y line. And point P(x, y) also lies on the straight line 3x + 5y = 15.
<br><br>Form the graph, you can see the point P can either be on 1<sup>st</sup> qudratant or 2<sup>nd</sup> qudratant. | mcq | jee-main-2019-online-8th-april-morning-slot |
QxqEuxDfNrfjec79WCjgy2xukfahc33t | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | If the perpendicular bisector of the line segment joining the points P(1 ,4) and Q(k, 3) has y-intercept equal to –4, then a value of k is : | [{"identifier": "A", "content": "$$\\sqrt {14} $$"}, {"identifier": "B", "content": "-4"}, {"identifier": "C", "content": "\u20132 "}, {"identifier": "D", "content": "$$\\sqrt {15} $$"}] | ["B"] | null | $${m_{PQ}} = {{4 - 3} \over {1 - k}} $$
<br><br>$$ \therefore $$ Slope of perpendicular bisector of PQ, $$ {m_ \bot } = k - 1$$<br><br>mid point of PQ = $$\left( {{{k + 1} \over 2},{7 \over 2}} \right)$$<br><br>equation of perpendicular bisector<br><br>$$y - {7 \over 2} = (k - 1)\left( {x - {{k + 1} \over 2}} \right)$$<br><br>for y intercept put x = 0<br><br>$$y = {7 \over 2} - \left( {{{{k^2} - 1} \over 2}} \right) = - 4$$<br><br>$${{{k^2} - 1} \over 2} = {{15} \over 2} \Rightarrow k = \pm 4$$ | mcq | jee-main-2020-online-4th-september-evening-slot |
jWtl9wTYgDrlkS0x00jgy2xukfuvjkza | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | A ray of light coming from the point (2, $$2\sqrt 3 $$) is incident at an angle 30<sup>o</sup> on the line x = 1 at the
point A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line AB
passes through the point : | [{"identifier": "A", "content": "(3, -$$\\sqrt 3 $$)"}, {"identifier": "B", "content": "(4, -$$\\sqrt 3 $$)"}, {"identifier": "C", "content": "$$\\left( {4, - {{\\sqrt 3 } \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {3, - {1 \\over {\\sqrt 3 }}} \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264527/exam_images/ysrpqzdcmjighfj5a4xy.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 77 English Explanation">
<br><br>Equation of reflected Ray P'B :
<br><br>y - 2$$\sqrt 3 $$ = tan 120<sup>o</sup> (x - 0)
<br><br>$$ \Rightarrow $$ $$\sqrt 3 $$x + y = 2$$\sqrt 3 $$
<br><br>(3, -$$\sqrt 3 $$) satisfy the line. | mcq | jee-main-2020-online-6th-september-morning-slot |
XUXGa7rHrUGBomtNuF1klrg4w6b | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | A man is walking on a straight line. The arithmetic mean
of the reciprocals of the intercepts of this line on the
coordinate axes is $${1 \over 4}$$. Three stones A, B and C are placed at the points
(1, 1), (2, 2) and (4, 4) respectively. Then, which of these stones is / are on the path of the man? | [{"identifier": "A", "content": "A only"}, {"identifier": "B", "content": "All the three"}, {"identifier": "C", "content": "C only"}, {"identifier": "D", "content": "B only"}] | ["D"] | null | Given, position of A = (1, 1)<br><br>Position of B = (2, 2)<br><br>Position of C = (4, 4)<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxoe9q9z/76596a6c-2785-45d0-8cbb-72347daf83fe/95cbec80-66eb-11ec-b4c9-97a7fc3f3aad/file-1kxoe9qa0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxoe9q9z/76596a6c-2785-45d0-8cbb-72347daf83fe/95cbec80-66eb-11ec-b4c9-97a7fc3f3aad/file-1kxoe9qa0.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2021 (Online) 24th February Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 75 English Explanation"><br>Let x-intercept be a and y-intercept be b.<br><br>Equation of line traced is<br><br>$${x \over a} + {y \over b} = 1$$<br><br>This is the equation of path, let a point (h, k) lie on this path.<br><br>Then, $${h \over a} + {k \over b} = 1$$<br><br>Also, AM of reciprocal of a and b = $${1 \over 4}$$<br><br>$$\therefore$$ $${{{1 \over a} + {1 \over b}} \over 2} = {1 \over 4}$$<br><br>$${1 \over a} + {1 \over b} = {1 \over 2}$$<br><br>On comparing Eqs. (i) and (ii), we get (h, k) = (2, 2)<br><br>Hence, the required stone is B(2, 2). | mcq | jee-main-2021-online-24th-february-morning-slot |
GnwOVu7cczs292GgmI1kluglgca | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | The intersection of three lines x $$-$$ y = 0, x + 2y = 3 and 2x + y = 6 is a : | [{"identifier": "A", "content": "Right angled triangle"}, {"identifier": "B", "content": "Equilateral triangle"}, {"identifier": "C", "content": "None of the above"}, {"identifier": "D", "content": "Isosceles triangle"}] | ["D"] | null | The given three lines are x $$-$$ y = 0, x + 2y = 3 and 2x + y = 6 then point of intersection,<br><br>lines x $$-$$ y = 0 and x + 2y = 3 is (1, 1)<br><br>lines x $$-$$ y = 0 and 2x + y = 6 is (2, 2)<br><br>and lines x + 2y = 3 and 2x + y = 0 is (3, 0)<br><br>The triangle ABC has vertices A(1, 1), B(2, 2) and C(3, 0)<br><br>$$ \therefore $$ AB = $$\sqrt 2 $$, BC = $$\sqrt 5 $$ and AC = $$\sqrt 5 $$<br><br>$$ \therefore $$ $$\Delta$$ABC is isosceles | mcq | jee-main-2021-online-26th-february-morning-slot |
9CTLjZBJ0UzByUARrD1kmjbnx0l | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | The maximum value of z in the following equation z = 6xy + y<sup>2</sup>, where 3x + 4y $$ \le $$ 100 and 4x + 3y $$ \le $$ 75 for x $$ \ge $$ 0 and y $$ \ge $$ 0 is __________. | [] | null | 904 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264667/exam_images/acnl84vwiakold3way0b.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 70 English Explanation">
<br>3x + 4y $$ \le $$ 100<br><br>4x + 3y $$ \le $$ 75<br><br>x $$ \ge $$ 0, y $$ \ge $$ 0<br><br>Feasible region is shown in the graph<br><br>Let maximum value of 6xy + y<sup>2</sup> = c<br><br>For a solution with feasible region,<br><br>6xy + y<sup>2</sup> = c and 4x + 3y = 75 must have at least one positive solution.<br><br>$${y^2} + 6y\left( {{{75 - 3y} \over 4}} \right) - c = 0 $$
<br><br>$$\Rightarrow {7 \over 2}{y^2} - {{225} \over 2}y + c = 0$$<br><br>$$ \Rightarrow {\left( {{{225} \over 2}} \right)^2} \ge 4.{7 \over 2}.c $$
<br><br>$$\Rightarrow c \le {{{{225}^2}} \over {56}} \approx 904$$ | integer | jee-main-2021-online-17th-march-morning-shift |
dmYotEs1QcgiGgtCOh1kmlhxh16 | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | The number of integral values of m so that the abscissa of point of intersection of lines 3x + 4y = 9 and y = mx + 1 is also an integer, is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "0"}] | ["B"] | null | 3x + 4(mx + 1) = 9<br><br>$$ \Rightarrow $$ x(3 + 4m) = 5<br><br>$$ \Rightarrow $$ $$x = {5 \over {(3 + 4m)}}$$<br><br>$$ \Rightarrow $$ (3 + 4m) = $$\pm$$1, $$\pm$$5<br><br>$$ \Rightarrow $$ 4m = $$-$$3 $$\pm$$ 1, $$-$$3 $$\pm$$ 5 <br><br>$$ \Rightarrow $$ 4m = $$-$$4, $$-$$2, $$-$$8, 2<br><br>$$ \Rightarrow $$ m = $$-$$1, $$-$$$${1 \over 2}$$, $$-$$2, $${1 \over 2}$$<br><br>$$ \therefore $$ Two integral value of m. | mcq | jee-main-2021-online-18th-march-morning-shift |
1krxgdh61 | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | The point P (a, b) undergoes the following three transformations successively :<br/><br/>(a) reflection about the line y = x.<br/><br/>(b) translation through 2 units along the positive direction of x-axis.<br/><br/>(c) rotation through angle $${\pi \over 4}$$ about the origin in the anti-clockwise direction.<br/><br/>If the co-ordinates of the final position of the point P are $$\left( { - {1 \over {\sqrt 2 }},{7 \over {\sqrt 2 }}} \right)$$, then the value of 2a + b is equal to : | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "7"}] | ["B"] | null | Image of A(a, b) along y = x is B(b, a). Translating it 2 units it becomes C(b + 2, a).<br><br>Now, applying rotation theorem<br><br>$$ - {1 \over {\sqrt 2 }} + {7 \over {\sqrt 2 }}i = \left( {(b + 2) + ai} \right)\left( {\cos {\pi \over 4} + i\sin {\pi \over 4}} \right)$$<br><br>$$ - {1 \over {\sqrt 2 }} + {7 \over {\sqrt 2 }}i = \left( {{{b + 2} \over {\sqrt 2 }} - {a \over {\sqrt 2 }}} \right) + i\left( {{{b + 2} \over {\sqrt 2 }} + {a \over {\sqrt 2 }}} \right)$$<br><br>$$\Rightarrow$$ b $$-$$ a + 2 = $$-$$1 ......(i)<br><br>and b + 2 + a = 7 ...... (ii)<br><br>$$\Rightarrow$$ a = 4; b = 1<br><br>$$\Rightarrow$$ 2a + b = 9 | mcq | jee-main-2021-online-27th-july-evening-shift |
1krxkmgz3 | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | Two sides of a parallelogram are along the lines 4x + 5y = 0 and 7x + 2y = 0. If the equation of one of the diagonals of the parallelogram is 11x + 7y = 9, then other diagonal passes through the point : | [{"identifier": "A", "content": "(1, 2)"}, {"identifier": "B", "content": "(2, 2)"}, {"identifier": "C", "content": "(2, 1)"}, {"identifier": "D", "content": "(1, 3)"}] | ["B"] | null | Both the lines pass through origin.<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266102/exam_images/h1fc0mfvmn1izevgxbxm.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 63 English Explanation"><br><br>point D is equal to intersection of 4x + 5y = 0 & 11x + 7y = 9<br><br>So, coordinates of point $$D = \left( {{5 \over 3}, - {4 \over 3}} \right)$$<br><br>Also, point B is point of intersection of 7x + 2y = 0 and 11x + 7y = 9<br><br>So, coordinates of point $$B = \left( { - {2 \over 3},{7 \over 3}} \right)$$<br><br>diagonals of parallelogram intersect at middle let middle point of B, D<br><br>$$ \Rightarrow \left( {{{{5 \over 3} - {2 \over 3}} \over 2},{{{{ - 4} \over 3} + {7 \over 3}} \over 2}} \right) = \left( {{1 \over 2},{1 \over 2}} \right)$$<br><br>equation of diagonal AC<br><br>$$ \Rightarrow (y - 0) = {{{1 \over \alpha } - 0} \over {{1 \over \alpha } - 0}}(x - 0)$$<br><br>$$y = x$$<br><br>diagonal AC passes through (2, 2) | mcq | jee-main-2021-online-27th-july-evening-shift |
1l6dw9dqv | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | <p>A line, with the slope greater than one, passes through the point $$A(4,3)$$ and intersects the line $$x-y-2=0$$ at the point B. If the length of the line segment $$A B$$ is $$\frac{\sqrt{29}}{3}$$, then $$B$$ also lies on the line :
</p> | [{"identifier": "A", "content": "$$2 x+y=9$$"}, {"identifier": "B", "content": "$$3 x-2 y=7$$"}, {"identifier": "C", "content": "$$ x+2 y=6$$"}, {"identifier": "D", "content": "$$2 x-3 y=3$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97rw5km/2f99cd4f-8200-402b-aaa3-f8d0d73a633d/21992d60-4b5a-11ed-bfde-e1cb3fafe700/file-1l97rw5kn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97rw5km/2f99cd4f-8200-402b-aaa3-f8d0d73a633d/21992d60-4b5a-11ed-bfde-e1cb3fafe700/file-1l97rw5kn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 45 English Explanation"><br><br>
Let inclination of required line is $\theta$,
<br><br>
So the coordinates of point $B$ can be assumed as<br><br>
$\left(4-\frac{\sqrt{29}}{3} \cos \theta, 3-\frac{\sqrt{29}}{3} \sin \theta\right)$
<br><br>
Which satisfices $x-y-2=0$
<br><br>
$4-\frac{\sqrt{29}}{3} \cos \theta-3+\frac{\sqrt{29}}{3} \sin \theta-2=0$
<br><br>
$\sin \theta-\cos \theta=\frac{3}{\sqrt{29}}$
<br><br>
By squaring
<br><br>
$\sin 2 \theta=\frac{20}{29}=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$
<br><br>
$\tan \theta=\frac{5}{2}$ only (because slope is greater than 1 )
<br><br>
$$
\sin \theta=\frac{5}{\sqrt{29}}, \cos \theta=\frac{2}{\sqrt{29}}
$$
<br><br>
Point $B:\left(\frac{10}{3}, \frac{4}{3}\right)$
<br><br>
Which also satisfies $x+2 y=6$ | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6rehhn1 | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | <p>Let $$m_{1}, m_{2}$$ be the slopes of two adjacent sides of a square of side a such that $$a^{2}+11 a+3\left(m_{1}^{2}+m_{2}^{2}\right)=220$$. If one vertex of the square is $$(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$$, where $$\alpha \in\left(0, \frac{\pi}{2}\right)$$ and the equation of one diagonal is $$(\cos \alpha-\sin \alpha) x+(\sin \alpha+\cos \alpha) y=10$$, then $$72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$$ is equal to :</p> | [{"identifier": "A", "content": "119"}, {"identifier": "B", "content": "128"}, {"identifier": "C", "content": "145"}, {"identifier": "D", "content": "155"}] | ["B"] | null | One vertex of square is
<br/><br/>
$(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$
<br/><br/>
and one of the diagonal is
<br/><br/>
$(\cos \alpha-\sin \alpha) x+(\sin \alpha+\cos \alpha) y=10$
<br/><br/>
So the other diagonal can be obtained as
<br/><br/>
$(\cos \alpha+\sin \alpha) x-(\cos \alpha-\sin \alpha) y=0$
<br/><br/>
So, the point of intersection of the diagonal will be
<br/><br/>
(5( $\cos \alpha-\sin \alpha), 5(\cos \alpha+\sin \alpha))$.
<br/><br/>
Therefore, the vertex opposite to the given vertex is $(0,0)$.
<br/><br/>
So, the diagonal length $=10 \sqrt{2}$
<br/><br/>
Side length $(a)=10$
<br/><br/>
It is given that
<br/><br/>
$a^{2}+11 a+3\left(m_{1}^{2}+m_{2}^{2}\right)=220$
<br/><br/>
$m_{1}^{2}+m_{2}^{2}=\frac{220-100-110}{3}=\frac{10}{3}$
<br/><br/>
and $m_{1} m_{2}=-1$
<br/><br/>
Slopes of the sides are tan $\alpha$ and $-\cot \alpha$
<br/><br/>
$\tan ^{2} \alpha=3$ or $\frac{1}{3}$
<br/><br/>
$72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$
<br/><br/>
$=72 \cdot \frac{\tan ^{4} \alpha+1}{\left(1+\tan ^{2} \alpha\right)^{2}}+a^{2}-3 a+13=128$ | mcq | jee-main-2022-online-29th-july-evening-shift |
1ldsuqxe4 | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | <p>A light ray emits from the origin making an angle 30$$^\circ$$ with the positive $$x$$-axis. After getting reflected by the line $$x+y=1$$, if this ray intersects $$x$$-axis at Q, then the abscissa of Q is :</p> | [{"identifier": "A", "content": "$${2 \\over {\\left( {\\sqrt 3 - 1} \\right)}}$$"}, {"identifier": "B", "content": "$${2 \\over {3 - \\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over {2\\left( {\\sqrt 3 + 1} \\right)}}$$"}, {"identifier": "D", "content": "$${2 \\over {3 + \\sqrt 3 }}$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ldtzo4m3/b85082b3-e539-44fe-bf5f-e63a3546f4f7/a7383ba0-a6c2-11ed-8d92-0101c7cb78f8/file-1ldtzo4m4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ldtzo4m3/b85082b3-e539-44fe-bf5f-e63a3546f4f7/a7383ba0-a6c2-11ed-8d92-0101c7cb78f8/file-1ldtzo4m4.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 34 English Explanation"></p>Let $Q(h, O)$<br><br>
$\because $ OP reflected by $x+y=1$.<br><br>
So, image of $Q$ lies on $y=\frac{x}{\sqrt{3}}$<br><br>
$$
\begin{aligned}
& \therefore \quad \frac{x-h}{1}=\frac{y}{1}=\frac{-2(h-1)}{2} \\\\
& \therefore \quad x=1, y=1-h
\end{aligned}
$$<br><br>
It lies on $y=\frac{x}{\sqrt{3}}$<br><br>
$$
\begin{aligned}
& \therefore \quad 1-h=\frac{1}{\sqrt{3}} \\\\
& \therefore \quad h=1-\frac{1}{\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}}=\frac{2}{3+\sqrt{3}}
\end{aligned}
$$ | mcq | jee-main-2023-online-29th-january-morning-shift |
1lh20btfj | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | <p>The straight lines $$\mathrm{l_{1}}$$ and $$\mathrm{l_{2}}$$ pass through the origin and trisect the line segment of the line L : $$9 x+5 y=45$$ between the axes. If $$\mathrm{m}_{1}$$ and $$\mathrm{m}_{2}$$ are the slopes of the lines $$\mathrm{l_{1}}$$ and $$\mathrm{l_{2}}$$, then the point of intersection of the line $$\mathrm{y=\left(m_{1}+m_{2}\right)}x$$ with L lies on :</p> | [{"identifier": "A", "content": "$$6 x-y=15$$"}, {"identifier": "B", "content": "$$6 x+y=10$$"}, {"identifier": "C", "content": "$$\\mathrm{y}-x=5$$"}, {"identifier": "D", "content": "$$y-2 x=5$$"}] | ["C"] | null | Given line $L: 9 x+5 y=45$ ..........(i)
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnmwb3jr/502d615e-8cf5-4478-9d73-0b9837e70b02/2ebbe560-68d6-11ee-a3a1-07c5c60fca90/file-6y3zli1lnmwb3js.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnmwb3jr/502d615e-8cf5-4478-9d73-0b9837e70b02/2ebbe560-68d6-11ee-a3a1-07c5c60fca90/file-6y3zli1lnmwb3js.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) 6th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 25 English Explanation">
<br>Also given that $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$, respectively.
