question stringlengths 79 9.83k | answer stringlengths 33 9.39k |
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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 :
Options:
[{"identifier": "A", "content": "4x \u2013 3y \u2013 4 = 0"}, {"identifier": "B", "content": "3x \u2013... | ["A"]
Explanation:
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 |
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 :
Options:
[{"identifier": "A", "content": "x \u2013 y + 7 = 0"}, {"identifier": "B", "content": "4x \u2013 3y + 24 = 0"}, {"identifier": "C", "content": "4x + 3... | ["B"]
Explanation:
<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 ... |
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 :
Options:
[{"identifier": "A", "content": "5x + 3y \u2013 11 = 0"}, {"identifier": "B", "content": "5x \u2013 3y + 1 = 0"}, {"identifier": "C", "content": "3x \u2013 5y + 7 = 0... | ["B"]
Explanation:
co-ordinates of point D are (4, 7)
<br><br>$$ \Rightarrow $$ line AD is 5x $$-$$ 3y + 1 = 0 |
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 :
Options:
[{"identifier": "A", "content": "(2, 1)"}, {"identifier": "B", "content": "(2, 6)"}, {"identifier": "C", "content": "(3, 5)"}, {"identifier": "D", "content": "(3, ... | ["D"]
Explanation:
<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 ... |
A point on the straight line, 3x + 5y = 15 which is equidistant from the coordinate axes will lie only
in :
Options:
[{"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... | ["A"]
Explanation:
<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"><so... |
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 :
Options:
[{"identifier": "A", "content": "$$\\sqrt {14} $$"}, {"identifier": "B", "content": "-4"}, {"identifier": "C", "content": "\u20132 "}, {"identifier": "D", "content": "$$... | ["B"]
Explanation:
$${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}... |
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 :
Options:
[{"identifier": "A", "content": "(3, -$$\\sqrt 3 $$)"}, {"iden... | ["A"]
Explanation:
<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 L... |
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?
Options:
[{"identif... | ["D"]
Explanation:
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-... |
The intersection of three lines x $$-$$ y = 0, x + 2y = 3 and 2x + y = 6 is a :
Options:
[{"identifier": "A", "content": "Right angled triangle"}, {"identifier": "B", "content": "Equilateral triangle"}, {"identifier": "C", "content": "None of the above"}, {"identifier": "D", "content": "Isosceles triangle"}] | ["D"]
Explanation:
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(... |
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 __________.
Options:
[] | 904
Explanation:
<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... |
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 :
Options:
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "0"}] | ["B"]
Explanation:
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>$$ \... |
The point P (a, b) undergoes the following three transformations successively :<br/><br/>(a) reflection about the line y = x.<br/><br/>(b) translation through 2 units along the positive direction of x-axis.<br/><br/>(c) rotation through angle $${\pi \over 4}$$ about the origin in the anti-clockwise direction.<br/><br/... | ["B"]
Explanation:
Image of A(a, b) along y = x is B(b, a). Translating it 2 units it becomes C(b + 2, a).<br><br>Now, applying rotation theorem<br><br>$$ - {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 {\s... |
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 :
Options:
[{"identifier": "A", "content": "(1, 2)"}, {"identifier": "B", "content": "(2, 2)"}, {"identifier": "C", "c... | ["B"]
Explanation:
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... |
<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>
Options:
[{"identifier": "A", "content": "$$2 x+y=9$$"}, {"identifier": "B", ... | ["C"]
Explanation:
<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-... |
<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 diago... | ["B"]
Explanation:
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-(\... |
<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>
Options:
[{"identifier": "A", "content": "$${2 \\over {\\left( {\\sqrt 3 - 1} \\right)}}$$"}, {"identif... | ["D"]
Explanation:
<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-a6c2-11... |
<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 ... | ["C"]
Explanation:
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/6y3zl... |
<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>
Options:
[] | 32
Explanation:
<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-c801-4445-8b27-227c5944b078/0... |
<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 $$2(\alpha+\beta+\gamma+\delta)$$... | ["A"]
Explanation:
<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-b6e9-9f078c18e4d... |
<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>
Options:
[{"identifier": "A", "content": "$$\... | ["D"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnkh52/7d853a24-3c49-423c-95b3-38a63c841203/38d89660-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnkh53.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqnkh52/7d853a24-3c49-423c-95b3-38a63c84120... |
<p>A ray of light coming from the point $$\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$$ is equal to:</p>
Options:
[{"identifier": "A", "content": "60"}, {... | ["B"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw37w9ht/b3e9aa4d-c4a4-4387-bd7f-990e7790d425/c5298810-1031-11ef-b980-477f779c8c59/file-1lw37w9hu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw37w9ht/b3e9aa4d-c4a4-4387-bd7f-990e7790d425/c5298810-1031-11... |
<p>Let a variable line of slope $$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$$ from the origin is</p>
Options:
[{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "15"}, {"id... | ["D"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd2p5t3/fdf80bf0-c51d-4c22-865c-bd4b3e3d59be/487f8460-159d-11ef-aabb-5de744d2ad81/file-1lwd2p5t4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd2p5t3/fdf80bf0-c51d-4c22-865c-bd4b3e3d59be/487f8460-159d-11... |
If 0 $$ \le $$ x < $${\pi \over 2}$$, then the number of values of x for which sin x $$-$$ sin 2x + sin 3x = 0, is :
Options:
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2"}] | ["D"]
Explanation:
sin x $$-$$ sin 2x + sin 3x = 0 $$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 $$ sin2x ... |
If $$\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]$$ is __________.
Options:
[] | 1
Explanation:
$$\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$$<br><br>$$ \Righ... |
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 :
Options:
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["A"]
Explanation:
<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>$$
4 - 4
\cos^2 x - 4
\cos^3 x + 9 -... |
<p>If $$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}$$ is equal to:</p>
Options:
[{"identifier": "A", "content": "$$\\frac{1}{2^{14}}\\left(2^{15}-15\\right)$$\n"}, {... | ["D"]
Explanation:
<p>$$\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{ali... |
<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>
Options:
[{"identifier": "A", "content": "$$\\frac{10-\\sqrt{10}}{12}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\sqrt{10}-10}{6}$$\n"}, {"identifier": "C", "c... | ["C"]
Explanation:
<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> |
<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>
Options:
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$-$$1"}] | ["D"]
Explanation:
<p>$$\begin{aligned}
& \frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6 \\
& \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 \\
& \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 \\
& \Rightar... |
<p>If $$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$$ belong to :</p>
Options:
[{"identifier": "A", "content": "$$(0, \\infty)$$\n"}, {"iden... | ["D"]
Explanation:
<p>$$\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 { ... |
<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>
Options:
[{"identifier": "A", "content": "$$6 \\pi$$\n"}, {"identifier": "B", "content": "$$9 \\pi$$\n"}, {"i... | ["A"]
Explanation:
<p>$$\begin{aligned}
& |\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8} \\
& \Rightarrow \frac{1}{4}|\cos 3 \theta| \leq \frac{1}{8} \\
& \cos 3 \theta \text { is max if } \cos 3 \theta=\frac{1}{2} \\
& \therefore \theta=\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{... |
<p>Let $$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)$... | 4
Explanation:
<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>$$\begin{aligned}
& \Rightarrow t \geq 4\left(1-\frac{t}{2}\right)\left(1-\frac{... |
<p>The number of solutions of $$\sin ^2 x+\left(2+2 x-x^2\right) \sin x-3(x-1)^2=0$$, where $$-\pi \leq x \leq \pi$$, is ________.</p>
Options:
[] | 2
Explanation:
<p>$$\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}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwey5iox/7c0cf8ef-bef6-44... |
Let $$\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}$$ :
Options:
[{"identifier": "A", "content": "$${{ ... | ["D"]
Explanation:
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>$$2\left( {1 + \cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)$$$$... |