<br><br>Let $A$ be the point of intersection of line $l_1$ and $L$, then <br><br>co-ordinates of $A$ are $\left(\frac{2 \times 5+1 \times 0}{2+1}, \frac{2 \times 0+1 \times 9}{2+1}\right)=\left(\frac{10}{3}, 3\right)$
<br><br>And let $B$ be the point of intersection of line $l_2$ and $L$, then <br><br>co-ordinates of $B$ are $$
\left(\frac{1 \times 5+2 \times 0}{1+2}, \frac{1 \times 0+2 \times 9}{1+2}\right)=\left(\frac{5}{3}, 6\right)
$$
<br><br>Now, slope of line $l_1,\left(m_1\right)=\frac{3-0}{\frac{10}{3}-0}=\frac{9}{10}$
<br><br>and slope of lines $l_2,\left(m_2\right)=\frac{6-0}{\frac{5}{3}-0}=\frac{18}{5}$
<br><br>$$
\begin{aligned}
\therefore \text { line } y & =\left(m_1+m_2\right) x \\\\
& =\left(\frac{9}{10}+\frac{18}{5}\right) x=\frac{45}{10} x=\frac{9}{2} x .......(ii)
\end{aligned}
$$
<br><br>Point of intersection of lines (i) and (ii)
<br><br>$$
\begin{aligned}
& 9 x+5\left(\frac{9}{2} x\right)=45 \\\\
& \Rightarrow x+\frac{5 x}{2}=5 \\\\
& \Rightarrow \frac{7 x}{2}=5 \Rightarrow x=\frac{10}{7} \\\\
& \text { and } y=\frac{9}{2} \times \frac{10}{7}=\frac{45}{7} \\\\
& \therefore \text { Point of intersection }=\left(\frac{10}{7}, \frac{45}{7}\right) \\\\
& \text { lies on line } y-x=5
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
jaoe38c1lsd505jf | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | <p>Let $$A(-2,-1), B(1,0), C(\alpha, \beta)$$ and $$D(\gamma, \delta)$$ be the vertices of a parallelogram $$A B C D$$. If the point $$C$$ lies on $$2 x-y=5$$ and the point $$D$$ lies on $$3 x-2 y=6$$, then the value of $$|\alpha+\beta+\gamma+\delta|$$ is equal to ___________.</p> | [] | null | 32 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwcdv7/6c17fcea-c801-4445-8b27-227c5944b078/0d75d830-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwcdv8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjwcdv7/6c17fcea-c801-4445-8b27-227c5944b078/0d75d830-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwcdv8.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) 31st January Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 18 English Explanation"></p>
<p>$$\begin{aligned}
& \mathrm{P} \equiv\left(\frac{\alpha-2}{2}, \frac{\beta-1}{2}\right) \equiv\left(\frac{\gamma+1}{2}, \frac{\delta}{2}\right) \\
& \frac{\alpha-2}{2}=\frac{\gamma+1}{2} \text { and } \frac{\beta-1}{2}=\frac{\delta}{2} \\
& \Rightarrow \alpha-\gamma=3 \ldots .(1), \beta-\delta=1 \ldots \ldots (2)
\end{aligned}$$</p>
<p>Also, $$(\gamma, \delta)$$ lies on $$3 x-2 y=6$$</p>
<p>$$3 \gamma-2 \delta=6$$ ..... (3)</p>
<p>and $$(\alpha, \beta)$$ lies on $$2 x-y=5$$</p>
<p>$$\Rightarrow 2 \alpha-\beta=5 \text {. }$$</p>
<p>Solving (1), (2), (3), (4)</p>
<p>$$\begin{aligned}
& \alpha=-3, \beta=-11, \gamma=-6, \delta=-12 \\
& |\alpha+\beta+\gamma+\delta|=32
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse4xwyu | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | <p>Let $$\alpha, \beta, \gamma, \delta \in \mathbb{Z}$$ and let $$A(\alpha, \beta), B(1,0), C(\gamma, \delta)$$ and $$D(1,2)$$ be the vertices of a parallelogram $$\mathrm{ABCD}$$. If $$A B=\sqrt{10}$$ and the points $$\mathrm{A}$$ and $$\mathrm{C}$$ lie on the line $$3 y=2 x+1$$, then $$2(\alpha+\beta+\gamma+\delta)$$ is equal to</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "10"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjikpx/f32556f0-5b8e-46de-b6e9-9f078c18e4de/05872950-cc2e-11ee-b20d-39b621d226e3/file-6y3zli1lsnjikpy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsnjikpx/f32556f0-5b8e-46de-b6e9-9f078c18e4de/05872950-cc2e-11ee-b20d-39b621d226e3/file-6y3zli1lsnjikpy.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) 31st January Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 17 English Explanation"></p>
<p>Let E is mid point of diagonals</p>
<p>$$\frac{\alpha+\gamma}{2}=\frac{1+1}{2}$$</p>
<p>$$\alpha+\gamma=2$$</p>
<p>& $$\frac{\beta+\delta}{2}=\frac{2+0}{2}$$</p>
<p>$$\beta+\delta=2$$</p>
<p>$$2(\alpha+\beta+\gamma+\delta)=2(2+2)=8$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
1lsg96zxo | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | <p>A line passing through the point $$\mathrm{A}(9,0)$$ makes an angle of $$30^{\circ}$$ with the positive direction of $$x$$-axis. If this line is rotated about A through an angle of $$15^{\circ}$$ in the clockwise direction, then its equation in the new position is :</p> | [{"identifier": "A", "content": "$$\\frac{y}{\\sqrt{3}+2}+x=9$$\n"}, {"identifier": "B", "content": "$$\\frac{x}{\\sqrt{3}+2}+y=9$$\n"}, {"identifier": "C", "content": "$$\\frac{x}{\\sqrt{3}-2}+y=9$$\n"}, {"identifier": "D", "content": "$$\\frac{y}{\\sqrt{3}-2}+x=9$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnkh52/7d853a24-3c49-423c-95b3-38a63c841203/38d89660-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnkh53.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqnkh52/7d853a24-3c49-423c-95b3-38a63c841203/38d89660-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnkh53.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 - Straight Lines and Pair of Straight Lines Question 12 English Explanation"></p>
<p>$$\mathrm{Eq}^{\mathrm{n}}: y-0=\tan 15^{\circ}(x-9) \Rightarrow y=(2-\sqrt{3})(x-9)$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
luy6z55g | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | <p>A ray of light coming from the point $$\mathrm{P}(1,2)$$ gets reflected from the point $$\mathrm{Q}$$ on the $$x$$-axis and then passes through the point $$R(4,3)$$. If the point $$S(h, k)$$ is such that $$P Q R S$$ is a parallelogram, then $$hk^2$$ is equal to:</p> | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "70"}, {"identifier": "C", "content": "80"}, {"identifier": "D", "content": "90"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw37w9ht/b3e9aa4d-c4a4-4387-bd7f-990e7790d425/c5298810-1031-11ef-b980-477f779c8c59/file-1lw37w9hu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw37w9ht/b3e9aa4d-c4a4-4387-bd7f-990e7790d425/c5298810-1031-11ef-b980-477f779c8c59/file-1lw37w9hu.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 Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 10 English Explanation"></p>
<p>$$\begin{aligned}
& P^{\prime} R: y+2=\frac{5}{3}(x-1) \\
& \text { For Point } Q \Rightarrow y=0 \\
& \frac{6}{5}=a-1 \Rightarrow a=\frac{11}{5}
\end{aligned}$$</p>
<p>Now, $$P Q R S$$ is parallelogram</p>
<p>$$\begin{aligned}
& \therefore \frac{h+a}{2}=\frac{4+1}{2} \Rightarrow h=5-\frac{11}{5}=\frac{14}{5} \\
& \text { and } \frac{2+3}{2}=\frac{k}{2} \Rightarrow K=5
\end{aligned}$$</p>
<p>Now $$h k^2=25 \times \frac{14}{5}=14 \times 5=70$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lvc57b6s | maths | straight-lines-and-pair-of-straight-lines | various-forms-of-straight-line | <p>Let a variable line of slope $$m>0$$ passing through the point $$(4,-9)$$ intersect the coordinate axes at the points $$A$$ and $$B$$. The minimum value of the sum of the distances of $$A$$ and $$B$$ from the origin is</p> | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "25"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd2p5t3/fdf80bf0-c51d-4c22-865c-bd4b3e3d59be/487f8460-159d-11ef-aabb-5de744d2ad81/file-1lwd2p5t4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd2p5t3/fdf80bf0-c51d-4c22-865c-bd4b3e3d59be/487f8460-159d-11ef-aabb-5de744d2ad81/file-1lwd2p5t4.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 - Straight Lines and Pair of Straight Lines Question 1 English Explanation"></p>
<p>$$\begin{aligned}
& L:(y+9)=m(x-4) \\
& A:\left(4+\frac{9}{m}, 0\right) \\
& B:(0,-9-4 m) \\
& O A+O B]_{\min } \\
& \Rightarrow E_{\min }=4+\frac{9}{m}+9+4 m \\
& E=13+\frac{9}{m}+4 m \\
& \frac{d E}{d M}=0 \Rightarrow-\frac{9}{m^2}+4=0 \Rightarrow m= \pm \frac{3}{2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{d^2 E}{d M^2}=\frac{18}{m^3}>0 \text { for } \quad m=\frac{3}{2} \\
& \therefore E_{\min }=4+6+9+6=25
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
IG2r7SisPd2bkNPztoE7U | maths | trigonometric-functions-and-equations | general-solution-and-principal-solution-of-the-equation | If 0 $$ \le $$ x < $${\pi \over 2}$$, then the number of values of x for which sin x $$-$$ sin 2x + sin 3x = 0, is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2"}] | ["D"] | null | sin x $$-$$ sin 2x + sin 3x = 0 $$x \in \left[ {0,{\pi \over 2}} \right)$$
<br><br>$$ \Rightarrow $$ (sin3x + sinx) $$-$$ sin2x = 0
<br><br>$$ \Rightarrow $$ 2sin2x.cos2x $$-$$ sin2x = 0
<br><br>$$ \Rightarrow $$ sin2x (2cosx $$-$$ 1) = 0
<br><br>sin 2x = 0
<br><br>x = 0
<br><br>and cos x = $${1 \over 2}$$
<br><br>and x = $${\pi \over 3}$$
<br><br>two solutions | mcq | jee-main-2019-online-9th-january-evening-slot |
EcINkQGDARFMVDjWTw1kluhk814 | maths | trigonometric-functions-and-equations | general-solution-and-principal-solution-of-the-equation | If $$\sqrt 3 ({\cos ^2}x) = (\sqrt 3 - 1)\cos x + 1$$, the number of solutions of the given equation when $$x \in \left[ {0,{\pi \over 2}} \right]$$ is __________. | [] | null | 1 | $$\sqrt 3 ({\cos ^2}x) = (\sqrt 3 - 1)\cos x + 1$$
<br><br>$$ \Rightarrow $$ $$\sqrt 3 {\cos ^2}x - \sqrt 3 \cos x + \cos x - 1 = 0$$<br><br>$$ \Rightarrow \sqrt 3 \cos x(\cos x - 1) + (\cos x - 1) = 0$$<br><br> $$ \Rightarrow (\cos x - 1)(\sqrt 3 \cos x + 1) = 0$$<br><br>$$\cos x = 1$$<br><br>$$ \Rightarrow x = 0$$ $$ [as x \in \left[ {0,{\pi \over 2}} \right]$$]<br><br>and $$\cos x = - {1 \over {\sqrt 3 }}$$ (not possible in $$x \in \left[ {0,{\pi \over 2}} \right]$$]
<br><br>$$ \therefore $$ Number of solution = 1 | integer | jee-main-2021-online-26th-february-morning-slot |
lsamiw6x | maths | trigonometric-functions-and-equations | solving-trigonometric-equations | The number of solutions of the equation $4 \sin ^2 x-4 \cos ^3 x+9-4 \cos x=0 ; x \in[-2 \pi, 2 \pi]$ is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["A"] | null | <p>We start by recognizing that $
\sin^2 x +
\cos^2 x = 1$. Substituting $
\sin^2 x = 1 -
\cos^2 x$ into the original equation gives:</p>
<p>$$
4(1 -
\cos^2 x) - 4
\cos^3 x + 9 - 4
\cos x = 0
$$</p>
<p>Rearranging and simplifying this equation, we have:</p>
<p>$$
4 - 4
\cos^2 x - 4
\cos^3 x + 9 - 4
\cos x = 0$$</p>
<p>$$
4
\cos^3 x + 4
\cos^2 x + 4
\cos x - 13 = 0$$</p>
<p>$4 \cos ^3 x+4 \cos ^2 x+4 \cos x=13$</p>
<p>Observing the bounds given, $x \in [-2\pi, 2\pi]$, we want to find how many solutions satisfy this cubic equation in terms of $
\cos x$. However, we note that the left-hand side (LHS) of the equation, representing a combination of cosines, could at most approach a maximum sum when $
\cos x = 1$, that being $4(1) + 4(1) + 4(1) = 12$. Yet, we have the equation set to equal 13, which is impossible given the maximum sum of the LHS can only be 12.</p>
<p>The above reasoning indicates that, within the domain specified, there is no value of $x$ for which the equation holds true. Therefore, the number of solutions to the equation is zero.</p> | mcq | jee-main-2024-online-1st-february-evening-shift |
jaoe38c1lscnfni9 | maths | trigonometric-functions-and-equations | solving-trigonometric-equations | <p>If $$2 \tan ^2 \theta-5 \sec \theta=1$$ has exactly 7 solutions in the interval $$\left[0, \frac{n \pi}{2}\right]$$, for the least value of $$n \in \mathbf{N}$$, then $$\sum_\limits{k=1}^n \frac{k}{2^k}$$ is equal to:</p> | [{"identifier": "A", "content": "$$\\frac{1}{2^{14}}\\left(2^{15}-15\\right)$$\n"}, {"identifier": "B", "content": "$$1-\\frac{15}{2^{13}}$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{2^{15}}\\left(2^{14}-14\\right)$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{2^{13}}\\left(2^{14}-15\\right)$$"}] | ["D"] | null | <p>$$\begin{aligned}
& 2 \tan ^2 \theta-5 \sec \theta-1=0 \\
& \Rightarrow 2 \sec ^2 \theta-5 \sec \theta-3=0 \\
& \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 \\
& \Rightarrow \sec \theta=-\frac{1}{2}, 3 \\
& \Rightarrow \cos \theta=-2, \frac{1}{3} \\
& \Rightarrow \cos \theta=\frac{1}{3}
\end{aligned}$$</p>
<p>For 7 solutions $$\mathrm{n}=13$$</p>
<p>$$\begin{aligned}
& \text { So, } \sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S} \text { (say) } \\
& \mathrm{S}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots .+\frac{13}{2^{13}} \\
& \frac{1}{2} \mathrm{~S}=\frac{1}{2^2}+\frac{1}{2^3}+\ldots .+\frac{12}{2^{13}}+\frac{13}{2^{14}} \\
& \Rightarrow \frac{\mathrm{S}}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow \mathrm{S}=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lseydb0d | maths | trigonometric-functions-and-equations | solving-trigonometric-equations | <p>If $$\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$$ is the solution of $$4 \cos \theta+5 \sin \theta=1$$, then the value of $$\tan \alpha$$ is</p> | [{"identifier": "A", "content": "$$\\frac{10-\\sqrt{10}}{12}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\sqrt{10}-10}{6}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{10}-10}{12}$$\n"}, {"identifier": "D", "content": "$$\\frac{10-\\sqrt{10}}{6}$$"}] | ["C"] | null | <p>$$4+5 \tan \theta=\sec \theta$$</p>
<p>Squaring : $$24 \tan ^2 \theta+40 \tan \theta+15=0$$</p>
<p>$$\tan \theta=\frac{-10 \pm \sqrt{10}}{12}$$</p>
<p>and $$\tan \theta=-\left(\frac{10+\sqrt{10}}{12}\right)$$ is Rejected.</p>
<p>(3) is correct.</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfkwdj6 | maths | trigonometric-functions-and-equations | solving-trigonometric-equations | <p>The sum of the solutions $$x \in \mathbb{R}$$ of the equation $$\frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6$$ is</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$-$$1"}] | ["D"] | null | <p>$$\begin{aligned}
& \frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6 \\
& \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 \\
& \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 \\
& \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(3+\cos ^2 2 x\right)}=x^3-x^2+6 \\
& x^3-x^2+2=0 \Rightarrow(x+1)\left(x^2-2 x+2\right)=0
\end{aligned}$$</p>
<p>so, sum of real solutions $$=-1$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsgaqmf3 | maths | trigonometric-functions-and-equations | solving-trigonometric-equations | <p>If $$2 \sin ^3 x+\sin 2 x \cos x+4 \sin x-4=0$$ has exactly 3 solutions in the interval $$\left[0, \frac{\mathrm{n} \pi}{2}\right], \mathrm{n} \in \mathrm{N}$$, then the roots of the equation $$x^2+\mathrm{n} x+(\mathrm{n}-3)=0$$ belong to :</p> | [{"identifier": "A", "content": "$$(0, \\infty)$$\n"}, {"identifier": "B", "content": "Z"}, {"identifier": "C", "content": "$$\\left(-\\frac{\\sqrt{17}}{2}, \\frac{\\sqrt{17}}{2}\\right)$$\n"}, {"identifier": "D", "content": "$$(-\\infty, 0)$$"}] | ["D"] | null | <p>$$\begin{aligned}
& 2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0 \\
& 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \\
& 6 \sin x-4=0 \\
& \sin x=\frac{2}{3} \\
& \mathbf{n}=5 \text { (in the given interval) } \\
& x^2+5 x+2=0 \\
& x=\frac{-5 \pm \sqrt{17}}{2} \\
& \text { Required interval }(-\infty, 0)
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
luy6z4wc | maths | trigonometric-functions-and-equations | solving-trigonometric-equations | <p>Let $$|\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8}, \theta \epsilon[0,2 \pi]$$. Then, the sum of all $$\theta \in[0,2 \pi]$$, where $$\cos 3 \theta$$ attains its maximum value, is :
</p> | [{"identifier": "A", "content": "$$6 \\pi$$\n"}, {"identifier": "B", "content": "$$9 \\pi$$\n"}, {"identifier": "C", "content": "$$18 \\pi$$\n"}, {"identifier": "D", "content": "$$15 \\pi$$"}] | ["A"] | null | <p>$$\begin{aligned}
& |\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8} \\
& \Rightarrow \frac{1}{4}|\cos 3 \theta| \leq \frac{1}{8} \\
& \cos 3 \theta \text { is max if } \cos 3 \theta=\frac{1}{2} \\
& \therefore \theta=\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9} \\
& \sum \theta_i=6 \pi
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lv2erh14 | maths | trigonometric-functions-and-equations | solving-trigonometric-equations | <p>Let $$S=\left\{\sin ^2 2 \theta:\left(\sin ^4 \theta+\cos ^4 \theta\right) x^2+(\sin 2 \theta) x+\left(\sin ^6 \theta+\cos ^6 \theta\right)=0\right.$$ has real roots $$\}$$. If $$\alpha$$ and $$\beta$$ be the smallest and largest elements of the set $$S$$, respectively, then $$3\left((\alpha-2)^2+(\beta-1)^2\right)$$ equals __________.</p> | [] | null | 4 | <p>For real roots</p>
<p>$$\begin{aligned}
& D \geq 0 \\
& \sin ^2 2 \theta \geq 4\left(\sin ^4 \theta+\cos ^4 \theta\right)\left(\sin ^6 \theta+\cos ^6 \theta\right)
\end{aligned}$$</p>
<p>Put $$\sin ^2 2 \theta=t$$</p>
<p>$$\begin{aligned}
& \Rightarrow t \geq 4\left(1-\frac{t}{2}\right)\left(1-\frac{3 t}{4}\right) \\
& 2 t \geq(2-t)(4-3 t) \\
& 3 t^2-12 t+8 \leq 0 \\
& t^2-4 t+\frac{8}{3} \leq 0 \\
& (t-2)^2+\frac{8}{3}-4 \leq 0 \\
& (t-2)^2 \leq \frac{4}{3} \\
& -\frac{2}{\sqrt{3}} \leq t-2 \leq \frac{2}{\sqrt{3}} \\
& 2-\frac{2}{\sqrt{3}} \leq t \leq 2+\frac{2}{\sqrt{3}} \\
& \because t \in[0,1] \\
& \Rightarrow 2-\frac{2}{\sqrt{3}} \leq t \leq 1 \\
& \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 \\
& \Rightarrow 3\left[(\alpha-2)^2+(\beta-1)^2\right]=4
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-evening-shift |
lv9s20oc | maths | trigonometric-functions-and-equations | solving-trigonometric-equations | <p>The number of solutions of $$\sin ^2 x+\left(2+2 x-x^2\right) \sin x-3(x-1)^2=0$$, where $$-\pi \leq x \leq \pi$$, is ________.</p> | [] | null | 2 | <p>$$\begin{aligned}
& \sin ^2 x+\left(3-(x-1)^2\right) \sin x-3(x-1)^2=0 \\
& \sin ^2 x+3 \sin x-(x-1)^2 \sin x-3(x-1)^2=0 \\
& \left.\sin x(\sin x+3)-(x-1)^2\right)[\sin x+3]=0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwey5iox/7c0cf8ef-bef6-4412-bab2-d9c808c7d020/130d0510-16a5-11ef-aafd-dd0edda8277a/file-1lwey5ioy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwey5iox/7c0cf8ef-bef6-4412-bab2-d9c808c7d020/130d0510-16a5-11ef-aafd-dd0edda8277a/file-1lwey5ioy.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) 5th April Evening Shift Mathematics - Trigonometric Equations Question 1 English Explanation"></p>
<p>There are two intersections between this graph.</p>
<p>So, Number of solution will be 2 .</p> | integer | jee-main-2024-online-5th-april-evening-shift |
3BKJ6l1wQpkLN7tB | maths | trigonometric-ratio-and-identites | addition-and-subtraction-formula | Let $$\alpha ,\,\beta $$ be such that $$\pi < \alpha - \beta < 3\pi $$.