Let <b>A</b> and <b>B</b> denote the statements
<p><b>A</b>: $$\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... | ["B"]
Explanation:
<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"> |
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/>
Options:
[{"identifier": "A", "content": "$${56 \\over 33}$$"}, {"identifier": "B", "content": "$$... | ["A"]
Explanation:
$$\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[ {\le... |
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 :
Options:
[{"identifier": "A", "content": "21/16"}, {"identifier": "B", "content": "63/52"}, {"identifier": "C", "content": "33/52"}, {"identifier": "D"... | ["D"]
Explanation:
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 $$ - {\pi \over 4} < \alpha - \beta < {\pi... |
If 0 < x, y < $$\pi$$ and cosx + cosy $$-$$ cos(x + y) = $${3 \over 2}$$, then sinx + cosy is equal to :
Options:
[{"identifier": "A", "content": "$${{1 + \\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$${{1 \\over 2}}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "... | ["A"]
Explanation:
$$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}} \... |
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 :
Options:
[{"identifier": "A", "content": "... | ["C"]
Explanation:
$$x = {1 \over 2}\left( {\tan {\pi \over 9} + \tan {{7\pi } \over {18}}} \right)$$<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/>$$x = \frac{1}{2} \left( \tan 20^\ci... |
<p>If cot$$\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$$ lies, respectively are :</p>
Options:
[{"identifier": "A", "cont... | ["A"]
Explanation:
$\because \cot \alpha=1, \quad \alpha \in\left(\pi, \frac{3 \pi}{2}\right)$<br/><br/> then $\tan \alpha=1$ <br/><br/>and $\sec \beta=-\frac{5}{3}, \quad \beta \in\left(\frac{\pi}{2}, \pi\right)$ <br/><br/>then $\tan \beta=-\frac{4}{3}$
<br/><br/>
$\therefore \tan (\alpha+\beta)=\frac{\tan \alpha+\ta... |
If $\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \mathrm{B}=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and <br/><br/>$\tan \mathrm{C}=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1 / 2}, 0<\mathrm{A}, \mathrm{B}, \mathrm{C}<\frac{\pi}{2}$, then $\mathrm{A}+\mathrm{B}$ is equal to :
Options:
[{"identifier": "A", "con... | ["A"]
Explanation:
<p>To find the sum of two angles in terms of tangent, we can use the tangent addition formula: </p>
<p>$$ \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)}} $$... |
<p>For $$\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$$ is equal to</p>
Options:
[{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "$$-$$2/3"}, {"identifier": "C", "conte... | ["C"]
Explanation:
<p>To find the value of $$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 +... |
<p>The number of solutions, of the equation $$e^{\sin x}-2 e^{-\sin x}=2$$, is :</p>
Options:
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "more than 2"}] | ["A"]
Explanation:
<p>Take $$e^{\sin x}=t(t>0)$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{t}-\frac{2}{\mathrm{t}}=2 \\
& \Rightarrow \frac{\mathrm{t}^2-2}{\mathrm{t}}=2 \\
& \Rightarrow \mathrm{t}^2-2 \mathrm{t}-2=0 \\
& \Rightarrow \mathrm{t}^2-2 \mathrm{t}+1=3 \\
& \Rightarrow(\mathrm{t}-1)^2=3 \\
& \Rightarro... |
If $$0 < x < \pi $$ and $$\cos x + \sin x = {1 \over 2},$$ then $$\tan x$$ is :
Options:
[{"identifier": "A", "content": "$${{\\left( {1 - \\sqrt 7 } \\right)} \\over 4}$$ "}, {"identifier": "B", "content": "$${{\\left( {4 - \\sqrt 7 } \\right)} \\over 3}$$ "}, {"identifier": "C", "content": "$$ - {{\\left( {4 +... | ["C"]
Explanation:
$$\cos x + \sin x = {1 \over 2}$$ <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>
$$\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin... |
The expression $${{\tan {\rm A}} \over {1 - \cot {\rm A}}} + {{\cot {\rm A}} \over {1 - \tan {\rm A}}}$$ can be written as:
Options:
[{"identifier": "A", "content": "$$\\sin {\\rm A}\\,\\cos {\\rm A} + 1$$ "}, {"identifier": "B", "content": "$$\\,\\sec {\\rm A}\\,\\cos ec{\\rm A} + 1$$ "}, {"identifier": "C", "conte... | ["B"]
Explanation:
Given expression can be written as
<br><br>$${{\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 \o... |
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 :
Options:
[{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$${1 \\over 12}$$"}, {"ident... | ["B"]
Explanation:
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>$$=$$ $${1 \over 4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {1 \over 6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$$
<... |
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 :
Options:
[{"identifier": "A", "content": "13 \u2013 4 cos<sup>2</sup>$$\\theta $$ + 6sin<sup>2</sup>... | ["B"]
Explanation:
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$$\theta $$... |
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 :
Options:
[{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "9"}, ... | ["C"]
Explanation:
$$
\begin{aligned}
& \log _{10} \sin x+\log _{10} \cos x=-1, x \in(0, \pi / 2) \\\\
& \log _{10}(\sin x \cos x)=-1 \\\\
& \Rightarrow \sin x \cos x=10^{-1}= {1 \over {10}} \\\\
& \log _{10}(\sin x+\cos x)={1 \over {2}}\left(\log _{10} n-1\right), n>0 \\\\
& 2 \log _{10}(\sin x+\cos x)=\left(\log _{... |
If 15sin<sup>4</sup>$$\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 :
Options:
[{"identifier": "A", "content": "500"}, {"identifier": "B", "content": "400"}, {"identifier": "C", "content": "250... | ["C"]
Explanation:
$$
\begin{aligned}
& \text { Given, } 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6 \\\\
& \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^2 \alpha+\cos ^2 \alpha\right)^2 \\\\
& \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^4 \alpha+\cos ^4 \alpha+2 \sin ^2 \alpha \c... |
<p>If $$\sin x=-\frac{3}{5}$$, where $$\pi< x <\frac{3 \pi}{2}$$, then $$80\left(\tan ^2 x-\cos x\right)$$ is equal to</p>
Options:
[{"identifier": "A", "content": "109"}, {"identifier": "B", "content": "108"}, {"identifier": "C", "content": "19"}, {"identifier": "D", "content": "18"}] | ["A"]
Explanation:
<p>$$\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}$$</p> |
If $$5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9$$,
<br/><br/>then the value of $$\cos 4x$$ is :
Options:
[{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 9}$$"}, {"identifier": "C", "content": "$$ - {7 \\over 9}$$"}, {"identifier": "D", "content": "$$ -... | ["C"]
Explanation:
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>$$5\left( {{{1 - t}... |
The value of cos<sup>2</sup>10° – cos10°cos50° + cos<sup>2</sup>50° is
Options:
[{"identifier": "A", "content": "$${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}$$"... | ["B"]
Explanation:
cos<sup>2</sup>10° – cos10°cos50° + cos<sup>2</sup>50°
<br><br>= $${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}$$ + cos20° + cos100° - cos40°]
<br><br>= $$... |
The equation y = sinx sin (x + 2) – sin<sup>2</sup>
(x + 1) represents a straight line lying in :
Options:
[{"identifier": "A", "content": "first, second and fourth quadrants"}, {"identifier": "B", "content": "first, third and fourth quadrants"}, {"identifier": "C", "content": "second and third quadrants only"}, {"i... | ["D"]
Explanation:
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 passe... |
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
_____.
Options:
[] | 1
Explanation:
$${{\sqrt 2 \sin \alpha } \over {\sqrt {1 + \cos 2\alpha } }} = {1 \over 7}$$
<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>$$ \Ri... |
The value of<br/>
$${\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 :
Options:
[{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over ... | ["D"]
Explanation:
$${\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>= $${\cos ^3}\left( {{\pi \over 8}} \right)\sin \left( {{\pi \over 8}} \right) + {\sin ^3}\left( {{\pi \over 8}} \r... |
If the equation cos<sup>4</sup> $$\theta $$ + sin<sup>4</sup> $$\theta $$ +
$$\lambda $$ = 0 has real
solutions for
$$\theta $$, then
$$\lambda $$ lies in the interval :
Options:
[{"identifier": "A", "content": "$$\\left[ { - {3 \\over 2}, - {5 \\over 4}} \\right]$$"}, {"identifier": "B", "content": "$$\\left( { - {1 ... | ["D"]
Explanation:
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> $$\theta $$ = -$$\lamb... |
If L = sin<sup>2</sup>$$\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)$$, then :
Options:
[{"identifier": "A", "content": "L = $$ - {1 \\over {2\\sqrt 2 }... | ["B"]
Explanation:
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>$$\left( {{\pi \over {8}}} \right)$$
<b... |
If $$\sin \theta + \cos \theta = {1 \over 2}$$, then 16(sin(2$$\theta$$) + cos(4$$\theta$$) + sin(6$$\theta$$)) is equal to :
Options:
[{"identifier": "A", "content": "23"}, {"identifier": "B", "content": "$$-$$27"}, {"identifier": "C", "content": "$$-$$23"}, {"identifier": "D", "content": "27"}] | ["C"]
Explanation:
$$\sin \theta + \cos \theta = {1 \over 2}$$<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><b... |
<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 $$f(\beta )$$... | ["C"]
Explanation:
$f(\theta)=3\left(\sin ^{4}\left(\frac{3 \pi}{2}-\theta\right)+\sin ^{4}(3 x+\theta)\right)-2\left(1-\sin ^{2} 2 \theta\right)$<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\r... |
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 ____________.