<br/>If $$sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}$$ and $$\cos \alpha + \cos \beta = - {{27} \over {65}}$$ then the value of $$\cos {{\alpha - \beta } \over 2}$$ : | [{"identifier": "A", "content": "$${{ - 6} \\over {65}}\\,\\,$$ "}, {"identifier": "B", "content": "$${3 \\over {\\sqrt {130} }}$$ "}, {"identifier": "C", "content": "$${6 \\over {65}}$$ "}, {"identifier": "D", "content": "$$ - {3 \\over {\\sqrt {130} }}$$"}] | ["D"] | null | Given $$sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}$$ .........(1) <br><br>and $$\cos \alpha + \cos \beta = - {{27} \over {65}}$$ ........(2)
<br><br>Square and add (1) and (2) you will get
<br><br>$$2\left( {1 + \cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)$$$$ = {{{{\left( {21} \right)}^2} + {{\left( {27} \right)}^2}} \over {{{\left( {65} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ $$2\left( {1 + \cos \left( {\alpha - \beta } \right)} \right) = {{1170} \over {{{\left( {65} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ $$4{\cos ^2}{{\alpha - \beta } \over 2}$$$$ = {{1170} \over {{{\left( {65} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ $${\cos ^2}{{\alpha - \beta } \over 2}$$$$ = {9 \over {130}}$$
<br><br>$$\therefore$$ $$\cos {{\alpha - \beta } \over 2} = \pm {3 \over {\sqrt {130} }}$$
<br><br>[ But $$\cos {{\alpha - \beta } \over 2} \ne + {3 \over {\sqrt {130} }}$$ <br><br>as $$\pi < \alpha - \beta < 3\pi $$
<br><br>$$ \Rightarrow $$ $${\pi \over 2} < {{\alpha - \beta } \over 2} < {{3\pi } \over 2}$$
<br><br>$$ \Rightarrow $$ $$\cos {{\alpha - \beta } \over 2} < 0$$ ]
<br><br>So $$\cos {{\alpha - \beta } \over 2} = - {3 \over {\sqrt {130} }}$$ | mcq | aieee-2004 |
hd48oCskZmQcbGFf | maths | trigonometric-ratio-and-identites | addition-and-subtraction-formula | Let <b>A</b> and <b>B</b> denote the statements
<p><b>A</b>: $$\cos \alpha + \cos \beta + \cos \gamma = 0$$</p>
<p><b>B</b>: $$\sin \alpha + \sin \beta + \sin \gamma = 0$$</p>
<p>If $$\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - {3 \over 2},$$ then:</p>
| [{"identifier": "A", "content": "<b>A</b> is false and <b>B</b> is true "}, {"identifier": "B", "content": "both <b>A</b> and <b>B</b> are true "}, {"identifier": "C", "content": "both <b>A</b> and <b>B</b> are false "}, {"identifier": "D", "content": "<b>A</b> is true and <b>B</b> is false"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265710/exam_images/vm4amwtny6bublz2trzy.webp" loading="lazy" alt="AIEEE 2009 Mathematics - Trigonometric Ratio and Identites Question 48 English Explanation"> | mcq | aieee-2009 |
qWyukXL9UGk1iVvA | maths | trigonometric-ratio-and-identites | addition-and-subtraction-formula | Let $$\cos \left( {\alpha + \beta } \right) = {4 \over 5}$$ and $$\sin \,\,\,\left( {\alpha - \beta } \right) = {5 \over {13}},$$ where $$0 \le \alpha ,\,\beta \le {\pi \over 4}.$$
<br/>Then $$tan\,2\alpha $$ =<br/> | [{"identifier": "A", "content": "$${56 \\over 33}$$"}, {"identifier": "B", "content": "$${19 \\over 12}$$ "}, {"identifier": "C", "content": "$${20 \\over 7}$$"}, {"identifier": "D", "content": "$${25 \\over 16}$$ "}] | ["A"] | null | $$\cos \left( {\alpha + \beta } \right) = {4 \over 5} \Rightarrow \tan \left( {\alpha + \beta } \right) = {3 \over 4}$$
<br><br>$$\sin \left( {\alpha - \beta } \right) = {5 \over {13}} \Rightarrow \tan \left( {\alpha - \beta } \right) = {5 \over {12}}$$
<br><br>$$\tan 2\alpha = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]$$
<br><br>$$ = {{{3 \over 4} + {5 \over {12}}} \over {1 - {3 \over 4}.{5 \over {12}}}} = {{56} \over {33}}$$ | mcq | aieee-2010 |
bphMWKhkG9CXaybI3b30Z | maths | trigonometric-ratio-and-identites | addition-and-subtraction-formula | If cos($$\alpha $$ + $$\beta $$) = 3/5 ,sin ( $$\alpha $$ - $$\beta $$) = 5/13 and
0 < $$\alpha , \beta$$ < $$\pi \over 4$$, then tan(2$$\alpha $$) is equal to : | [{"identifier": "A", "content": "21/16"}, {"identifier": "B", "content": "63/52"}, {"identifier": "C", "content": "33/52"}, {"identifier": "D", "content": "63/16"}] | ["D"] | null | Given $$0 < \alpha < {\pi \over 4}$$
<br><br>and $$0 < \beta < {\pi \over 4}$$
<br><br>$$ \therefore $$ $$0 > - \beta > - {\pi \over 4}$$
<br><br>$$ \therefore $$ $$0 < \alpha + \beta < {\pi \over 2}$$
<br><br>and $$ - {\pi \over 4} < \alpha - \beta < {\pi \over 4}$$
<br><br>As cos($$\alpha $$ + $$\beta $$) = 3/5
<br><br>so $${\tan \left( {\alpha + \beta } \right) = {4 \over 3}}$$
<br><br>As sin( $$\alpha $$ - $$\beta $$) = 5/13
<br><br>so $${\tan \left( {\alpha - \beta } \right) = {5 \over {12}}}$$
<br><br>Now tan(2$$\alpha $$) = tan($$\alpha $$ + $$\beta $$ + $$\alpha $$ - $$\beta $$)
<br><br>= $${{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)} \over {1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}$$
<br><br>= $${{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3} \times {5 \over {12}}}}$$ = $${{63} \over {16}}$$ | mcq | jee-main-2019-online-8th-april-morning-slot |
BsuLEX4QMoaT0JZXqP1klt7sbsa | maths | trigonometric-ratio-and-identites | addition-and-subtraction-formula | If 0 < x, y < $$\pi$$ and cosx + cosy $$-$$ cos(x + y) = $${3 \over 2}$$, then sinx + cosy is equal to : | [{"identifier": "A", "content": "$${{1 + \\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$${{1 \\over 2}}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "D", "content": "$${{1 - \\sqrt 3 } \\over 2}$$"}] | ["A"] | null | $$2\cos \left( {{{x + y} \over 2}} \right)\cos \left( {{{x - y} \over 2}} \right) - \left[ {2{{\cos }^2}\left( {{{x + y} \over 2}} \right) - 1} \right] = {3 \over 2}$$<br><br>$$2\cos \left( {{{x + y} \over 2}} \right)\left[ {\cos \left( {{{x - y} \over 2}} \right) - \cos \left( {{{x + y} \over 2}} \right)} \right] = {1 \over 2}$$<br><br>$$2\cos \left( {{{x + y} \over 2}} \right)\left[ {2\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right)} \right] = {1 \over 2}$$<br><br>$$\cos \left( {{{x + y} \over 2}} \right).\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right) = {1 \over 8}$$<br><br>Possible when $${x \over 2} = 30^\circ $$ & $${y \over 2} = 30^\circ $$<br><br>$$x = y = 60^\circ $$<br><br>$$\sin x + \cos y = {{\sqrt 3 } \over 2} + {1 \over 2} = {{\sqrt 3 + 1} \over 2}$$ | mcq | jee-main-2021-online-25th-february-evening-slot |
1krxj7yrx | maths | trigonometric-ratio-and-identites | addition-and-subtraction-formula | If $$\tan \left( {{\pi \over 9}} \right),x,\tan \left( {{{7\pi } \over {18}}} \right)$$ are in arithmetic progression and $$\tan \left( {{\pi \over 9}} \right),y,\tan \left( {{{5\pi } \over {18}}} \right)$$ are also in arithmetic progression, then $$|x - 2y|$$ is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $$x = {1 \over 2}\left( {\tan {\pi \over 9} + \tan {{7\pi } \over {18}}} \right)$$<br><br>and $$2y = \tan {\pi \over 9} + \tan {{5\pi } \over {18}}$$<br><br>If we interpret the angles in degrees (as suggested by the numbers 20, 50, and 70), we have :
<br/><br/>$$x = \frac{1}{2} \left( \tan 20^\circ + \tan 70^\circ \right),$$
<br/><br/>and
<br/><br/>$$2y = \tan 20^\circ + \tan 50^\circ.$$
<br/><br/>The expression for $|x - 2y|$ is then :
<br/><br/>$$|x - 2y| = \left|\frac{\tan 20^\circ + \tan 70^\circ}{2} - \left( \tan 20^\circ + \tan 50^\circ \right)\right|.$$
<br/><br/>$$|x - 2y| = \left|\frac{\tan 20^\circ + \tan 70^\circ - 2 \tan 20^\circ - 2 \tan 50^\circ}{2}\right|,$$
<br/><br/>which simplifies to :
<br/><br/>$$|x - 2y| = \left|\frac{\tan 70^\circ - \tan 20^\circ - 2 \tan 50^\circ}{2}\right|.$$
<br/><br/>We know,
<br/><br/>$$
\begin{aligned}
& \tan 70=\frac{\tan 20+\tan 50}{1-\tan 20 \tan 50} \\\\
&\Rightarrow \tan 70-\tan 70 \cdot \tan 20 \tan 50=\tan 20+\tan 50 \\\\
&\Rightarrow \tan 70-\tan 50-\tan 20-\tan 50=0 \\\\
&\Rightarrow \tan 70-\tan 20-2 \tan 50=0
\end{aligned}
$$
<br/><br/>$$ \therefore $$ $$|x - 2y| = \left|\frac{\tan 70^\circ - \tan 20^\circ - 2 \tan 50^\circ}{2}\right|$$ = $$\left| {{0 \over 2}} \right|$$ = 0 | mcq | jee-main-2021-online-27th-july-evening-shift |
1l55iwg5a | maths | trigonometric-ratio-and-identites | addition-and-subtraction-formula | <p>If cot$$\alpha$$ = 1 and sec$$\beta$$ = $$ - {5 \over 3}$$, where $$\pi < \alpha < {{3\pi } \over 2}$$ and $${\pi \over 2} < \beta < \pi $$, then the value of $$\tan (\alpha + \beta )$$ and the quadrant in which $$\alpha$$ + $$\beta$$ lies, respectively are :</p> | [{"identifier": "A", "content": "$$ - {1 \\over 7}$$ and IV<sup>th</sup> quadrant"}, {"identifier": "B", "content": "7 and I<sup>st</sup> quadrant"}, {"identifier": "C", "content": "$$-$$7 and IV<sup>th</sup> quadrant"}, {"identifier": "D", "content": "$$ {1 \\over 7}$$ and I<sup>st</sup> quadrant"}] | ["A"] | null | $\because \cot \alpha=1, \quad \alpha \in\left(\pi, \frac{3 \pi}{2}\right)$<br/><br/> then $\tan \alpha=1$ <br/><br/>and $\sec \beta=-\frac{5}{3}, \quad \beta \in\left(\frac{\pi}{2}, \pi\right)$ <br/><br/>then $\tan \beta=-\frac{4}{3}$
<br/><br/>
$\therefore \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}$
<br/><br/>
$$
\begin{aligned}
&=\frac{1-\frac{4}{3}}{1+\frac{4}{3}} \\\\
&=-\frac{1}{7}
\end{aligned}
$$
<br/><br/>
$$
\alpha+\beta \in\left(\frac{3 \pi}{2}, 2 \pi\right) \text { i.e. fourth quadrant }
$$ | mcq | jee-main-2022-online-28th-june-evening-shift |
lsaodfiz | maths | trigonometric-ratio-and-identites | addition-and-subtraction-formula | If $\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \mathrm{B}=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and <br/><br/>$\tan \mathrm{C}=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1 / 2}, 0<\mathrm{A}, \mathrm{B}, \mathrm{C}<\frac{\pi}{2}$, then $\mathrm{A}+\mathrm{B}$ is equal to : | [{"identifier": "A", "content": "$\\mathrm{C}$"}, {"identifier": "B", "content": "$\\pi-C$"}, {"identifier": "C", "content": "$2 \\pi-C$"}, {"identifier": "D", "content": "$\\frac{\\pi}{2}-\\mathrm{C}$"}] | ["A"] | null | <p>To find the sum of two angles in terms of tangent, we can use the tangent addition formula: </p>
<p>$$ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} $$</p>
<p>Let's compute $\tan (A + B)$ using the given $\tan A$ and $\tan B$: </p>
<p>$$ \tan A = \frac{1}{\sqrt{x(x^2+x+1)}} $$</p>
<p>$$ \tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}} $$</p>
<p>Now, we can apply the addition formula: </p>
<p>$$ \tan (A + B) = \frac{\frac{1}{\sqrt{x(x^2+x+1)}} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1 - \frac{1}{\sqrt{x(x^2+x+1)}} \cdot \frac{\sqrt{x}}{\sqrt{x^2+x+1}}} $$</p>
<p>$$ \tan (A + B) = \frac{\frac{1 + \sqrt{x^2}}{\sqrt{x(x^2+x+1)}}}{1 - \frac{1}{(x^2+x+1)}} $$</p>
<p>$$ \tan (A + B) = \frac{\frac{\sqrt{x^2} + 1}{\sqrt{x(x^2+x+1)}}}{\frac{(x^2+x+1) - 1}{(x^2+x+1)}} $$</p>
<p>$$ \tan (A + B) = \frac{(x + 1)\sqrt{(x^2+x+1)}}{(x^2 + x){\sqrt{x}}} $$</p>
<p>$$ \tan (A + B) = \frac{{(x + 1)}\sqrt{(x^2+x+1)}}{x{(x + 1)}{\sqrt{x}}} $$</p>
<p>$$ \tan (A + B) = \frac{\sqrt{x^2+x+1}}{x^{3/2}} $$</p>
Let us first simplify $\tan \mathrm{C}$:
<p>$$\tan \mathrm{C} = \sqrt{x^{-3}+x^{-2}+x^{-1}} = \sqrt{\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}} = \sqrt{\frac{1+x+x^2}{x^3}} = \frac{\sqrt{x^2+x+1}}{x^{3/2}}.$$</p>
<p>We see that: </p>
<p>$$ \tan (A + B) = \tan C $$</p>
<p>Since tangent is positive and all the angles $A$, $B$, and $C$ are in the first quadrant, we can say that: </p>
<p>$$ A + B = C $$</p>
<p>Hence, the answer is: </p>
<p>Option A : $C$</p> | mcq | jee-main-2024-online-1st-february-morning-shift |
1lsg3vc3r | maths | trigonometric-ratio-and-identites | addition-and-subtraction-formula | <p>For $$\alpha, \beta \in(0, \pi / 2)$$, let $$3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$$ and a real number $$k$$ be such that $$\tan \alpha=k \tan \beta$$. Then, the value of $$k$$ is equal to</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "$$-$$2/3"}, {"identifier": "C", "content": "$$-$$5"}, {"identifier": "D", "content": "2/3"}] | ["C"] | null | <p>To find the value of $$k$$, the given conditions are:</p>
<p>$$3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$$</p>
<p>And $$\tan \alpha = k \tan \beta$$</p>
<p>For the first equation, using the sum and difference formulas for sine, we can rewrite the equation as:</p>
<p>$$3(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2(\sin \alpha \cos \beta - \cos \alpha \sin \beta)$$</p>
<p>Simplifying this, we get:</p>
<p>$$3 \sin \alpha \cos \beta + 3 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta - 2 \cos \alpha \sin \beta$$</p>
<p>Rearranging the terms, we obtain:</p>
<p>$$5 \sin \beta \cos \alpha = - \sin \alpha \cos \beta$$</p>
<p>Dividing both sides by $$\sin \alpha \cos \beta$$, we get:</p>
<p>$$\frac{5 \sin \beta \cos \alpha}{\sin \alpha \cos \beta} = -1$$</p>
<p>Which simplifies to:</p>
<p>$$5 \tan \beta = - \tan \alpha$$</p>
<p>So, taking the reciprocal, we have:</p>
<p>$$\tan \alpha = -5 \tan \beta$$</p>
<p>Therefore, by comparing this equation with the given $$\tan \alpha = k \tan \beta$$, we find that $$k = -5$$.</p>
<p>Thus, the value of $$k$$ is <strong>$$-5$$</strong>.</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
jaoe38c1lsd4e0v7 | maths | trigonometric-ratio-and-identites | basic-definition-of-trigonometric-function | <p>The number of solutions, of the equation $$e^{\sin x}-2 e^{-\sin x}=2$$, is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "more than 2"}] | ["A"] | null | <p>Take $$e^{\sin x}=t(t>0)$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{t}-\frac{2}{\mathrm{t}}=2 \\
& \Rightarrow \frac{\mathrm{t}^2-2}{\mathrm{t}}=2 \\
& \Rightarrow \mathrm{t}^2-2 \mathrm{t}-2=0 \\
& \Rightarrow \mathrm{t}^2-2 \mathrm{t}+1=3 \\
& \Rightarrow(\mathrm{t}-1)^2=3 \\
& \Rightarrow \mathrm{t}=1 \pm \sqrt{3} \\
& \Rightarrow \mathrm{t}=1 \pm 1.73 \\
& \Rightarrow \mathrm{t}=2.73 \text { or }-0.73 \text { (rejected as } \mathrm{t}>0) \\
& \Rightarrow \mathrm{e}^{\sin \mathrm{x}}=2.73 \\
& \Rightarrow \log _{\mathrm{e}} \mathrm{e}^{\sin \mathrm{x}}=\log _{\mathrm{e}} 2.73 \\
& \Rightarrow \sin \mathrm{x}=\log _{\mathrm{e}} 2.73>1
\end{aligned}$$</p>
<p>So no solution.</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
9L50ljSVtdEZAmnd | maths | trigonometric-ratio-and-identites | fundamental-identities | If $$0 < x < \pi $$ and $$\cos x + \sin x = {1 \over 2},$$ then $$\tan x$$ is : | [{"identifier": "A", "content": "$${{\\left( {1 - \\sqrt 7 } \\right)} \\over 4}$$ "}, {"identifier": "B", "content": "$${{\\left( {4 - \\sqrt 7 } \\right)} \\over 3}$$ "}, {"identifier": "C", "content": "$$ - {{\\left( {4 + \\sqrt 7 } \\right)} \\over 3}$$ "}, {"identifier": "D", "content": "$${{\\left( {1 + \\sqrt 7 } \\right)} \\over 4}$$ "}] | ["C"] | null | $$\cos x + \sin x = {1 \over 2}$$ <br><br>
$$ \Rightarrow {\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)^2} = {1 \over 4}$$<br><br>
$$ \Rightarrow {\cos ^2}x + {\sin ^2}x + 2\cos x\sin x = {1 \over 4}$$<br>
$$\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin 2x} \right]$$
<br><br>$$ \Rightarrow 1 + \sin 2x = {1 \over 4}$$
<br><br>$$ \Rightarrow \sin 2x = - {3 \over 4},$$ so $$x$$ is obtuse and
<br><br>$${{2\tan x} \over {1 + {{\tan }^2}x}} = - {3 \over 4}$$
<br><br>$$ \Rightarrow 3{\tan ^2}x + 8\tan x + 3 = 0$$
<br><br>$$\therefore$$ $$\tan x = {{ - 8 \pm \sqrt {64 - 36} } \over 6}$$
<br><br>$$ = {{ - 4 \pm \sqrt 7 } \over 3}$$
<br><br>as $$\tan x < 0\,$$
<br><br>$$\therefore$$ $$\tan x = {{ - 4 - \sqrt 7 } \over 3}$$ | mcq | aieee-2006 |
i5tjd3WubiuGMhss | maths | trigonometric-ratio-and-identites | fundamental-identities | The expression $${{\tan {\rm A}} \over {1 - \cot {\rm A}}} + {{\cot {\rm A}} \over {1 - \tan {\rm A}}}$$ can be written as: | [{"identifier": "A", "content": "$$\\sin {\\rm A}\\,\\cos {\\rm A} + 1$$ "}, {"identifier": "B", "content": "$$\\,\\sec {\\rm A}\\,\\cos ec{\\rm A} + 1$$ "}, {"identifier": "C", "content": "$$\\tan {\\rm A} + \\cot {\\rm A}$$ "}, {"identifier": "D", "content": "$$\\sec {\\rm A} + \\cos ec{\\rm A}$$ "}] | ["B"] | null | Given expression can be written as
<br><br>$${{\sin A} \over {\cos A}} \times {{sin\,A} \over {\sin A - \cos A}} + {{\cos A} \over {\sin A}} \times {{\cos A} \over {\cos A - sin\,A}}$$
<br><br>(As $$\tan A = {{\sin A} \over {\cos A}}$$ and $$\cot A = {{\cos A} \over {\sin A}}$$ )
<br><br>$$ = {1 \over {\sin A - \cos A}}\left\{ {{{{{\sin }^3}A - {{\cos }^3}A} \over {\cos A\sin A}}} \right\}$$
<br><br>$$ = {{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}\,A} \over {\sin A\cos A}}$$
<br><br>$$ = 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A$$ | mcq | jee-main-2013-offline |
T8c6mk5UF9taxkOB | maths | trigonometric-ratio-and-identites | fundamental-identities | Let $$f_k\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$ where $$x \in R$$ and $$k \ge \,1.$$
<br/>Then $${f_4}\left( x \right) - {f_6}\left( x \right)\,\,$$ equals : | [{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$${1 \\over 12}$$"}, {"identifier": "C", "content": "$${1 \\over 6}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["B"] | null | Let $${f_k}\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}.x} \right)$$
<br><br>Consider
<br><br>$${f_4}\left( x \right) - {f_6}\left( x \right) $$
<br><br>$$=$$ $${1 \over 4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {1 \over 6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$$
<br><br>$$ = {1 \over 4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - {1 \over 6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]$$
<br><br>$$ = {1 \over 4} - {1 \over 6} = {1 \over {12}}$$ | mcq | jee-main-2014-offline |
NehdCaCp32Un7xttPHdPb | maths | trigonometric-ratio-and-identites | fundamental-identities | For any $$\theta \in \left( {{\pi \over 4},{\pi \over 2}} \right)$$, the expression
<br/><br/> $$3{(\cos \theta - \sin \theta )^4}$$$$ + 6{(\sin \theta + \cos \theta )^2} + 4{\sin ^6}\theta $$
<br/><br/>equals : | [{"identifier": "A", "content": "13 \u2013 4 cos<sup>2</sup>$$\\theta $$ + 6sin<sup>2</sup>$$\\theta $$cos<sup>2</sup>$$\\theta $$"}, {"identifier": "B", "content": "13 \u2013 4 cos<sup>6</sup>$$\\theta $$"}, {"identifier": "C", "content": "13 \u2013 4 cos<sup>2</sup>$$\\theta $$ + 6cos<sup>2</sup>$$\\theta $$"}, {"identifier": "D", "content": "13 \u2013 4 cos<sup>4</sup>$$\\theta $$ + 2sin<sup>2</sup>$$\\theta $$cos<sup>2</sup>$$\\theta $$"}] | ["B"] | null | Given,
<br><br>3(sin$$\theta $$ $$-$$ cos$$\theta $$)<sup>4</sup> + 6(sin$$\theta $$ + cos$$\theta $$)<sup>2</sup> + 4sin<sup>6</sup>$$\theta $$
<br><br>= 3[(sin$$\theta $$ $$-$$ cos$$\theta $$)<sup>2</sup>]<sup>2</sup> + 6 (sin<sup>2</sup>$$\theta $$ + cos<sup>2</sup>$$\theta $$ + 2sin$$\theta $$cos$$\theta $$) + 4sin<sup>6</sup>$$\theta $$
<br><br>= 3[sin<sup>2</sup>$$\theta $$ + cos<sup>2</sup>$$\theta $$ $$-$$2sin$$\theta $$cos$$\theta $$]<sup>2</sup> + 6(1 + sin2$$\theta $$) + 4sin<sup>6</sup>$$\theta $$
<br><br>= 3(1 $$-$$ sin2$$\theta $$)<sup>2</sup> + 6(1 + sin2$$\theta $$) + 4sin<sup>6</sup>$$\theta $$
<br><br>= 3 (1 $$-$$ 2 sin2$$\theta $$ + sin<sup>2</sup>2$$\theta $$) + 6 + 6sin2$$\theta $$ + 4sin<sup>6</sup>$$\theta $$
<br><br>= 3 $$-$$ 6sin2$$\theta $$ + 3sin<sup>2</sup>2$$\theta $$ + 6 + 6sin2$$\theta $$ + 4sin<sup>6</sup>$$\theta $$
<br><br>= 9 + 3sin<sup>2</sup>2$$\theta $$ + 4 sin<sup>6</sup>$$\theta $$
<br><br>= 9 + 3(2sin$$\theta $$cos$$\theta $$)<sup>2</sup> + 4(1 $$-$$ cos<sup>2</sup>$$\theta $$)<sup>3</sup>
<br><br>= 9 + 12sin<sup>2</sup>$$\theta $$ cos<sup>2</sup>$$\theta $$ + 4 (1 $$-$$ cos<sup>6</sup>$$\theta $$ $$-$$ 3cos<sup>2</sup>$$\theta $$ + 3cos<sup>4</sup>$$\theta $$)
<br><br>= 13 + 12 (1 $$-$$ cos<sup>2</sup>$$\theta $$ $$-$$ 4cos<sup>6</sup>$$\theta $$ $$-$$ 12cos$$\theta $$ + 12 cos<sup>4</sup>$$\theta $$
<br><br>= 13 + 12 cos<sup>2</sup>$$\theta $$ $$-$$ 12 cos<sup>4</sup>$$\theta $$ $$-$$ 4cos<sup>6</sup>$$\theta $$ $$-$$ 12 cos<sup>2</sup>$$\theta $$ + 12 cos<sup>4</sup>$$\theta $$
<br><br>= 13 $$-$$ 4 cos<sup>6</sup>$$\theta $$ | mcq | jee-main-2019-online-9th-january-morning-slot |
mS2BpguXsYP3SeqGbW1kmhz5a04 | maths | trigonometric-ratio-and-identites | fundamental-identities | If for x $$\in$$ $$\left( {0,{\pi \over 2}} \right)$$, log<sub>10</sub>sinx + log<sub>10</sub>cosx = $$-$$1 and log<sub>10</sub>(sinx + cosx) = $${1 \over 2}$$(log<sub>10</sub> n $$-$$ 1), n > 0, then the value of n is equal to : | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "20"}] | ["C"] | null | $$
\begin{aligned}
& \log _{10} \sin x+\log _{10} \cos x=-1, x \in(0, \pi / 2) \\\\
& \log _{10}(\sin x \cos x)=-1 \\\\
& \Rightarrow \sin x \cos x=10^{-1}= {1 \over {10}} \\\\
& \log _{10}(\sin x+\cos x)={1 \over {2}}\left(\log _{10} n-1\right), n>0 \\\\
& 2 \log _{10}(\sin x+\cos x)=\left(\log _{10} n-\log _{10} 10\right) \\\\
& \Rightarrow \log _{10}(\sin x+\cos x)^2=\log _{10}({n \over {10}}) \\\\
& \Rightarrow (\sin x+\cos x)^2={n \over {10}} \\\\
& \Rightarrow \sin ^2 x+\cos ^2 x+2 \sin x \cos x=\frac{n}{10} \\\\
& \Rightarrow 1+2({1 \over {10}})={n \over {10}} \Rightarrow {12 \over {10}}={n \over {10}} \\\\
& \therefore n=12
\end{aligned}
$$ | mcq | jee-main-2021-online-16th-march-morning-shift |
UnK3tRJPrakUcdlvs61kmm2o767 | maths | trigonometric-ratio-and-identites | fundamental-identities | If 15sin<sup>4</sup>$$\alpha$$ + 10cos<sup>4</sup>$$\alpha$$ = 6, for some $$\alpha$$$$\in$$R, then the value of <br/><br/>27sec<sup>6</sup>$$\alpha$$ + 8cosec<sup>6</sup>$$\alpha$$ is equal to : | [{"identifier": "A", "content": "500"}, {"identifier": "B", "content": "400"}, {"identifier": "C", "content": "250"}, {"identifier": "D", "content": "350"}] | ["C"] | null | $$
\begin{aligned}
& \text { Given, } 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6 \\\\
& \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^2 \alpha+\cos ^2 \alpha\right)^2 \\\\
& \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^4 \alpha+\cos ^4 \alpha+2 \sin ^2 \alpha \cos ^2 \alpha\right) \\\\
& \Rightarrow 9 \sin ^4 \alpha+4 \cos ^4 \alpha-12 \sin ^2 \alpha \cos ^2 \alpha=0 \\\\
& \Rightarrow \quad\left(3 \sin ^2 \alpha-2 \cos ^2 \alpha\right)^2=0 \\\\
& \Rightarrow \quad 3 \sin ^2 \alpha-2 \cos ^2 \alpha=0 \\\\
& \Rightarrow \quad 3 \sin ^2 \alpha=2 \cos ^2 \alpha \\\\
& \Rightarrow \quad \tan ^2 \alpha=2 / 3 \\\\
& \therefore \quad \cot ^2 \alpha=3 / 2 \\\\
& \text { Now, } 27 \sec ^6 \alpha+8 \operatorname{cosec}^6 \alpha=27\left(\sec ^2 \alpha\right)^3+8\left(\operatorname{cosec}^2 \alpha\right)^3 \\\\
& =27\left(1+\tan ^2 \alpha\right)^3+8\left(+\cot ^2 \alpha\right)^3 \\\\
& =27\left(1+\frac{2}{3}\right)^3+8\left(1+\frac{3}{2}\right)^3=250
\end{aligned}
$$ | mcq | jee-main-2021-online-18th-march-evening-shift |
lv5grwfx | maths | trigonometric-ratio-and-identites | fundamental-identities | <p>If $$\sin x=-\frac{3}{5}$$, where $$\pi< x <\frac{3 \pi}{2}$$, then $$80\left(\tan ^2 x-\cos x\right)$$ is equal to</p> | [{"identifier": "A", "content": "109"}, {"identifier": "B", "content": "108"}, {"identifier": "C", "content": "19"}, {"identifier": "D", "content": "18"}] | ["A"] | null | <p>$$\begin{aligned}
& \sin x=-\frac{3}{5} \text { where } \pi < x < \frac{3 \pi}{2} \\
& \qquad \tan x=\frac{3}{4}, \cos x=\frac{-4}{5} \\
& \therefore 80\left(\tan ^2 x-\cos x\right) \\
& =80\left(\frac{9}{16}+\frac{4}{5}\right) \\
& =80\left(\frac{45+64}{80}\right) \\
& =109
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
SeQofKYPFzNpqJwL | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | If $$5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9$$,
<br/><br/>then the value of $$\cos 4x$$ is : | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 9}$$"}, {"identifier": "C", "content": "$$ - {7 \\over 9}$$"}, {"identifier": "D", "content": "$$ - {3 \\over 5}$$"}] | ["C"] | null | Given that,
<br><br>$$5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9$$
<br><br>$$ \Rightarrow 5\left( {{{{{\sin }^2}x} \over {{{\cos }^2}x}} - {{\cos }^2}x} \right) = 2\left( {2{{\cos }^2}x - 1} \right) + 9$$
<br><br>Let $${\cos ^2}x = t,$$ then we have
<br><br>$$5\left( {{{1 - t} \over t} - t} \right) = 2\left( {2t - 1} \right) + 9$$
<br><br>$$ \Rightarrow 5\left( {{{1 - t - {t^2}} \over t}} \right) = 4t - 2 + 9$$
<br><br>$$ \Rightarrow 5 - 5t - 5{t^2} = 4{t^2} + 7t$$
<br><br>$$ \Rightarrow 9{t^2} + 12t - 5 = 0$$
<br><br>$$ \Rightarrow 9{t^2} + 15t - 3t - 5 = 0$$
<br><br>$$ \Rightarrow 3t\left( {3t + 5} \right) - 1\left( {3t + 5} \right) = 0$$
<br><br>$$ \Rightarrow \left( {3t + 5} \right)\left( {3t - 1} \right) = 0$$
<br><br>$$\therefore$$ $$t = {1 \over 3}$$ and $$t = - {5 \over 3}$$
<br><br>If $$t = - {5 \over 3}$$ then $${\cos ^2}x$$ is negative
<br><br>So , $$t$$ can not be $$ - {5 \over 3}$$.