Options:
[] | 48
Explanation:
<p>$$\begin{aligned}
& \cos 2 x+a \cdot \sin x=2 a-7 \\
& a(\sin x-2)=2(\sin x-2)(\sin x+2) \\
& \sin x=2, a=2(\sin x+2) \\
& \Rightarrow a \in[2,6] \\
& p=2 \quad q=6 \\
& r=\tan 9^{\circ}+\cot 9^{\circ}-\tan 27-\cot 27 \\
& r=\frac{1}{\sin 9 \cdot \cos 9}-\frac{1}{\sin 27 \cdot \cos 27} \\
& =2\left[... |
If $$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}$$ is given by : </br>
Options:
[{"identifier": "A", "content": "$${\\left( {a - b} \\right)^2}$$ "}, {"ide... | ["A"]
Explanation:
Given $$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>  ... |
If $$A = {\sin ^2}x + {\cos ^4}x,$$ then for all real $$x$$:
Options:
[{"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": "$${... | ["D"]
Explanation:
$$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) ... |
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 :
Options:
[{"identifier": "A", "content": "$${{15} \\over 4}$$ "}, {"identifier": "B", "content": "$${{9} \\over... | ["B"]
Explanation:
Given,
<br><br>4 + $${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 $$ 4 sin<sup>2</sup>x cos<sup>2</sup>x $$-$$ 2cos<sup>4</sup> x
<br><br>= 4 + 2 (1 $$-$$ cos<sup>2</su... |
The maximum value of 3cos$$\theta $$ + 5sin $$\left( {\theta - {\pi \over 6}} \right)$$ for any real value of $$\theta $$ is :
Options:
[{"identifier": "A", "content": "$$\\sqrt {34} $$ "}, {"identifier": "B", "content": "$$\\sqrt {31} $$"}, {"identifier": "C", "content": "$$\\sqrt {19} $$"}, {"identifier": "D", "... | ["C"]
Explanation:
y = 3cos$$\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} $$ |
The number of integral values of 'k' for which the equation $$3\sin x + 4\cos x = k + 1$$ has a solution, k$$\in$$R is ___________.
Options:
[] | 11
Explanation:
We know,
<br><br>$$ - \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 $$ Set of integers = $$... |
<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>
Options:
[{"identifier": "A", "content": "$$\\left[ { - 2, - 1} \\right]$$"}, {"identifier": "B", "content": "$$\\left[ { - {3 \\over 2}, - 1} \\right]$$"}, {"identi... | ["B"]
Explanation:
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/>$$\lambda = (2 \cos ^2 x - 1)^2 - 2(1 - \cos ... |
The value of <br/><br/>$$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)$$ is :
Options:
[{"identifier": "A", "content": "$${1 \\over ... | ["C"]
Explanation:
$$2\sin \left( {{\pi \over 8}} \right)\sin \left( {{{2\pi } \over 8}} \right)\sin \left( {{{3\pi } \over 8}} \right)\sin \left( {{{5\pi } \over 8}} \right)\sin \left( {{{6\pi } \over 8}} \right)\sin \left( {{{7\pi } \over 8}} \right)$$<br><br>$$2{\sin ^2}{\pi \over 8}{\sin ^2}{{2\pi } \over 8}{\si... |
If $$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}$$, then :
Options:
[{"identifier": "A", "content": "x(1 + y) = 1"}, {"identifier": "B", "content": "y(1 \u2... | ["B"]
Explanation:
$$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 {{{\c... |
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 :
Options:
[{"identifier": "A", "content": "$$\\sqr... | ["D"]
Explanation:
$${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... |
The value of sin 10º sin30º sin50º sin70º is :-
Options:
[{"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"]
Explanation:
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>= $${1 \over 2}$$ [ $${1 \over 4} \times {1 \o... |
<p>If $${\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}}$$ is equal to __________.</p>
Options:
[] | 80
Explanation:
<p>$$(\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^\ci... |
<p>$$16\sin (20^\circ )\sin (40^\circ )\sin (80^\circ )$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "$$\\sqrt 3 $$"}, {"identifier": "B", "content": "2$$\\sqrt 3 $$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4$$\\sqrt 3 $$"}] | ["B"]
Explanation:
<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> |
<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>
Options:
[{"identifier": "A", "content": "$$\\frac{3}{16}$$"}, {"identifier": "B", "content": "$$\\frac{1}{16}$$"}, {"i... | ["B"]
Explanation:
<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>$$ = 2\sin \left( {{{11\pi - 10\pi } \over {22}}} \right)\sin \left( {{{11\pi - 8\pi } \over {22}}} \right)\sin \left( {{{11\pi - 6\pi } \over {22}}} \right)\s... |
<p>$$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}}$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["D"]
Explanation:
Let
<br/><br/>$$
\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}\ri... |
<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>
Options:
[{"identifier": "A", "content": "$$\\frac{6-\\sqrt{6}}{(3 \\sqrt{6}-2)}$$\n"}, {"identifier": "B", "content": "$$\\frac{4}{(3 \\sqrt{6}+2)}$$\n"}, {"identifier": "C... | ["D"]
Explanation:
<p>$$\begin{aligned}
& 4 \cos \theta-3 \sin \theta=1 \\
& 4 \cos \theta-1=3 \sin \theta \\
& 16 \cos ^2 \theta+1-8 \cos \theta=9\left(1-\cos ^2 \theta\right) \\
& \Rightarrow 25 \cos ^2 \theta-8 \cos \theta-8=0 \\
& \Rightarrow \cos \theta=\frac{8 \pm \sqrt{64+4 \times 25 \times 8}}{2.25} \\
& =\fra... |
The value of $$\cot {\pi \over {24}}$$ is :
Options:
[{"identifier": "A", "content": "$$\\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 $$"}, {"identifier": "D", "content": "$... | ["B"]
Explanation:
$$\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(... |
<p>$$\alpha = \sin 36^\circ $$ is a root of which of the following equation?</p>
Options:
[{"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": "... | ["C"]
Explanation:
<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... |
<p>The value of 2sin (12$$^\circ$$) $$-$$ sin (72$$^\circ$$) is :</p>
Options:
[{"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",... | ["D"]
Explanation:
<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>$$ = - 2\left( {{{\sqrt 5 - 1} \over 4}... |
<p>If $$\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)$$ is :</p>
Options:
[{"identifier": "A", "content": "$$5 - {3 \\over 2}\\sqrt 3 $$"}, {"identifier": "B", "content": "$$4 - 2\\sqrt 3 $$"}, {"identifier": "C"... | ["D"]
Explanation:
<p>$$\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>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>
Options:
[{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "54"}, {"identifier": ... | ["B"]
Explanation:
$$
\begin{aligned}
& 4 \cos ^2 \theta-1=4\left(1-\sin ^2 \theta\right)-1 \\\\
& =3-4 \sin ^2 \theta \\\\
& =\frac{3 \sin \theta-4 \sin ^3 \theta}{\sin \theta} \\\\
& =\frac{\sin 3 \theta}{\sin \theta}
\end{aligned}
$$
<br/><br/>$$36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\r... |
<p>The value of $$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$$ is __________.</p>
Options:
[] | 4
Explanation:
$\begin{aligned} & \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} \\\\ & =\tan 9^{\circ}+\tan \left(90^{\circ}-9^{\circ}\right)-\tan 27^{\circ}-\tan \left(90^{\circ}-27^{\circ}\right) \\\\ & =\tan 9^{\circ}+\cot 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ} \\\\ & =\frac{\sin 9^{\circ}}{\cos... |
<p>If the value of $$\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$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "54"}, {"identifier": "B", "content":... | ["B"]
Explanation:
<p>To find the value of $$\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... |
The value of $$\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$$ is -
Options:
[{"identifier": "A", "content": "$${1 \\over {256}}$$"}, {"identifier": "B", "content": "$${1 \\over {2}}$$"}, {"identifier": "C", "content": "$${1 \\over {1024}}$$"}, {"id... | ["D"]
Explanation:
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> $${\pi \over {{2^8}}}\, = \,{2^2}\theta $$
<b... |
<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>
Options:
[{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "$$-$$$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$-$$$${1 \\over... | ["B"]
Explanation:
<p>$$\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... |
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 t... | ["C"]
Explanation:
$$\left| {\matrix{
a & {{a^2}} & {1 + {a^3}} \cr
b & {{b^2}} & {1 + {b^3}} \cr
c & {{c^2}} & {1 + {c^3}} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \left| {\matrix{
a & {{a^2}} & 1 \cr
b & {{b^2}} & 1 \cr
c & {{c^2}} & 1... |
Consider points $$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$$ is a :</br>