<br><br>So, correct value of $$t = {1 \over 3}$$ then $$\cos {}^2x = t = {1 \over 3}$$
<br><br>$$\therefore\,\,\,$$ $$\cos 4x$$
<br><br>$$ = 2{\cos ^2}2x - 1$$
<br><br>$$ = 2{\left[ {2{{\cos }^2}x - 1} \right]^2} - 1$$
<br><br>$$ = 2.{\left[ {2.{1 \over 3} - 1} \right]^2} - 1$$
<br><br>$$ = 2.{\left( { - {1 \over 3}} \right)^2} - 1$$
<br><br>$$ = {2 \over 9} - 1$$
<br><br>$$ = - {7 \over 9}$$ | mcq | jee-main-2017-offline |
96UfPZHX8yI3nr1Mzp18hoxe66ijvwpjdul | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | The value of cos<sup>2</sup>10° – cos10°cos50° + cos<sup>2</sup>50° is | [{"identifier": "A", "content": "$${3 \\over 2} + \\cos {20^o}$$"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "$${3 \\over 2}(1 + \\cos {20^o})$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["B"] | null | cos<sup>2</sup>10° – cos10°cos50° + cos<sup>2</sup>50°
<br><br>= $${1 \over 2}$$[ 2cos<sup>2</sup>10° – 2cos10°cos50° + 2cos<sup>2</sup>50°]
<br><br>= $${1 \over 2}$$[ 1 + cos20° - cos60° - cos40° + 1 + cos100°]
<br><br>= $${1 \over 2}$$[ 2 - $${1 \over 2}$$ + cos20° + cos100° - cos40°]
<br><br>= $${1 \over 2}$$[ $${3 \over 2}$$ + 2cos60°cos40° - cos40°]
<br><br>= $${1 \over 2}$$[ $${3 \over 2}$$ + cos40° - cos40°]
<br><br>= $${3 \over 4}$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
e4bRKSVACJYj70SZqn3rsa0w2w9jx65v8fb | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | The equation y = sinx sin (x + 2) – sin<sup>2</sup>
(x + 1) represents a straight line lying in :
| [{"identifier": "A", "content": "first, second and fourth quadrants"}, {"identifier": "B", "content": "first, third and fourth quadrants"}, {"identifier": "C", "content": "second and third quadrants only"}, {"identifier": "D", "content": "third and fourth quadrants only "}] | ["D"] | null | y = sinx.sin(x+2) - sin<sup>2</sup>(x+1)<br><br>
$$ \Rightarrow {1 \over 2}\left\{ {2\sin \left( {x + 2} \right)\sin x - 2{{\sin }^2}(x + 1)} \right\}$$<br><br>
$$ \Rightarrow {1 \over 2}\left\{ {\cos 2 - \cos (2x + 2) + cos(2x + 2) - 1} \right\} = - {\sin ^2}1 < 0$$<br><br>
Hence the line passes through III and IV quadrant | mcq | jee-main-2019-online-12th-april-morning-slot |
rPyf3XUC2TZFVtTerz7k9k2k5hkb2j0 | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | If $${{\sqrt 2 \sin \alpha } \over {\sqrt {1 + \cos 2\alpha } }} = {1 \over 7}$$ and $$\sqrt {{{1 - \cos 2\beta } \over 2}} = {1 \over {\sqrt {10} }}$$<br/><br/>
$$\alpha ,\beta \in \left( {0,{\pi \over 2}} \right)$$ then tan($$\alpha $$ + 2$$\beta $$) is equal to
_____. | [] | null | 1 | $${{\sqrt 2 \sin \alpha } \over {\sqrt {1 + \cos 2\alpha } }} = {1 \over 7}$$
<br><br>$$ \Rightarrow $$ $${{\sqrt 2 \sin \alpha } \over {\sqrt {2{{\cos }^2}\alpha } }}$$ = $${1 \over 7}$$
<br><br>$$ \Rightarrow $$ $${{\sqrt 2 \sin \alpha } \over {\sqrt 2 \cos \alpha }}$$ = $${1 \over 7}$$
<br><br>$$ \Rightarrow $$ tan$$\alpha $$ = $${1 \over 7}$$
<br><br>Also given $$\sqrt {{{1 - \cos 2\beta } \over 2}} = {1 \over {\sqrt {10} }}$$
<br><br>$$ \Rightarrow $$ $${{\sqrt 2 \sin \beta } \over {\sqrt 2 }}$$ = $${1 \over {\sqrt {10} }}$$
<br><br>$$ \Rightarrow $$ sin $$\beta $$ = $${1 \over {\sqrt {10} }}$$
<br><br>$$ \therefore $$ tan $$\beta $$ = $${1 \over 3}$$
<br><br>$$\tan 2\beta = {{2\tan \beta } \over {1 - {{\tan }^2}\beta }}$$
<br><br>= $${{2\left( {{1 \over 3}} \right)} \over {1 - {1 \over 9}}}$$ = $${3 \over 4}$$
<br><br>$$ \therefore $$ tan($$\alpha $$ + 2$$\beta $$) = $${{\tan \alpha + \tan 2\beta } \over {1 - \tan \alpha .\tan 2\beta }}$$
<br><br>= $${{{1 \over 7} + {3 \over 4}} \over {1 - {1 \over 7}.{3 \over 4}}}$$
<br><br>= $${{25} \over {25}}$$ = 1 | integer | jee-main-2020-online-8th-january-evening-slot |
7j1f4iHiihhSRLgPEH7k9k2k5itrr6k | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | The value of<br/>
$${\cos ^3}\left( {{\pi \over 8}} \right)$$$${\cos}\left( {{3\pi \over 8}} \right)$$+$${\sin ^3}\left( {{\pi \over 8}} \right)$$$${\sin}\left( {{3\pi \over 8}} \right)$$<br/>
is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$${1 \\over 2{\\sqrt 2 }}$$"}] | ["D"] | null | $${\cos ^3}\left( {{\pi \over 8}} \right)$$$${\cos}\left( {{3\pi \over 8}} \right)$$+$${\sin ^3}\left( {{\pi \over 8}} \right)$$$${\sin}\left( {{3\pi \over 8}} \right)$$
<br><br>= $${\cos ^3}\left( {{\pi \over 8}} \right)\sin \left( {{\pi \over 8}} \right) + {\sin ^3}\left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)$$
<br><br>= $$\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)\left[ {{{\cos }^2}\left( {{\pi \over 8}} \right) + {{\sin }^2}\left( {{\pi \over 8}} \right)} \right]$$
<br><br>= $$\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)$$ $$ \times $$ 1
<br><br>= $${1 \over 2} \times 2\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)$$
<br><br>= $${1 \over 2}\sin \left( {{\pi \over 4}} \right)$$
<br><br>= $${1 \over {2\sqrt 2 }}$$ | mcq | jee-main-2020-online-9th-january-morning-slot |
FrL0CoCOTxGqqgxgm1jgy2xukez5prmu | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | If the equation cos<sup>4</sup> $$\theta $$ + sin<sup>4</sup> $$\theta $$ +
$$\lambda $$ = 0 has real
solutions for
$$\theta $$, then
$$\lambda $$ lies in the interval : | [{"identifier": "A", "content": "$$\\left[ { - {3 \\over 2}, - {5 \\over 4}} \\right]$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over 2}, - {1 \\over 4}} \\right]$$"}, {"identifier": "C", "content": "$$\\left( { - {5 \\over 4}, - 1} \\right]$$"}, {"identifier": "D", "content": "$$\\left[ { - 1, - {1 \\over 2}} \\right]$$"}] | ["D"] | null | cos<sup>4</sup> $$\theta $$ + sin<sup>4</sup> $$\theta $$ +
$$\lambda $$ = 0
<br><br>$$ \Rightarrow $$ 1 – 2sin<sup>2</sup> $$\theta $$ cos<sup>2</sup> $$\theta $$ = -$$\lambda $$
<br><br>$$ \Rightarrow $$ 1 - $$\frac{1}{2} \times 4$$sin<sup>2</sup> $$\theta $$ cos<sup>2</sup> $$\theta $$ = -$$\lambda $$
<br><br>$$ \Rightarrow $$ 1 - $$\frac{\sin^{2} 2\theta }{2} $$ = -$$\lambda $$
<br><br>$$ \Rightarrow $$2($$\lambda $$ + 1) = sin<sup>2</sup> 2$$\theta $$
<br><br> 0 $$ \le $$ 2 ($$\lambda $$ + 1) $$ \le $$ 1
<br><br> 0 $$ \le $$ ($$\lambda $$ + 1) $$ \le $$ $$\frac{1}{2} $$
<br><br>-1 $$ \le $$ $$\lambda $$ $$ \le $$ -$$\frac{1}{2} $$ | mcq | jee-main-2020-online-2nd-september-evening-slot |
ob7w7Pqmgg7gJ5rZurjgy2xukfqdnbzg | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | If L = sin<sup>2</sup>$$\left( {{\pi \over {16}}} \right)$$ - sin<sup>2</sup>$$\left( {{\pi \over {8}}} \right)$$ and
<br/>M = cos<sup>2</sup>$$\left( {{\pi \over {16}}} \right)$$ - sin<sup>2</sup>$$\left( {{\pi \over {8}}} \right)$$, then : | [{"identifier": "A", "content": "L = $$ - {1 \\over {2\\sqrt 2 }} + {1 \\over 2}\\cos {\\pi \\over 8}$$"}, {"identifier": "B", "content": "M = $${1 \\over {2\\sqrt 2 }} + {1 \\over 2}\\cos {\\pi \\over 8}$$"}, {"identifier": "C", "content": "M = $${1 \\over {4\\sqrt 2 }} + {1 \\over 4}\\cos {\\pi \\over 8}$$"}, {"identifier": "D", "content": "L = $${1 \\over {4\\sqrt 2 }} - {1 \\over 4}\\cos {\\pi \\over 8}$$"}] | ["B"] | null | We will use here those two formulas,
<br><br>sin<sup>2</sup> $$\theta $$ = $${{1 - \cos 2\theta } \over 2}$$ and cos<sup>2</sup> $$\theta $$ = $${{1 + \cos 2\theta } \over 2}$$
<br><br>L = sin<sup>2</sup>$$\left( {{\pi \over {16}}} \right)$$ - sin<sup>2</sup>$$\left( {{\pi \over {8}}} \right)$$
<br><br>$$ \Rightarrow $$ L = $$\left( {{{1 - \cos \left( {{\pi \over 8}} \right)} \over 2}} \right)$$ - $$\left( {{{1 - \cos \left( {{\pi \over 4}} \right)} \over 2}} \right)$$
<br><br>$$ \Rightarrow $$ L = $${1 \over 2}\left( {\cos \left( {{\pi \over 4}} \right) - \cos \left( {{\pi \over 8}} \right)} \right)$$
<br><br>$$ \Rightarrow $$ L = $${1 \over {2\sqrt 2 }} - {1 \over 2}\cos \left( {{\pi \over 8}} \right)$$
<br><br>M = cos<sup>2</sup>$$\left( {{\pi \over {16}}} \right)$$ - sin<sup>2</sup>$$\left( {{\pi \over {8}}} \right)$$
<br><br>$$ \Rightarrow $$ M = $$\left( {{{1 + \cos \left( {{\pi \over 8}} \right)} \over 2}} \right)$$ - $$\left( {{{1 - \cos \left( {{\pi \over 4}} \right)} \over 2}} \right)$$
<br><br>$$ \Rightarrow $$ M = $${1 \over {2\sqrt 2 }} + {1 \over 2}\cos {\pi \over 8}$$ | mcq | jee-main-2020-online-5th-september-evening-slot |
1ks082nke | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | If $$\sin \theta + \cos \theta = {1 \over 2}$$, then 16(sin(2$$\theta$$) + cos(4$$\theta$$) + sin(6$$\theta$$)) is equal to : | [{"identifier": "A", "content": "23"}, {"identifier": "B", "content": "$$-$$27"}, {"identifier": "C", "content": "$$-$$23"}, {"identifier": "D", "content": "27"}] | ["C"] | null | $$\sin \theta + \cos \theta = {1 \over 2}$$<br><br>$${\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = {1 \over 4}$$<br><br>$$\sin 2\theta = - {3 \over 4}$$<br><br>Now :<br><br>$$\cos 4\theta = 1 - 2{\sin ^2}2\theta $$<br><br>$$ = 1 - 2{\left( { - {3 \over 4}} \right)^2}$$<br><br>$$ = 1 - 2 \times {9 \over {16}} = - {1 \over 8}$$<br><br>$$\sin 6\theta = 3\sin 2\theta - 4{\sin ^3}2\theta $$<br><br>$$ = (3 - 4{\sin ^2}2\theta ).\sin 2\theta $$<br><br>$$ = \left[ {3 - 4\left( {{9 \over {16}}} \right)} \right].\left( { - {3 \over 4}} \right)$$<br><br>$$ \Rightarrow \left[ {{3 \over 4}} \right] \times \left( { - {3 \over 4}} \right) = - {9 \over {16}}$$<br><br>$$16[\sin 2\theta + \cos 4\theta + \sin 6\theta ]$$<br><br>= $$16\left( { - {3 \over 4} - {1 \over 8} - {9 \over {16}}} \right) = - 23$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ldsud15f | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | <p>Let $$f(\theta ) = 3\left( {{{\sin }^4}\left( {{{3\pi } \over 2} - \theta } \right) + {{\sin }^4}(3\pi + \theta )} \right) - 2(1 - {\sin ^2}2\theta )$$ and $$S = \left\{ {\theta \in [0,\pi ]:f'(\theta ) = - {{\sqrt 3 } \over 2}} \right\}$$. If $$4\beta = \sum\limits_{\theta \in S} \theta $$, then $$f(\beta )$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{9}{8}$$"}, {"identifier": "B", "content": "$$\\frac{3}{2}$$"}, {"identifier": "C", "content": "$$\\frac{5}{4}$$"}, {"identifier": "D", "content": "$$\\frac{11}{8}$$"}] | ["C"] | null | $f(\theta)=3\left(\sin ^{4}\left(\frac{3 \pi}{2}-\theta\right)+\sin ^{4}(3 x+\theta)\right)-2\left(1-\sin ^{2} 2 \theta\right)$<br/><br/> $S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\}$ <br/><br/>$\Rightarrow \mathrm{f}(\theta)=3\left(\cos ^{4} \theta+\sin ^{4} \theta\right)-2 \cos ^{2} 2 \theta$
<br/><br/>
$\Rightarrow \mathrm{f}(\theta)=3\left(1-\frac{1}{2} \sin ^{2} 2 \theta\right)-2 \cos ^{2} 2 \theta$
<br/><br/>
$\Rightarrow \mathrm{f}(\theta)=3-\frac{3}{2} \sin ^{2} 2 \theta-2 \cos ^{2} \theta$<br/><br/> $=\frac{3}{2}-\frac{1}{2} \cos ^{2} 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right)$
<br/><br/>
$f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4}$
<br/><br/>
$f^{\prime}(\theta)=\sin 4 \theta$
<br/><br/>
$\Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2}$
<br/><br/>
$\Rightarrow 4 \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{3}$
<br/><br/>
$\Rightarrow \theta=\frac{\mathrm{n} \pi}{4}+(-1)^{\mathrm{n}} \frac{\pi}{12}$
<br/><br/>
$$
\begin{aligned}
& \Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right) \\\\
& \Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2} \\\\
& \Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f(\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4}
\end{aligned}
$$ | mcq | jee-main-2023-online-29th-january-morning-shift |
lsblfc2j | maths | trigonometric-ratio-and-identites | multiple-and-sub-multiple-angle-formula | Let the set of all $a \in \mathbf{R}$ such that the equation $\cos 2 x+a \sin x=2 a-7$ has a solution be $[p, q]$ and $r=\tan 9^{\circ}-\tan 27^{\circ}-\frac{1}{\cot 63^{\circ}}+\tan 81^{\circ}$, then pqr is equal to ____________. | [] | null | 48 | <p>$$\begin{aligned}
& \cos 2 x+a \cdot \sin x=2 a-7 \\
& a(\sin x-2)=2(\sin x-2)(\sin x+2) \\
& \sin x=2, a=2(\sin x+2) \\
& \Rightarrow a \in[2,6] \\
& p=2 \quad q=6 \\
& r=\tan 9^{\circ}+\cot 9^{\circ}-\tan 27-\cot 27 \\
& r=\frac{1}{\sin 9 \cdot \cos 9}-\frac{1}{\sin 27 \cdot \cos 27} \\
& =2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}\right] \\
& r=4 \\
& p \cdot q \cdot r=2 \times 6 \times 4=48
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-morning-shift |
y9ojKGeV1MXuOiW6 | maths | trigonometric-ratio-and-identites | range-of-trigonometric-functions | If $$u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $$
<br/><br>then the difference between the maximum and minimum values of $${u^2}$$ is given by : </br> | [{"identifier": "A", "content": "$${\\left( {a - b} \\right)^2}$$ "}, {"identifier": "B", "content": "$$2\\sqrt {{a^2} + {b^2}} $$ "}, {"identifier": "C", "content": "$${\\left( {a + b} \\right)^2}$$"}, {"identifier": "D", "content": "$$2\\left( {{a^2} + {b^2}} \\right)$$ "}] | ["A"] | null | Given $$u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } $$$$+ \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $$
<br><br>$$\therefore$$ $${u^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta + {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta $$
<br> $$ + 2\sqrt {\left( {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } \right)} \times \sqrt {\left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right)} $$
<br><br>$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)} $$
<br><br>$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {1 - 2{{\cos }^2}\theta {{\sin }^2}\theta } \right)} $$
<br><br>$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {\left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}} $$
<br><br>$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\left( {2\cos \theta \sin \theta } \right)}^2}} \over 4} + {a^2}{b^2}} $$
<br><br>$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}} $$
<br><br>We know $$0 \le {\sin ^2}2\theta \le 1$$
<br><br>$$\therefore$$ $$0 \le {\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} \le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4}$$
<br><br>$$ \Rightarrow $$ $${a^2}{b^2} \le $$ $${\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}$$$$ \le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}$$
<br><br>$$\therefore$$ Min value of $${u^2} = {a^2} + {b^2}$$ $$ + 2\sqrt {{a^2}{b^2}} $$ = $${\left( {a + b} \right)^2}$$
<br><br>and Max value of $${u^2} = {a^2} + {b^2}$$ $$ + 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}} $$
<br><br>$$= {a^2} + {b^2}$$ $$ + 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2} + 4{a^2}{b^2}} \over 4}} $$
<br><br>$$= {a^2} + {b^2}$$ $$ + 2\sqrt {{{{{\left( {{a^2} + {b^2}} \right)}^2}} \over 4}} $$
<br><br>$$= {a^2} + {b^2}$$ $$+\, {a^2} + {b^2}$$
<br><br>$${ = 2\left( {{a^2} + {b^2}} \right)}$$
<br><br>Max of $${u^2}$$ - Min of $${u^2}$$ = $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {a + b} \right)^2}$$
<br><br>= $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {{a^2} + {b^2} + 2ab} \right)}$$
<br><br>= $$\sqrt {{{ = 2\left( {{a^2} + {b^2} - 2ab} \right)} \over 4}} $$
<br><br>= $${\left( {a - b} \right)^2}$$ | mcq | aieee-2004 |
CKkZX75qLg5aUvAv | maths | trigonometric-ratio-and-identites | range-of-trigonometric-functions | If $$A = {\sin ^2}x + {\cos ^4}x,$$ then for all real $$x$$: | [{"identifier": "A", "content": "$${{13} \\over {16}} \\le A \\le 1$$ "}, {"identifier": "B", "content": "$$1 \\le A \\le 2$$ "}, {"identifier": "C", "content": "$${3 \\over 4} \\le A \\le {{13} \\over {16}}$$ "}, {"identifier": "D", "content": "$${{3} \\over {4}} \\le A \\le 1$$"}] | ["D"] | null | $$A = {\sin ^2}x + {\cos ^4}x$$
<br><br>$$ = {\sin ^2}x + {\cos ^2}x\left( {1 - {{\sin }^2}x} \right)$$
<br><br>$$ = {\sin ^2}x + {\cos ^2}x - {1 \over 4}{\left( {2\sin x.\cos x} \right)^2}$$
<br><br>$$ = 1 - {1 \over 4}{\sin ^2}\left( {2x} \right)$$
<br><br>Now $$0 \le {\sin ^2}\left( {2x} \right) \le 1$$
<br><br>$$ \Rightarrow 0 \ge - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge - {1 \over 4}$$
<br><br>$$ \Rightarrow 1 \ge 1 - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge 1 - {1 \over 4}$$
<br><br>$$ \Rightarrow 1 \ge A \ge {3 \over 4}$$ | mcq | aieee-2011 |
vvonxCPLDXK1IQm4XaIMW | maths | trigonometric-ratio-and-identites | range-of-trigonometric-functions | If <i>m</i> and <i>M</i> are the minimum and the maximum values of
<br/><br/>4 + $${1 \over 2}$$ sin<sup>2</sup> 2x $$-$$ 2cos<sup>4</sup> x, x $$ \in $$ <b>R</b>, then <i>M</i> $$-$$ <i>m</i> is equal to : | [{"identifier": "A", "content": "$${{15} \\over 4}$$ "}, {"identifier": "B", "content": "$${{9} \\over 4}$$"}, {"identifier": "C", "content": "$${{7} \\over 4}$$"}, {"identifier": "D", "content": "$${{1} \\over 4}$$"}] | ["B"] | null | Given,
<br><br>4 + $${1 \over 2}$$ sin<sup>2</sup> 2x $$-$$ 2cos<sup>4</sup> x
<br><br>= 4 + $${1 \over 2}$$ (2sinx cosx)<sup>2</sup> $$-$$ 2cos<sup>4</sup>x
<br><br>= 4 + $${1 \over 2}$$ $$ \times $$ 4 sin<sup>2</sup>x cos<sup>2</sup>x $$-$$ 2cos<sup>4</sup> x
<br><br>= 4 + 2 (1 $$-$$ cos<sup>2</sup>x) cos<sup>2</sup>x $$-$$ 2cos<sup>4</sup> x
<br><br>= 4 + 2 cos<sup>2</sup>x $$-$$ 4cos<sup>4</sup>x
<br><br>= $$-$$ 4 $$\left\{ {\cos } \right.$$<sup>4</sup>x $$-$$ $$\left. {{{{{\cos }^2}x} \over 2} - 1} \right\}$$
<br><br>= $$-$$ 4 $$\left\{ {\cos } \right.$$<sup>4</sup>x $$-$$ 2 . $${1 \over 4}$$ . cos<sup>2</sup>x + $$\left. {{1 \over {16}} - {1 \over {16}} - 1} \right\}$$
<br><br>= $$-$$ 4 $$\left\{ {{{\left( {{{\cos }^2}x - {1 \over 4}} \right)}^2} - {{17} \over {16}}} \right\}$$
<br><br>We know,
<br><br>O $$ \le $$ cos<sup>2</sup>x $$ \le $$ 1
<br><br>$$ \Rightarrow $$ $$ - {1 \over 4}$$ $$ \le $$cos<sup>2</sup>x $$ - {1 \over 4}$$ $$ \le $$ $${3 \over 4}$$
<br><br>$$ \Rightarrow $$ O $$ \le $$ $${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$$ $$ \le $$ $${9 \over {16}}$$
<br><br>$$ \Rightarrow $$ $$-$$ $${17 \over {16}}$$ $$ \le $$ $${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$$ $$-$$ $${{17} \over {16}}$$ $$ \le $$ $${9 \over {16}}$$ $$-$$ $${{17} \over {16}}$$
<br><br>$$ \Rightarrow $$ $$-$$ $${{17} \over {16}}$$ $$ \le $$ $${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$$ $$-$$ $${{17} \over {16}}$$ $$ \le $$ $$-$$ $${{1} \over {2}}$$
<br><br>$$ \Rightarrow $$ $${{17} \over {4}}$$ $$ \ge $$ $$-$$ 4 $$\left\{ {{{\left( {{{\cos }^2}x - {1 \over 4}} \right)}^2} - {{17} \over {16}}} \right\} \ge 2$$
<br><br>$$ \therefore $$ Maximum value, M = $${{17} \over 4}$$
<br><br> Minimum value, m = $$2$$
<br><br>$$ \therefore $$ M $$-$$ m = $${{17} \over 4}$$ $$-$$ $$2$$ = $${{9} \over 4}$$ | mcq | jee-main-2016-online-9th-april-morning-slot |
nvMc3eovpat0DlEbPzFtJ | maths | trigonometric-ratio-and-identites | range-of-trigonometric-functions | The maximum value of 3cos$$\theta $$ + 5sin $$\left( {\theta - {\pi \over 6}} \right)$$ for any real value of $$\theta $$ is : | [{"identifier": "A", "content": "$$\\sqrt {34} $$ "}, {"identifier": "B", "content": "$$\\sqrt {31} $$"}, {"identifier": "C", "content": "$$\\sqrt {19} $$"}, {"identifier": "D", "content": "$${{\\sqrt {79} } \\over 2}$$"}] | ["C"] | null | y = 3cos$$\theta $$ + 5 $$\left( {\sin \theta {{\sqrt 3 } \over 2} - \cos \theta {1 \over 2}} \right)$$
<br><br>$${{5\sqrt 3 } \over 2}$$ sin$$\theta $$ + $${1 \over 2}$$cos$$\theta $$
<br><br>y<sub>max</sub> = $$\sqrt {{{75} \over 4} + {1 \over 4}} $$ = $$\sqrt {19} $$ | mcq | jee-main-2019-online-12th-january-morning-slot |
x224fLDpPMTj1APtsv1kluhwa2j | maths | trigonometric-ratio-and-identites | range-of-trigonometric-functions | The number of integral values of 'k' for which the equation $$3\sin x + 4\cos x = k + 1$$ has a solution, k$$\in$$R is ___________. | [] | null | 11 | We know,
<br><br>$$ - \sqrt {{a^2} + {b^2}} \le a\cos x + b\sin x \le \sqrt {{a^2} + {b^2}} $$<br><br>$$ \therefore $$ $$ - \sqrt {{3^2} + {4^2}} \le 3\cos x + 4\sin x \le \sqrt {{3^2} + {4^2}} $$<br><br>$$ - 5 \le k + 1 \le 5$$<br><br>$$ - 6 \le k \le 4$$<br><br>$$ \therefore $$ Set of integers = $$ - 6, - 5, - 4, - 3, - 2, - 1,0,1,2,3,4$$ = Total 11 intergers. | integer | jee-main-2021-online-26th-february-morning-slot |
1ldsf366t | maths | trigonometric-ratio-and-identites | range-of-trigonometric-functions | <p>The set of all values of $$\lambda$$ for which the equation $${\cos ^2}2x - 2{\sin ^4}x - 2{\cos ^2}x = \lambda $$ has a real solution $$x$$, is :</p> | [{"identifier": "A", "content": "$$\\left[ { - 2, - 1} \\right]$$"}, {"identifier": "B", "content": "$$\\left[ { - {3 \\over 2}, - 1} \\right]$$"}, {"identifier": "C", "content": "$$\\left[ { - 2, - {3 \\over 2}} \\right]$$"}, {"identifier": "D", "content": "$$\\left[ { - 1, - {1 \\over 2}} \\right]$$"}] | ["B"] | null | The given equation is
<br/><br/>$$\cos ^2 2x - 2 \sin ^4 x - 2 \cos ^2 x = \lambda$$
<br/><br/>Using the trigonometric identities $$\cos^2x = 1 - \sin^2x$$ and $$\cos^22x = 1 - 2\sin^2x$$, we can rewrite the equation in terms of $$\cos^2x$$:
<br/><br/>$$\lambda = (2 \cos ^2 x - 1)^2 - 2(1 - \cos ^2 x)^2 - 2 \cos ^2 x$$
<br/><br/>Simplify this equation :
<br/><br/>$$\lambda = 4 \cos ^4 x - 4 \cos ^2 x + 1 - 2(1 - 2 \cos ^2 x + \cos ^4 x) - 2 \cos ^2 x$$
<br/><br/>$$\lambda = 2 \cos ^4 x - 2 \cos ^2 x - 1$$
<br/><br/>We can factor out a 2 and rewrite this as :
<br/><br/>$$\lambda = 2(\cos ^4 x - \cos ^2 x - \frac{1}{2})$$
<br/><br/>$$\lambda = 2[(\cos ^2 x - \frac{1}{2})^2 - \frac{3}{4}]$$
<br/><br/>So $\lambda_{\max }=2\left[\frac{1}{4}-\frac{3}{4}\right]=2 \times\left(\frac{-2}{4}\right)=-1$ (maximum value) and
<br/><br/>$\lambda_{\min }=2\left[0-\frac{3}{4}\right]=-\frac{3}{2}($ minimum value $)$
<br/><br/>So range of the value of $x$ is $\left[\frac{-3}{2},-1\right]$. | mcq | jee-main-2023-online-29th-january-evening-shift |
1ktd1gy9c | maths | trigonometric-ratio-and-identites | reduction-formula | The value of <br/><br/>$$2\sin \left( {{\pi \over 8}} \right)\sin \left( {{{2\pi } \over 8}} \right)\sin \left( {{{3\pi } \over 8}} \right)\sin \left( {{{5\pi } \over 8}} \right)\sin \left( {{{6\pi } \over 8}} \right)\sin \left( {{{7\pi } \over 8}} \right)$$ is : | [{"identifier": "A", "content": "$${1 \\over {4\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${1 \\over {8\\sqrt 2 }}$$"}] | ["C"] | null | $$2\sin \left( {{\pi \over 8}} \right)\sin \left( {{{2\pi } \over 8}} \right)\sin \left( {{{3\pi } \over 8}} \right)\sin \left( {{{5\pi } \over 8}} \right)\sin \left( {{{6\pi } \over 8}} \right)\sin \left( {{{7\pi } \over 8}} \right)$$<br><br>$$2{\sin ^2}{\pi \over 8}{\sin ^2}{{2\pi } \over 8}{\sin ^2}{{3\pi } \over 8}$$<br><br>$${\sin ^2}{\pi \over 8}{\sin ^2}{{3\pi } \over 8}$$<br><br>$${\sin ^2}{\pi \over 8}{\cos ^2}{\pi \over 8}$$<br><br>$${1 \over 4}{\sin ^2}\left( {{\pi \over 4}} \right) = {1 \over 8}$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
2IgmuZ8GACutFW67uv7k9k2k5kh6qs3 | maths | trigonometric-ratio-and-identites | summation-series-problems | If $$x = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta } $$ and $$y = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } $$<br/><br/> for
0 < $$\theta $$ < $${\pi \over 4}$$, then : | [{"identifier": "A", "content": "x(1 + y) = 1"}, {"identifier": "B", "content": "y(1 \u2013 x) = 1"}, {"identifier": "C", "content": "y(1 + x) = 1"}, {"identifier": "D", "content": "x(1 \u2013 y) = 1"}] | ["B"] | null | $$x = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta } $$
<br><br>= 1 – tan<sup>2</sup>$$\theta $$ + tan<sup>2</sup> 4$$\theta $$ + ...
<br><br>= $${1 \over {1 + {{\tan }^2}\theta }}$$ = cos<sup>2</sup> $$\theta $$ ....(1)
<br><br>$$y = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } $$
<br><br>= 1 + cos<sup>2</sup>
$$\theta $$ + cos<sup>4</sup>
$$\theta $$ + cos<sup>6</sup>
$$\theta $$ + ....
<br><br>= $${1 \over {1 - {{\cos }^2}\theta }}$$ = $${1 \over {{{\sin }^2}\theta }}$$
<br><br>$$ \Rightarrow $$ sin<sup>2</sup> $$\theta $$ = $${1 \over y}$$ ...(2)
<br><br>Adding (1) and (2), we get,
<br><br>x + $${1 \over y}$$ = sin<sup>2</sup> $$\theta $$ + cos<sup>2</sup> $$\theta $$
<br><br>$$ \Rightarrow $$ x + $${1 \over y}$$ = 1
<br><br>$$ \Rightarrow $$ y(1 – x) = 1 | mcq | jee-main-2020-online-9th-january-evening-slot |
TmaIWZqCcYAXnrHmjJ1klrh57ue | maths | trigonometric-ratio-and-identites | summation-series-problems | If
$${e^{\left( {{{\cos }^2}x + {{\cos }^4}x + {{\cos }^6}x + ...\infty } \right){{\log }_e}2}}$$
satisfies the equation t<sup>2</sup> - 9t + 8 = 0, then the value of
<br/>$${{2\sin x} \over {\sin x + \sqrt 3 \cos x}}\left( {0 < x < {\pi \over 2}} \right)$$ is : | [{"identifier": "A", "content": "$$\\sqrt 3 $$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "2$$\\sqrt 3 $$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["D"] | null | $${e^{({{\cos }^2}x + {{\cos }^4}x + ...........\infty )\ln 2}} = {2^{{{\cos }^2}x + {{\cos }^4}x + ...........\infty }}$$
<br><br>= $${2^{{{{{\cos }^2}x} \over {1 - {{\cos }^2}x}}}}$$
<br><br>$$ = {2^{{{\cot }^2}x}}$$<br><br>Given, $${t^2} - 9t + 8 = 0 \Rightarrow t = 1,8$$<br><br>$$ \Rightarrow {2^{{{\cot }^2}x}} = 1,8 \Rightarrow co{t^2}x = 0,3$$<br><br>$$0 < x < {\pi \over 2} \Rightarrow \cot x = \sqrt 3 $$<br><br>$$ \therefore $$ $$ {{2\sin x} \over {\sin x + \sqrt 3 \cos x}} = {2 \over {1 + \sqrt 3 \cot x}} = {2 \over 4} = {1 \over 2}$$ | mcq | jee-main-2021-online-24th-february-morning-slot |
G4svy42K3MikqAuXaU18hoxe66ijvwuh9my | maths | trigonometric-ratio-and-identites | transformation-formula | The value of sin 10º sin30º sin50º sin70º is :- | [{"identifier": "A", "content": "$${1 \\over {36}}$$"}, {"identifier": "B", "content": "$${1 \\over {16}}$$"}, {"identifier": "C", "content": "$${1 \\over {32}}$$"}, {"identifier": "D", "content": "$${1 \\over {18}}$$"}] | ["B"] | null | sin 10º sin30º sin50º sin70º
<br><br>= sin30º sin50º sin 10º sin70º
<br><br>= $${1 \over 2}$$ [ sin50º sin 10º sin70º ]
<br><br>= $${1 \over 2}$$ [ sin(60º - 10º) sin 10º sin(60º + 10º) ]
<br><br>= $${1 \over 2}$$ [ $${1 \over 4}\sin $$3(10º) ]
<br><br>= $${1 \over 2}$$ [ $${1 \over 4} \times {1 \over 2}$$]
<br><br>= $${1 \over {16}}$$
<br><br><b>Note :</b>
<br><br> $$\sin \left( {60^\circ - A} \right)\sin A\sin \left( {60^\circ - A} \right) = {1 \over 4}\sin 3A$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
1l58aokp9 | maths | trigonometric-ratio-and-identites | transformation-formula | <p>If $${\sin ^2}(10^\circ )\sin (20^\circ )\sin (40^\circ )\sin (50^\circ )\sin (70^\circ ) = \alpha - {1 \over {16}}\sin (10^\circ )$$, then $$16 + {\alpha ^{ - 1}}$$ is equal to __________.</p> | [] | null | 80 | <p>$$(\sin 10^\circ \,.\,\sin 50^\circ \,.\,\sin 70^\circ )\,.\,(\sin 10^\circ \,.\,\sin 20^\circ \,.\,\sin 40^\circ )$$</p>
<p>$$ = \left( {{1 \over 4}\sin 30^\circ } \right)\,.\,\left[ {{1 \over 2}\sin 10^\circ (\cos 20^\circ - \cos 60^\circ )} \right]$$</p>
<p>$$ = {1 \over {16}}\left[ {\sin 10^\circ \left( {\cos 20^\circ - {1 \over 2}} \right)} \right]$$</p>
<p>$$ = {1 \over {32}}[2\sin 10^\circ \,.\,\cos 20^\circ - \sin 10^\circ ]$$</p>
<p>$$ = {1 \over {32}}[\sin 30^\circ - \sin 10^\circ - \sin 10^\circ ]$$</p>
<p>$$ = {1 \over {64}} - {1 \over {64}}\sin 10^\circ $$</p>
<p>Clearly, $$\alpha = {1 \over {64}}$$</p>
<p>Hence $$16 + {\alpha ^{ - 1}} = 80$$</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l58glche | maths | trigonometric-ratio-and-identites | transformation-formula | <p>$$16\sin (20^\circ )\sin (40^\circ )\sin (80^\circ )$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\sqrt 3 $$"}, {"identifier": "B", "content": "2$$\\sqrt 3 $$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4$$\\sqrt 3 $$"}] | ["B"] | null | <p>$$16\sin 20^\circ \,.\,\sin 40^\circ \,.\,\sin 80^\circ $$</p>
<p>$$ = 4\sin 60^\circ $$ {$$\because$$ $$4\sin \theta \,.\,\sin (60^\circ - \theta )\,.\,\sin (60^\circ + \theta ) = \sin 3\theta $$}</p>
<p>$$ = 2\sqrt 3 $$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l6f2znok | maths | trigonometric-ratio-and-identites | transformation-formula | <p>$$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{3}{16}$$"}, {"identifier": "B", "content": "$$\\frac{1}{16}$$"}, {"identifier": "C", "content": "$$\\frac{1}{32}$$"}, {"identifier": "D", "content": "$$\\frac{9}{32}$$"}] | ["B"] | null | <p>$$2\sin {\pi \over {22}}\sin {{3\pi } \over {22}}\sin {{5\pi } \over {22}}\sin {{7\pi } \over {22}}\sin {{9\pi } \over {22}}$$</p>
<p>$$ = 2\sin \left( {{{11\pi - 10\pi } \over {22}}} \right)\sin \left( {{{11\pi - 8\pi } \over {22}}} \right)\sin \left( {{{11\pi - 6\pi } \over {22}}} \right)\sin \left( {{{11\pi - 4\pi } \over {22}}} \right)\sin \left( {{{11\pi - 2\pi } \over {22}}} \right)$$</p>
<p>$$ = 2\cos {\pi \over {11}}\cos {{2\pi } \over {11}}\cos {{3\pi } \over {11}}\cos {{4\pi } \over {11}}\cos {{5\pi } \over {11}}$$</p>
<p>$$ = {{2\sin {{32\pi } \over {11}}} \over {{2^5}\sin {\pi \over {11}}}}$$</p>
<p>$$ = {1 \over {16}}$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1lgxtayey | maths | trigonometric-ratio-and-identites | transformation-formula | <p>$$96\cos {\pi \over {33}}\cos {{2\pi } \over {33}}\cos {{4\pi } \over {33}}\cos {{8\pi } \over {33}}\cos {{16\pi } \over {33}}$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Let
<br/><br/>$$
\begin{aligned}
& A=96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33} \\\\
& \Rightarrow 2 A=96 \times 2\left(\cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\
& \Rightarrow 2 A \times \sin \frac{\pi}{33} =96 \times\left(2 \sin \frac{\pi}{33} \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cdot \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\
& \Rightarrow 2 A \times \sin \frac{\pi}{33}=6 \times \sin \frac{32 \pi}{33}=6 \times \sin \frac{\pi}{33} \\\\
& \Rightarrow 2 A=6 \Rightarrow A=3
\end{aligned}
$$
<br/><br/>Thus, the required answer is 3 . | mcq | jee-main-2023-online-10th-april-morning-shift |
lv7v3v7b | maths | trigonometric-ratio-and-identites | transformation-formula | <p>Suppose $$\theta \in\left[0, \frac{\pi}{4}\right]$$ is a solution of $$4 \cos \theta-3 \sin \theta=1$$. Then $$\cos \theta$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{6-\\sqrt{6}}{(3 \\sqrt{6}-2)}$$\n"}, {"identifier": "B", "content": "$$\\frac{4}{(3 \\sqrt{6}+2)}$$\n"}, {"identifier": "C", "content": "$$\\frac{6+\\sqrt{6}}{(3 \\sqrt{6}+2)}$$\n"}, {"identifier": "D", "content": "$$\\frac{4}{(3 \\sqrt{6}-2)}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& 4 \cos \theta-3 \sin \theta=1 \\
& 4 \cos \theta-1=3 \sin \theta \\
& 16 \cos ^2 \theta+1-8 \cos \theta=9\left(1-\cos ^2 \theta\right) \\
& \Rightarrow 25 \cos ^2 \theta-8 \cos \theta-8=0 \\
& \Rightarrow \cos \theta=\frac{8 \pm \sqrt{64+4 \times 25 \times 8}}{2.25} \\
& =\frac{8 \pm 4 \sqrt{4+50}}{2.25}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& =\frac{4 \pm 2 \sqrt{54}}{25} \\
& \text { As } \theta \in\left[0, \frac{\pi}{4}\right] \\
& \Rightarrow \cos \theta=\frac{4+6 \sqrt{6}}{25}=\frac{4}{3 \sqrt{6}-2}
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
1krzm14od | maths | trigonometric-ratio-and-identites | trigonometric-ratio-of-standard-angles | The value of $$\cot {\pi \over {24}}$$ is : | [{"identifier": "A", "content": "$$\\sqrt 2 + \\sqrt 3 + 2 - \\sqrt 6 $$"}, {"identifier": "B", "content": "$$\\sqrt 2 + \\sqrt 3 + 2 + \\sqrt 6 $$"}, {"identifier": "C", "content": "$$\\sqrt 2 - \\sqrt 3 - 2 + \\sqrt 6 $$"}, {"identifier": "D", "content": "$$3\\sqrt 2 - \\sqrt 3 - \\sqrt 6 $$"}] | ["B"] | null | $$\cot \theta = {{1 + \cos 2\theta } \over {\sin 2\theta }} = {{1 + \left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right)} \over {\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right)}}$$<br><br>$$\theta = {\pi \over {24}}$$<br><br>$$ \Rightarrow \cot \left( {{\pi \over {24}}} \right) = {{1 + \left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right)} \over {\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right)}}$$<br><br>$$ = {{\left( {2\sqrt 2 + \sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 - 1} \right)}} \times {{\left( {\sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 + 1} \right)}}$$<br><br>$$ = {{2\sqrt 6 + 2\sqrt 2 + 3 + \sqrt 3 + \sqrt 3 + 1} \over 2}$$<br><br>$$ = \sqrt 6 + \sqrt 2 + \sqrt 3 + 2$$ | mcq | jee-main-2021-online-25th-july-evening-shift |
1l56rn0ah | maths | trigonometric-ratio-and-identites | trigonometric-ratio-of-standard-angles | <p>$$\alpha = \sin 36^\circ $$ is a root of which of the following equation?</p> | [{"identifier": "A", "content": "$$16{x^4} - 10{x^2} - 5 = 0$$"}, {"identifier": "B", "content": "$$16{x^4} + 20{x^2} - 5 = 0$$"}, {"identifier": "C", "content": "$$16{x^4} - 20{x^2} + 5 = 0$$"}, {"identifier": "D", "content": "$$4{x^4} - 10{x^2} + 5 = 0$$"}] | ["C"] | null | <p>Given that $\alpha = \sin 36^\circ$, we need to determine which equation it is a root of.</p>
<p>We start with the known relationship for $\cos 72^\circ$:</p>
<p>$ \cos 72^\circ = \frac{\sqrt{5}-1}{4} $</p>
<p>Using the double-angle formula for cosine:</p>
<p>$ \cos 72^\circ = 1 - 2 \sin^2 36^\circ $</p>
<p>Substitute $\alpha$ for $\sin 36^\circ$:</p>
<p>$ 1 - 2\alpha^2 = \frac{\sqrt{5}-1}{4} $</p>
<p>Multiply both sides by 4:</p>
<p>$ 4 - 8\alpha^2 = \sqrt{5} - 1 $</p>
<p>Add 1 to both sides:</p>
<p>$ 5 - 8\alpha^2 = \sqrt{5} $</p>
<p>Square both sides to eliminate the radical:</p>
<p>$ (5 - 8\alpha^2)^2 = 5 $</p>
<p>Expand the left side:</p>
<p>$ 25 + 64\alpha^4 - 80\alpha^2 = 5 $</p>
<p>Simplify by subtracting 5 from both sides:</p>
<p>$ 64\alpha^4 - 80\alpha^2 + 20 = 0 $</p>
<p>Divide the entire equation by 4:</p>
<p>$ 16\alpha^4 - 20\alpha^2 + 5 = 0 $</p>
<p>Thus, the equation $16\alpha^4 - 20\alpha^2 + 5 = 0$ is the one for which $\alpha = \sin 36^\circ$ is a root.</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l59komxs | maths | trigonometric-ratio-and-identites | trigonometric-ratio-of-standard-angles | <p>The value of 2sin (12$$^\circ$$) $$-$$ sin (72$$^\circ$$) is :</p> | [{"identifier": "A", "content": "$${{\\sqrt 5 (1 - \\sqrt 3 )} \\over 4}$$"}, {"identifier": "B", "content": "$${{1 - \\sqrt 5 } \\over 8}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 (1 - \\sqrt 5 )} \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 (1 - \\sqrt 5 )} \\over 4}$$"}] | ["D"] | null | <p>$$2\sin 12^\circ - \sin 72^\circ $$</p>
<p>$$ = \sin 12^\circ + ( - 2\cos 42^\circ \,.\,\sin 30^\circ )$$</p>
<p>$$ = \sin 12^\circ - \cos 42^\circ $$</p>
<p>$$ = \sin 12^\circ - \sin 48^\circ $$</p>
<p>$$ = 2\sin 18^\circ \,.\,\cos 30^\circ $$</p>
<p>$$ = - 2\left( {{{\sqrt 5 - 1} \over 4}} \right)\,.\,{{\sqrt 3 } \over 2}$$</p>
<p>$$ = {{\sqrt 3 \left( {1 - \sqrt 5 } \right)} \over 4}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1ldr7eiav | maths | trigonometric-ratio-and-identites | trigonometric-ratio-of-standard-angles | <p>If $$\tan 15^\circ + {1 \over {\tan 75^\circ }} + {1 \over {\tan 105^\circ }} + \tan 195^\circ = 2a$$, then the value of $$\left( {a + {1 \over a}} \right)$$ is :</p> | [{"identifier": "A", "content": "$$5 - {3 \\over 2}\\sqrt 3 $$"}, {"identifier": "B", "content": "$$4 - 2\\sqrt 3 $$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <p>$$\tan 15^\circ + \tan 15^\circ - \tan 15^\circ + \tan 15^\circ $$</p>
<p>$$ = 2\tan 15^\circ $$</p>
<p>$$ = 2\left( {2 - \sqrt 3 } \right) = 2a \Rightarrow a = 2 - \sqrt 3 $$</p>
<p>$$\therefore$$ $${1 \over a} + a \Rightarrow \left( {2 + \sqrt 3 } \right) + \left( {2 - \sqrt 3 } \right) = 4$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1lgylyp94 | maths | trigonometric-ratio-and-identites | trigonometric-ratio-of-standard-angles | <p>The value of $$36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\right)\left(4 \cos ^{2} 81^{\circ}-1\right)\left(4 \cos ^{2} 243^{\circ}-1\right)$$ is :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "54"}, {"identifier": "D", "content": "27"}] | ["B"] | null | $$
\begin{aligned}
& 4 \cos ^2 \theta-1=4\left(1-\sin ^2 \theta\right)-1 \\\\
& =3-4 \sin ^2 \theta \\\\
& =\frac{3 \sin \theta-4 \sin ^3 \theta}{\sin \theta} \\\\
& =\frac{\sin 3 \theta}{\sin \theta}
\end{aligned}
$$
<br/><br/>$$36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\right)\left(4 \cos ^{2} 81^{\circ}-1\right)\left(4 \cos ^{2} 243^{\circ}-1\right)$$
<br/><br/>$$
\begin{aligned}
& =36\left[\frac{\sin 27^{\circ}}{\sin 9^{\circ}} \times \frac{\sin 81^{\circ}}{\sin 27^{\circ}} \times \frac{\sin 243^{\circ}}{\sin 81^{\circ}} \times \frac{\sin 729^{\circ}}{\sin 243^{\circ}}\right] \\\\
& =36\left[\frac{\sin 729^{\circ}}{\sin 9^{\circ}}\right]=36 \times 1=36
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh2ythlt | maths | trigonometric-ratio-and-identites | trigonometric-ratio-of-standard-angles | <p>The value of $$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$$ is __________.</p> | [] | null | 4 | $\begin{aligned} & \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} \\\\ & =\tan 9^{\circ}+\tan \left(90^{\circ}-9^{\circ}\right)-\tan 27^{\circ}-\tan \left(90^{\circ}-27^{\circ}\right) \\\\ & =\tan 9^{\circ}+\cot 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ} \\\\ & =\frac{\sin 9^{\circ}}{\cos 9^{\circ}}+\frac{\cos 9^{\circ}}{\sin 9^{\circ}}-\left(\frac{\sin 27^{\circ}}{\cos 27^{\circ}}+\frac{\cos 27^{\circ}}{\sin 27^{\circ}}\right) \\\\ & =\frac{\sin ^2 9^{\circ}+\cos ^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}-\left(\frac{\sin ^2 27^{\circ}+\cos ^2 27^{\circ}}{\cos 27^{\circ} \sin 27^{\circ}}\right) \\\\ & =\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}} \\\\ & =\frac{2 \times 4}{\sqrt{5}-1}-\frac{2 \times 4}{\sqrt{5}+1} \\\\ & =8\left(\frac{\sqrt{5}+1-\sqrt{5}+1}{5-1}\right)=2(2)=4\end{aligned}$ | integer | jee-main-2023-online-6th-april-evening-shift |