Options:
[{"identifier": "A", "c... | ["D"]
Explanation:
$$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... |
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 :
Options:
[{"identifier": "A", "content": "$$\\sqrt {288} $$ "}, {"identifier": "B", "content": "... | ["D"]
Explanation:
<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 $$\,\,\overrightarrow {AD} $$ $$ = {{\left( {3 + ... |
Let $$\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 $$ is collinear with $$\overrighta... | ["A"]
Explanation:
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 $\overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}$ ... |
If $$C$$ is the mid point of $$AB$$ and $$P$$ is any point outside $$AB,$$ then :
Options:
[{"identifier": "A", "content": "$$\\overrightarrow {PA} + \\overrightarrow {PB} = 2\\overrightarrow {PC} $$ "}, {"identifier": "B", "content": "$$\\overrightarrow {PA} + \\overrightarrow {PB} = \\overrightarrow {PC} $$ "},... | ["A"]
Explanation:
$$\overrightarrow {PA} + \overrightarrow {AP = 0} $$ and $$\overrightarrow {PC} + \overrightarrow {CP} = 0$$
<br><br>$$ \Rightarrow \overrightarrow {PA} + \overrightarrow {AC} + \overrightarrow {CP} = 0$$
<br><br>and
<br><br>$$\overrightarrow {PB} + \overrightarrow {BC} + \overrightarrow {... |
Let $$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$$ is :
Options:
[{"identifier": "A", "content": "the Geometric Mean of $$a$$ and $$b$$"}, {"identifi... | ["A"]
Explanation:
Vector $$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 &... |
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 :
Options:
[{"identi... | ["D"]
Explanation:
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 $$\pi .$$ |
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... | ["D"]
Explanation:
<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 ... |
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 :
Options:
[{"identifier": "A", "content": "$$\\sqrt {18} $$ "}, {"identifier": "B", "content"... | ["C"]
Explanation:
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>$$ = \sqrt {16 + 1 + 1... |
Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$$ respectively, then the position vector of the orthocentre of this triangle, is :... | ["D"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264248/exam_images/rtiinzvghawcvhtcjcms.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Mathematics - Vector Algebra Question 183 English Ex... |
If the position vectors of the vertices A, B and C of a $$\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 $$A meets BC is :
... | ["C"]
Explanation:
Suppose angular bisector of A meets BC at D(x, , z)
<br><br>Using angular bisector theorem,
<br><br>$${{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 {{... |
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 $$\... | ["D"]
Explanation:
$$\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\la... |
The lines
<br/>$$\overrightarrow r = \left( {\widehat i - \widehat j} \right) + l\left( {2\widehat i + \widehat k} \right)$$ and
<br/>$$\overrightarrow r = \left( {2\widehat i - \widehat j} \right) + m\left( {\widehat i + \widehat j + \widehat k} \right)$$
Options:
[{"identifier": "A", "content": "do not intersect ... | ["A"]
Explanation:
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> = $$\overrightarrow r = \left( {2\wideh... |
If $$\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|$$ is_____.</br>
Options:
[] | 4
Explanation:
Let angle between $$\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( {\... |
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 :
Options:
[{"identifier": "A", "content": "$${1 \\o... | ["A"]
Explanation:
$$\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 - ... |
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 equa... | ["B"]
Explanation:
<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 Exp... |
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 ... | ["D"]
Explanation:
$${\left| {\overrightarrow a } \right|_{old}} = {\left| {\overrightarrow a } \right|_{new}}$$<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>... |
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