lv3ve3z9 | maths | trigonometric-ratio-and-identites | trigonometric-ratio-of-standard-angles | <p>If the value of $$\frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}}$$ is $$\frac{a \sqrt{5}-b}{c}$$, where $$a, b, c$$ are natural numbers and $$\operatorname{gcd}(a, c)=1$$, then $$a+b+c$$ is equal to :</p> | [{"identifier": "A", "content": "54"}, {"identifier": "B", "content": "52"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "40"}] | ["B"] | null | <p>To find the value of $$\frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}}$$ in the form $$\frac{a \sqrt{5}-b}{c}$$, we need to simplify the given expression. Let's start by using some fundamental trigonometric identities.</p>
<p>We know that:</p>
<p>$$\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$$</p>
<p>$$\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$$</p>
<p>First, substitute these values into the expression:</p>
<p>$$\frac{3 \cdot \frac{\sqrt{5} + 1}{4} + 5 \cdot \frac{\sqrt{5} - 1}{4}}{5 \cdot \frac{\sqrt{5} + 1}{4} - 3 \cdot \frac{\sqrt{5} - 1}{4}}$$</p>
<p>Simplify the numerator and the denominator:</p>
<p>Numerator: $$3 \cdot \frac{\sqrt{5} + 1}{4} + 5 \cdot \frac{\sqrt{5} - 1}{4} = \frac{3(\sqrt{5} + 1) + 5(\sqrt{5} - 1)}{4} = \frac{3\sqrt{5} + 3 + 5\sqrt{5} - 5}{4} = \frac{8\sqrt{5} - 2}{4} = 2 \sqrt{5} - \frac{1}{2}$$</p>
<p>Denominator: $$5 \cdot \frac{\sqrt{5} + 1}{4} - 3 \cdot \frac{\sqrt{5} - 1}{4} = \frac{5(\sqrt{5} + 1) - 3(\sqrt{5} - 1)}{4} = \frac{5\sqrt{5} + 5 - 3\sqrt{5} + 3}{4} = \frac{2\sqrt{5} + 8}{4} = \frac{\sqrt{5}+4}{2}$$</p>
<p>Now combine the simplified numerator and denominator:</p>
<p>$$\frac{2 \sqrt{5} - \frac{1}{2}}{\frac{\sqrt{5}+4}{2}} = \frac{(2 \sqrt{5} - \frac{1}{2}) \cdot 2}{\sqrt{5}+4} = \frac{4 \sqrt{5} - 1}{\sqrt{5}+4}$$</p>
<p>Rationalizing the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator:</p>
<p>$$\frac{(4 \sqrt{5} - 1)(\sqrt{5}-4)}{(\sqrt{5}+4)(\sqrt{5}-4)}$$</p>
<p>The denominator simplifies to:</p>
<p>$$ \sqrt{5}^2 - 4^2 = 5 - 16 = -11$$</p>
<p>The numerator simplifies to:</p>
<p>$$ (4 \sqrt{5} - 1)(\sqrt{5}-4) = (4 \sqrt{5} \cdot \sqrt{5} - 4 \cdot 4 \sqrt{5} - 1 \cdot \sqrt{5} + 1 \cdot 4) = (20 - 16 \sqrt{5} - \sqrt{5} + 4) = 24 - 17 \sqrt{5}$$</p>
<p>Combining them, we get:</p>
<p>$$\frac{24 - 17\sqrt{5}}{-11} = \frac{17\sqrt{5}-24}{11}$$</p>
<p>Thus, $$a = 17, b = 24, c = 11$$.</p>
<p>Therefore, $$a + b + c = 17 + 24 + 11 = 52$$.</p>
<p>So the correct option is:</p>
<p><strong>Option B: 52</strong></p> | mcq | jee-main-2024-online-8th-april-evening-shift |
G7kjRdgRTJ4I6DyRbBaKI | maths | trigonometric-ratio-and-identites | trigonometric-series | The value of $$\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$$ is - | [{"identifier": "A", "content": "$${1 \\over {256}}$$"}, {"identifier": "B", "content": "$${1 \\over {2}}$$"}, {"identifier": "C", "content": "$${1 \\over {1024}}$$"}, {"identifier": "D", "content": "$${1 \\over {512}}$$"}] | ["D"] | null | Given $$\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$$
<br><br>Let $${\pi \over {{2^{10}}}}\, = \,\theta $$
<br><br>$$ \therefore $$ $${\pi \over {{2^9}}}\, = \,2\theta $$
<br><br> $${\pi \over {{2^8}}}\, = \,{2^2}\theta $$
<br><br> $${\pi \over {{2^7}}}\, = \,{2^3}\theta $$
<br>.
<br>.
<br><br>$${\pi \over {{2^2}}}\, = \,{2^8}\theta $$
<br><br>So given term becomes,
<br><br>$$\cos {2^8}\theta .\cos {2^7}\theta .....\cos \theta $$$$.\sin {\pi \over {{2^{10}}}}$$
<br><br>= $$(\cos \theta .\cos 2\theta ......\cos {2^8}\theta )\sin {\pi \over {{2^{10}}}}$$
<br><br>= $${{\sin {2^9}\theta } \over {{2^9}\sin \theta }}.\sin {\pi \over {{2^{10}}}}$$
<br><br>= $${{\sin {2^9}\left( {{\pi \over {{2^{10}}}}} \right)} \over {{2^9}\sin {\pi \over {{2^{10}}}}}}.\sin {\pi \over {{2^{10}}}}$$
<br><br>= $${{\sin \left( {{\pi \over 2}} \right)} \over {{2^9}}}$$
<br><br>= $${1 \over {{2^9}}}$$ = $${1 \over {512}}$$
<br><br><b>Note :</b>
<br><br/>$$(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )$$ = $${{\sin {2^n}\theta } \over {{2^n}\sin \theta }}$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
1l57oq18q | maths | trigonometric-ratio-and-identites | trigonometric-series | <p>The value of $$\cos \left( {{{2\pi } \over 7}} \right) + \cos \left( {{{4\pi } \over 7}} \right) + \cos \left( {{{6\pi } \over 7}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "$$-$$$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$-$$$${1 \\over 3}$$"}, {"identifier": "D", "content": "$$-$$$${1 \\over 4}$$"}] | ["B"] | null | <p>$$\cos {{2\pi } \over 7} + \cos {{4\pi } \over 7} + \cos {{6\pi } \over 7} = {{\sin 3\left( {{\pi \over 7}} \right)} \over {\sin {\pi \over 7}}}\cos {{\left( {{{2\pi } \over 7} + {{6\pi } \over 7}} \right)} \over 2}$$</p>
<p>$$ = {{\sin \left( {{{3\pi } \over 7}} \right)\,.\,\cos \left( {{{4\pi } \over 7}} \right)} \over {\sin \left( {{\pi \over 7}} \right)}}$$</p>
<p>$$ = {{2\sin {{4\pi } \over 7}\cos {{4\pi } \over 7}} \over {2\sin {\pi \over 7}}}$$</p>
<p>$$ = {{\sin \left( {{{8\pi } \over 7}} \right)} \over {2\sin {\pi \over 7}}} = {{ - \sin {\pi \over 7}} \over {2\sin {\pi \over 7}}} = {{ - 1} \over 2}$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
ksVbg1zWUG2kWR6p | maths | vector-algebra | algebra-and-modulus-of-vectors | If $$\left| {\matrix{
a & {{a^2}} & {1 + {a^3}} \cr
b & {{b^2}} & {1 + {b^3}} \cr
c & {{c^2}} & {1 + {c^3}} \cr
} } \right| = 0$$ and vectors $$\left( {1,a,{a^2}} \right),\,\,$$
<br/><br>$$\left( {1,b,{b^2}} \right)$$ and $$\left( {1,c,{c^2}} \right)\,$$ are non-coplanar, then the product $$abc$$ equals :</br> | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$2$$ "}, {"identifier": "C", "content": "$$-1$$ "}, {"identifier": "D", "content": "$$1$$"}] | ["C"] | null | $$\left| {\matrix{
a & {{a^2}} & {1 + {a^3}} \cr
b & {{b^2}} & {1 + {b^3}} \cr
c & {{c^2}} & {1 + {c^3}} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \left| {\matrix{
a & {{a^2}} & 1 \cr
b & {{b^2}} & 1 \cr
c & {{c^2}} & 1 \cr
} } \right| + \left| {\matrix{
a & {{a^2}} & {{a^3}} \cr
b & {{b^2}} & {{b^3}} \cr
c & {{c^2}} & {{c^3}} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \left( {1 + abc} \right)\left| {\matrix{
1 & a & {{a^2}} \cr
1 & b & {{b^2}} \cr
1 & c & {{c^2}} \cr
} } \right| = 0$$
<br><br>As $$\,\,\,\left| {\matrix{
1 & a & {{a^2}} \cr
1 & b & {{b^2}} \cr
1 & c & {{c^2}} \cr
} } \right| \ne 0$$ (given condition)
<br><br>$$\therefore$$ $$abc=-1$$ | mcq | aieee-2003 |
ZBaAeUoELX0dxgSK | maths | vector-algebra | algebra-and-modulus-of-vectors | Consider points $$A, B, C$$ and $$D$$ with position
<br/><br>vectors $$7\widehat i - 4\widehat j + 7\widehat k,\widehat i - 6\widehat j + 10\widehat k, - \widehat i - 3\widehat j + 4\widehat k$$ and $$5\widehat i - \widehat j + 5\widehat k$$ respectively. Then $$ABCD$$ is a :</br> | [{"identifier": "A", "content": "parallelogram but not a rhombus "}, {"identifier": "B", "content": "square "}, {"identifier": "C", "content": "rhombus "}, {"identifier": "D", "content": "None"}] | ["D"] | null | $$A = \left( {7, - 4,7} \right),B = \left( {1, - 6,10} \right),$$
<br><br>$$C = \left( { - 1, - 3,4} \right)$$ and $$D = \left( {5, - 1,5} \right)$$
<br><br>$$AB = \sqrt {{{\left( {7 - 1} \right)}^2} + {{\left( { - 4 + 6} \right)}^2} + {{\left( {7 - 10} \right)}^2}} $$
<br><br>$$ = \sqrt {36 + 4 + 9} = 7$$
<br><br>Similarly $$BC = 7,\,\,CD = \sqrt {41} ,\,\,DA = \sqrt {17} $$
<br><br>$$\therefore$$ None of the options is satisfied | mcq | aieee-2003 |
nsIbnkrvsOGyXaLY | maths | vector-algebra | algebra-and-modulus-of-vectors | The vectors $$\overrightarrow {AB} = 3\widehat i + 4\widehat k\,\,\& \,\,\overrightarrow {AC} = 5\widehat i - 2\widehat j + 4\widehat k$$ are the sides of triangle $$ABC.$$ The length of the median through $$A$$ is : | [{"identifier": "A", "content": "$$\\sqrt {288} $$ "}, {"identifier": "B", "content": "$$\\sqrt {18} $$"}, {"identifier": "C", "content": "$$\\sqrt {72} $$"}, {"identifier": "D", "content": "$$\\sqrt {33} $$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264103/exam_images/a6qatzrn0wehdulviizi.webp" loading="lazy" alt="AIEEE 2003 Mathematics - Vector Algebra Question 223 English Explanation">
<br><br>$$PV\,\,$$ of $$\,\,\overrightarrow {AD} $$ $$ = {{\left( {3 + 5} \right)i + \left( {0 - 2} \right)j + \left( {4 + 4} \right)k} \over 2}$$
<br><br>$$ = 4i - j + 4k\,\,$$
<br><br>or $$\,\,\,\left| {\overrightarrow {AD} } \right| = \sqrt {16 + 16 + 1} = \sqrt {33} $$ | mcq | aieee-2003 |
p4AOH1crTPg4bgPR | maths | vector-algebra | algebra-and-modulus-of-vectors | Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be three non-zero vectors such that no two of these are collinear. If the vector $$\overrightarrow a + 2\overrightarrow b $$ is collinear with $$\overrightarrow c $$ and $$\overrightarrow b + 3\overrightarrow c $$ is collinear with $$\overrightarrow a $$ ($$\lambda $$ being some non-zero scalar) then $$\overrightarrow a + 2\overrightarrow b + 6\overrightarrow c $$ equals to : | [{"identifier": "A", "content": "$\\overrightarrow{0}$"}, {"identifier": "B", "content": "$$\\lambda \\overrightarrow b $$ "}, {"identifier": "C", "content": "$$\\lambda \\overrightarrow c $$"}, {"identifier": "D", "content": "$$\\lambda \\overrightarrow a $$"}] | ["A"] | null | If $\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}$ is collinear with $\overrightarrow{\mathbf{c}}$, then
<br/><br/>$$
\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}=t \overrightarrow{\mathbf{c}}
$$
<br/><br/>Also, if $\overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}$ is collinear with $\overrightarrow{\mathbf{a}}$, then
<br/><br/>$$
\begin{aligned}
& \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}=\lambda \overrightarrow{\mathbf{a}} \\\\
& \Rightarrow \overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}}
\end{aligned}
$$
<br/><br/>On putting this value in Eq. (i), we get
<br/><br/>$$
\overrightarrow{\mathbf{a}}+2(\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}}
$$
<br/><br/>$$
\Rightarrow \overrightarrow{\mathbf{a}}+2 \lambda \overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}}=t \overrightarrow{\mathbf{c}} $$
<br/><br/>$$
\Rightarrow (\overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}}-2 \lambda \overrightarrow{\mathbf{a}}
$$
<br/><br/>On comparing, we get
<br/><br/>and $-6=t $
<br/><br/>$\Rightarrow t=-6$
<br/><br/>From Eq. (i),
<br/><br/>$$\vec{a}+2 \vec{b}=-6 \vec{c} $$
<br/><br/>$$\Rightarrow \vec{a}+2 \vec{b}+6 \vec{c}=\overrightarrow{0}$$ | mcq | aieee-2004 |
iFtmeeyyKqw8k9qd | maths | vector-algebra | algebra-and-modulus-of-vectors | If $$C$$ is the mid point of $$AB$$ and $$P$$ is any point outside $$AB,$$ then : | [{"identifier": "A", "content": "$$\\overrightarrow {PA} + \\overrightarrow {PB} = 2\\overrightarrow {PC} $$ "}, {"identifier": "B", "content": "$$\\overrightarrow {PA} + \\overrightarrow {PB} = \\overrightarrow {PC} $$ "}, {"identifier": "C", "content": "$$\\overrightarrow {PA} + \\overrightarrow {PB} = 2\\overrightarrow {PC} = \\overrightarrow 0 $$ "}, {"identifier": "D", "content": "$$\\overrightarrow {PA} + \\overrightarrow {PB} = \\overrightarrow {PC} = \\overrightarrow 0 $$ "}] | ["A"] | null | $$\overrightarrow {PA} + \overrightarrow {AP = 0} $$ and $$\overrightarrow {PC} + \overrightarrow {CP} = 0$$
<br><br>$$ \Rightarrow \overrightarrow {PA} + \overrightarrow {AC} + \overrightarrow {CP} = 0$$
<br><br>and
<br><br>$$\overrightarrow {PB} + \overrightarrow {BC} + \overrightarrow {CP} = 0$$
<br><br>Adding, we get
<br><br>$$\overrightarrow {PA} + \overrightarrow {PB} + \overrightarrow {AC} + \overrightarrow {BC} + 2\overrightarrow {CP} = 0.$$
<br><br>Since
<br><br>$$\overrightarrow {AC} = - \overrightarrow {BC} $$ $$\,\,\,\,\,\,$$ & $$\,\,\,\,\,\,$$ $$\overrightarrow {CP} = - \overrightarrow {PC} $$
<br><br>$$ \Rightarrow \overrightarrow {PA} + \overrightarrow {PB} - 2\overrightarrow {PC} = 0.$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264283/exam_images/duu7crwpjpg3itzy9gd0.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Vector Algebra Question 213 English Explanation"> | mcq | aieee-2005 |
Kup36fmEjHBf8paP | maths | vector-algebra | algebra-and-modulus-of-vectors | Let $$a, b$$ and $$c$$ be distinct non-negative numbers. If the vectors $$a\widehat i + a\widehat j + c\widehat k,\,\,\widehat i + \widehat k$$ and $$c\widehat i + c\widehat j + b\widehat k$$ lie in a plane, then $$c$$ is : | [{"identifier": "A", "content": "the Geometric Mean of $$a$$ and $$b$$"}, {"identifier": "B", "content": "the Arithmetic Mean of $$a$$ and $$b$$"}, {"identifier": "C", "content": "equal to zero "}, {"identifier": "D", "content": "the Harmonic Mean of $$a$$ and $$b$$"}] | ["A"] | null | Vector $$a\overrightarrow i + a\overrightarrow j + c\overrightarrow k ,\,\,\overrightarrow i + \overrightarrow k $$
<br><br>and $$c\overrightarrow i + c\overrightarrow j + b\overrightarrow k $$ are coplanar
<br><br>$$\left| {\matrix{
a & a & c \cr
1 & 0 & 1 \cr
c & c & b \cr
} } \right| = 0 \Rightarrow {c^2} = ab$$
<br><br>$$ \Rightarrow c = \sqrt {ab} $$
<br><br>$$\therefore$$ $$c$$ is $$G.M.$$ of $$a$$ and $$b.$$ | mcq | aieee-2005 |
4qsvyU90C1dWTbNp | maths | vector-algebra | algebra-and-modulus-of-vectors | The non-zero vectors are $${\overrightarrow a ,\overrightarrow b }$$ and $${\overrightarrow c }$$ are related by $${\overrightarrow a = 8\overrightarrow b }$$ and $${\overrightarrow c = - 7\overrightarrow b \,\,.}$$ Then the angle between $${\overrightarrow a }$$ and $${\overrightarrow c }$$ is : | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$ "}, {"identifier": "D", "content": "$$\\pi $$ "}] | ["D"] | null | Clearly $$\overrightarrow a = - {8 \over 7}\overrightarrow c $$
<br><br>$$ \Rightarrow \overrightarrow a ||\overrightarrow c $$ and are opposite in direction
<br><br>$$\therefore$$ Angle between $$\overrightarrow a $$ and $$\overrightarrow c $$ is $$\pi .$$ | mcq | aieee-2008 |
lkt203b6 | maths | vector-algebra | algebra-and-modulus-of-vectors | Let $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$ be three non-zero vectors which are pairwise non-collinear. If $\overrightarrow a+3 \overrightarrow b$ is collinear with $\overrightarrow c$ and $\overrightarrow b+2 \overrightarrow c$ is collinear with $\overrightarrow a$, then $\overrightarrow a+\overrightarrow b+6 \overrightarrow c$ is : | [{"identifier": "A", "content": "$\\overrightarrow a+\\overrightarrow c$"}, {"identifier": "B", "content": "$\\overrightarrow c$"}, {"identifier": "C", "content": "$\\overrightarrow a$"}, {"identifier": "D", "content": "$\\overrightarrow 0$"}] | ["D"] | null | <p>We are given that $\overrightarrow a + 3 \overrightarrow b$ is collinear with $\overrightarrow c$, and $\overrightarrow b + 2 \overrightarrow c$ is collinear with $\overrightarrow a$. This means we can write:</p>
<ol>
<li>$\overrightarrow a + 3 \overrightarrow b = \lambda \overrightarrow c \quad ...(i)$</li>
<li>$\overrightarrow b + 2 \overrightarrow c = \mu \overrightarrow a \quad ...(ii)$</li>
</ol>
<p>for some scalars $\lambda$ and $\mu$.</p>
<p>We are trying to find $\overrightarrow a + \overrightarrow b + 6\overrightarrow c$ in terms of $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$. We can also express this as :</p>
<p>$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = (\lambda + 6) \overrightarrow c \quad ...(iii)$</p>
<p>by adding $6\overrightarrow c$ to both sides of equation (i).</p>
<p>Now, from equation (ii), multiplying by 3 gives us :</p>
<p>$3\overrightarrow b + 6 \overrightarrow c = 3\mu \overrightarrow a \quad ...(iv)$</p>
<p>Adding $\overrightarrow a$ to both sides of equation (iv) gives :</p>
<p>$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = (1 + 3\mu) \overrightarrow a \quad ...(v)$</p>
<p>Now, we have two expressions for $\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c$, one in terms of $\overrightarrow c$ (from equation iii) and one in terms of $\overrightarrow a$ (from equation v). Setting these equal to each other gives :</p>
<p>$(\lambda + 6) \overrightarrow c = (1 + 3\mu) \overrightarrow a \quad ...(vi)$</p>
<p>Since $\overrightarrow a$ and $\overrightarrow c$ are not collinear, this equation can only hold if the coefficients on both sides are zero, hence :</p>
<p>$\lambda + 6 = 0$ and $1 + 3\mu = 0$</p>
<p>This gives $\lambda = -6$ and $\mu = -\frac{1}{3}$.</p>
<p>Finally, substituting $\lambda = -6$ into equation (iii) gives :</p>
<p>$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = 0$</p>
<p>So, $\overrightarrow a + \overrightarrow b + 6\overrightarrow c = \overrightarrow 0$.</p>
<p>Therefore, the correct answer is Option D : $\overrightarrow 0$.</p>
| mcq | aieee-2011 |
1moQH12BR5I24J9y | maths | vector-algebra | algebra-and-modulus-of-vectors | If the vectors $$\overrightarrow {AB} = 3\widehat i + 4\widehat k$$ and $$\overrightarrow {AC} = 5\widehat i - 2\widehat j + 4\widehat k$$ are the sides of a triangle $$ABC,$$ then the length of the median through $$A$$ is : | [{"identifier": "A", "content": "$$\\sqrt {18} $$ "}, {"identifier": "B", "content": "$$\\sqrt {72} $$"}, {"identifier": "C", "content": "$$\\sqrt {33} $$"}, {"identifier": "D", "content": "$$\\sqrt {45} $$ "}] | ["C"] | null | As $$M$$ is mid point of $$BC$$
<br><br>$$\therefore$$ $$\overrightarrow {AM} = {1 \over 2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)$$
<br><br>$$ = 4\overrightarrow i + \overrightarrow j + 4\overrightarrow k $$
<br><br>Length of median $$AM$$
<br><br>$$ = \sqrt {16 + 1 + 16} = \sqrt {33} $$ | mcq | jee-main-2013-offline |
GzUDojnFZg8u9lKQ6BPEk | maths | vector-algebra | algebra-and-modulus-of-vectors | Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$$ respectively, then the position vector of the orthocentre of this triangle, is : | [{"identifier": "A", "content": "$${\\overrightarrow a + \\overrightarrow b + \\overrightarrow c }$$ "}, {"identifier": "B", "content": "$$ - \\left( {{{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\overrightarrow 0 $$ "}, {"identifier": "D", "content": "$$\\left( {{{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 2}} \\right)$$ "}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264248/exam_images/rtiinzvghawcvhtcjcms.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Mathematics - Vector Algebra Question 183 English Explanation">
<br><br>Given,
<br><br>Position vector of circumcentre, $$\overrightarrow C = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$$
<br><br>We know, position vector of centroid, $$\overrightarrow G = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 3}$$
<br><br>Now, let $$\overrightarrow R $$ be the orthocentre of the triangle.
<br><br>We know, $$\overrightarrow G $$ $$ = {{2\overrightarrow C + \overrightarrow R } \over 3}$$
<br><br>$$ \Rightarrow $$ 3$$\overrightarrow G $$ $$ = 2\overrightarrow C + \overrightarrow R $$
<br><br>$$ \Rightarrow $$ $$\overrightarrow R = 3\overrightarrow G - 2\overrightarrow C $$
<br><br>= $$\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) - 2\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}} \right)$$
<br><br>= $${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}$$ | mcq | jee-main-2016-online-10th-april-morning-slot |
6n5pJMAxl9mPBSplixN9e | maths | vector-algebra | algebra-and-modulus-of-vectors | If the position vectors of the vertices A, B and C of a $$\Delta $$ ABC are respectively $$4\widehat i + 7\widehat j + 8\widehat k,$$ $$2\widehat i + 3\widehat j + 4\widehat k,$$ and $$2\widehat i + 5\widehat j + 7\widehat k,$$ then the position vectors of the point, where the bisector of $$\angle $$A meets BC is : | [{"identifier": "A", "content": "$${1 \\over 2}\\left( {4\\widehat i + 8\\widehat j + 11\\widehat k} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over 3}\\left( {6\\widehat i + 11\\widehat j + 15\\widehat k} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over 3}\\left( {6\\widehat i + 13\\widehat j + 18\\widehat k} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 4}\\left( {8\\widehat i + 14\\widehat j + 19\\widehat k} \\right)$$"}] | ["C"] | null | Suppose angular bisector of A meets BC at D(x, , z)
<br><br>Using angular bisector theorem,
<br><br>$${{AB} \over {AC}}$$ = $${{BD} \over {DC}}$$
<br><br> $${{BD} \over {DC}}$$ = $${{\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {8 - 4} \right)}^2}} } \over {\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 5} \right)}^2} + {{\left( {8 - 7} \right)}^2}} }}$$
<br><br>= $${{\sqrt {{2^2} + {4^2} + {4^2}} } \over {\sqrt {{2^2} + {2^2} + {1^2}} }}$$ = $${6 \over 3}$$ = 2
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267236/exam_images/mxsolnud9xur1rrhm47n.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15th April Evening Slot Mathematics - Vector Algebra Question 188 English Explanation">
<br><br>So, D(x, y, z) $$ \equiv $$ $$\left( {{{\left( 2 \right)\left( 2 \right) + \left( 1 \right)\left( 2 \right)} \over {2 + 1}},{{\left( 2 \right)\left( 5 \right) + \left( 1 \right)\left( 3 \right)} \over {2 + 1}}} \right.$$, $$\left. {{{\left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 4 \right)} \over {2 + 1}}} \right)$$
<br><br>D(x, y, z) $$ \equiv $$ $$\left( {{6 \over 3},{{13} \over 3},{{18} \over 3}} \right)$$
<br><br>Therefore, position vector of point p = $${1 \over 3}$$ (6i + 13j + 18k) | mcq | jee-main-2018-online-15th-april-evening-slot |
udRepuDAf9PhtRCVX4vWm | maths | vector-algebra | algebra-and-modulus-of-vectors | If $$\overrightarrow \alpha $$ = $$\left( {\lambda - 2} \right)\overrightarrow a + \overrightarrow b $$ and $$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow a + 3\overrightarrow b $$ be two given vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ are non-collinear. The value of $$\lambda $$ for which vectors $$\overrightarrow \alpha $$ and $$\overrightarrow \beta $$ are collinear, is - | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "$$-$$3"}, {"identifier": "D", "content": "$$-$$4"}] | ["D"] | null | $$\overrightarrow \alpha = \left( {\lambda - 2} \right)\overrightarrow \alpha + \overrightarrow b $$
<br><br>$$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow \alpha + 3\overrightarrow b $$
<br><br>$${{\lambda - 2} \over {4\lambda - 2}} = {1 \over 3}$$
<br><br>$$3\lambda - 6 = 4\lambda - 2$$
<br><br>$$\lambda = - 4$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
fMrt9F3WxfZGa87KLbjgy2xukezm4q3d | maths | vector-algebra | algebra-and-modulus-of-vectors | The lines
<br/>$$\overrightarrow r = \left( {\widehat i - \widehat j} \right) + l\left( {2\widehat i + \widehat k} \right)$$ and
<br/>$$\overrightarrow r = \left( {2\widehat i - \widehat j} \right) + m\left( {\widehat i + \widehat j + \widehat k} \right)$$ | [{"identifier": "A", "content": "do not intersect for any values of $$l$$ and m"}, {"identifier": "B", "content": "intersect for all values of $$l$$ and m"}, {"identifier": "C", "content": "intersect when $$l$$ = 2 and m = $${1 \\over 2}$$"}, {"identifier": "D", "content": "intersect when $$l$$ = 1 and m = 2"}] | ["A"] | null | L<sub>1</sub> = $$\overrightarrow r = \left( {\widehat i - \widehat j} \right) + l\left( {2\widehat i + \widehat k} \right)$$
<br><br>= $$\widehat i\left( {1 + 2l} \right) + \widehat j\left( { - 1} \right) + \widehat k\left( l \right)$$
<br><br>L<sub>2</sub> = $$\overrightarrow r = \left( {2\widehat i - \widehat j} \right) + m\left( {\widehat i + \widehat j + \widehat k} \right)$$
<br><br>= $$\widehat i\left( {2 + m} \right) + \widehat j\left( {m - 1} \right) + \widehat k\left( { - m} \right)$$
<br><br>Equating coefficient of $$\widehat i$$, $$\widehat j$$ and $$\widehat k$$ of L<sub>1</sub>
and L<sub>2</sub>
<br><br>2l + 1 = m + 2 ... (1)
<br><br>–1 = –1 + m ...(2)
<br><br>l = –m ...(3)
<br><br>from (ii) m = 0
<br><br>from (iii) $$l$$ = 0
<br><br>These values of m and $$l$$ do not satisfy equation (1).
<br><br>Hence the two lines do not intersect for any values of $$l$$ and m. | mcq | jee-main-2020-online-3rd-september-morning-slot |
vyIvUkH6pDxHx9TP8Pjgy2xukfw15iow | maths | vector-algebra | algebra-and-modulus-of-vectors | If $$\overrightarrow a $$
and $$\overrightarrow b $$
are unit vectors, then the greatest value of
<br/><br>$$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$$ is_____.</br> | [] | null | 4 | Let angle between $$\overrightarrow a $$ and $$\overrightarrow b $$
be $$\theta $$.
<br><br>$$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$$
<br><br>= $$\sqrt 3 \left( {\sqrt {1 + 1 + 2\cos \theta } } \right)$$ + $$\left( {\sqrt {1 + 1 - 2\cos \theta } } \right)$$
<br><br>= $$\sqrt 3 \left( {\sqrt {2 + 2\cos \theta } } \right)$$ + $$\left( {\sqrt {2 - 2\cos \theta } } \right)$$
<br><br>= $$\sqrt 6 \left( {\sqrt {1 + \cos \theta } } \right)$$ + $$\sqrt 2 \left( {\sqrt {1 - \cos \theta } } \right)$$
<br><br>= $$\sqrt 6 \left( {\sqrt {2{{\cos }^2}{\theta \over 2}} } \right)$$ + $$\sqrt 2 \left( {\sqrt {2{{\sin }^2}{\theta \over 2}} } \right)$$
<br><br>= $$2\sqrt 3 \left| {\cos {\theta \over 2}} \right|$$ + 2$$\left| {\sin {\theta \over 2}} \right|$$
<br><br>$$ \le $$ $$\sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2}} $$ = 4
<br><br><b>Note :</b> |x| = $$\sqrt {{x^2}} $$
<br><br>|x - 1| = $$\sqrt {{{\left( {x - 1} \right)}^2}} $$
<br><br>|sin x| = $$\sqrt {{{\sin }^2}x} $$
<br><br>That is why $${\sqrt {{{\sin }^2}{\theta \over 2}} }$$ = $$\left| {\sin {\theta \over 2}} \right|$$ and $${\sqrt {{{\cos }^2}{\theta \over 2}} }$$ = $$\left| {\cos {\theta \over 2}} \right|$$ | integer | jee-main-2020-online-6th-september-morning-slot |
WE3BfzabLGO8CzgZL41kluxsedn | maths | vector-algebra | algebra-and-modulus-of-vectors | If vectors $$\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k$$ and $$\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k$$ are collinear, then a possible unit vector parallel to the vector $$x\widehat i + y\widehat j + z\widehat k$$ is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 3 }}\\left( {\\widehat i - \\widehat j + \\widehat k} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 2 }}\\left( { - \\widehat j + \\widehat k} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}\\left( {\\widehat i - \\widehat j} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 3 }}\\left( {\\widehat i + \\widehat j - \\widehat k} \\right)$$"}] | ["A"] | null | $$\overrightarrow {{a_2}} = \lambda \overrightarrow {{a_1}} $$<br><br>$$\widehat i + y\widehat j + z\widehat k = \lambda (x\widehat i - \widehat j + \widehat k)$$<br><br>$$1 = \lambda x,y = - \lambda ,z = \lambda $$<br><br>$$x\widehat i + y\widehat j + z\widehat k = {1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k$$<br><br>Unit vector $$ = {{{1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k} \over {\sqrt {{1 \over {{\lambda ^2}}} + {\lambda ^2} + {\lambda ^2}} }}$$<br><br>$$ = {{\widehat i - {\lambda ^2}\widehat j + {\lambda ^2}\widehat k} \over {\sqrt {1 + 2{\lambda ^4}} }}$$<br><br>Let $${\lambda ^2} = 1$$, possible unit vector $$ = {{\widehat i - \widehat j + \widehat k} \over {\sqrt 3 }}$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
PKaCFkWCxapnmisq1O1kmhvprkw | maths | vector-algebra | algebra-and-modulus-of-vectors | Let a vector $$\alpha \widehat i + \beta \widehat j$$ be obtained by rotating the vector $$\sqrt 3 \widehat i + \widehat j$$ by an angle 45$$^\circ$$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices ($$\alpha$$, $$\beta$$), (0, $$\beta$$) and (0, 0) is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2$${\\sqrt 2 }$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266760/exam_images/elmya39omzdt3xm8r3m7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Mathematics - Vector Algebra Question 145 English Explanation">
<br>($$\alpha$$, $$\beta$$) $$ \equiv $$ (2 cos 75$$^\circ$$, 2 sin 75$$^\circ$$)<br><br>Area = $${1 \over 2}$$ (2 cos 75$$^\circ$$) (2 sin 75$$^\circ$$)<br><br>= sin(150$$^\circ$$) = $${1 \over 2}$$ square unit | mcq | jee-main-2021-online-16th-march-morning-shift |
HlfRsGRwqx3bVkQvd41kmlijvdt | maths | vector-algebra | algebra-and-modulus-of-vectors | A vector $$\overrightarrow a $$ has components 3p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, $$\overrightarrow a $$ has components p + 1 and $$\sqrt {10} $$, then the value of p is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$ - {5 \\over 4}$$"}, {"identifier": "C", "content": "$${4 \\over 5}$$"}, {"identifier": "D", "content": "$$-$$1"}] | ["D"] | null | $${\left| {\overrightarrow a } \right|_{old}} = {\left| {\overrightarrow a } \right|_{new}}$$<br><br>(3p)<sup>2</sup> + 1 = (p + 1)<sup>2</sup> + 10<br><br>$$ \Rightarrow $$ 9p<sup>2</sup> $$-$$ p<sup>2</sup> $$-$$ 2p $$-$$ 10 = 0<br><br>$$ \Rightarrow $$ 8p<sup>2</sup> $$-$$ 2p $$-$$ 10 = 0<br><br>$$ \Rightarrow $$ 4p<sup>2</sup> $$-$$ p $$-$$ 5 = 0<br><br>$$ \Rightarrow $$ 4p<sup>2</sup> $$-$$ 5p + 4p $$-$$ 5 = 0<br><br>$$ \Rightarrow $$ (4p $$-$$ 5) (p + 1) = 0<br><br>$$ \Rightarrow $$ p = $${5 \over 4}$$, $$-$$ 1 | mcq | jee-main-2021-online-18th-march-morning-shift |
Subsets and Splits