1981
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1981
The general solution of the trigonometric equation sin x+cos x=1 is given by:
A.
$2n\pi ;\,n = 0,\, \pm 1,\, \pm 2....$
B.
$x = 2n\pi + \pi /2;\,n = 0,\, \pm 1,\, \pm 2....$
C.
$x = n\pi + {\left( { - 1} \right)^n}\,\,\,\,\,\,\,{\pi \over 4} - {\pi \over 4}$ ; $n = 0,\, \pm 1,\, \pm 2..$
Show Answer
1980
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1980
The equation $\,2{\cos ^2}{x \over 2}{\sin ^2}x = {x^2} + {x^{ - 2}};\,0 < x \le {\pi \over 2}$ has
C.
more than one solution
Show Answer
1980
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1980
Given $A = {\sin ^2}\theta + {\cos ^4}\theta $ then for all real values of $\theta $
B.
${3 \over 4} \le A \le 1$
C.
${13\over 16} \le A \le 1$
D.
${3 \over 4} \le A \le {{13} \over {16}}$
Show Answer
1979
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1979
If $\alpha + \beta + \gamma = 2\pi ,$ then
A.
$tan{\alpha \over 2} + \tan {\beta \over 2} + \tan {\gamma \over 2} = \tan {\alpha \over 2}\tan {\beta \over 2}\tan {\gamma \over 2}$
B.
$\tan {\alpha \over 2}\tan {\beta \over 2} + \tan {\beta \over 2}\tan {\gamma \over 2} + \tan {\gamma \over 2}\tan {\alpha \over 2} = 1$
C.
$tan{\alpha \over 2} + \tan {\beta \over 2} + \tan {\gamma \over 2} = - \tan {\alpha \over 2}\tan {\beta \over 2}\tan {\gamma \over 2}$
Show Answer
1979
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1979
If $\tan \theta = - {4 \over 3},then\sin \theta \,is\,$
A.
$ - {4 \over 5}\,but\,not\,{4 \over 5}$
B.
$ - {4 \over 5}\,or\,{4 \over 5}$
C.
${4 \over 5}\,\,but\,not\, - {4 \over 5}$
Show Answer
2025
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2025 Paper 2 Online
Let
$ \alpha=\frac{1}{\sin 60^{\circ} \sin 61^{\circ}}+\frac{1}{\sin 62^{\circ} \sin 63^{\circ}}+\cdots+\frac{1}{\sin 118^{\circ} \sin 119^{\circ}} $
Then the value of
$ \left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2 $
is _____________.
Show Answer
Correct Answer: 3
Explanation:
To solve for the value of $\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2$, we begin by considering the expression for $\alpha$:
$ \alpha = \frac{1}{\sin 60^{\circ} \cdot \sin 61^{\circ}} + \frac{1}{\sin 62^{\circ} \cdot \sin 63^{\circ}} + \cdots + \frac{1}{\sin 118^{\circ} \cdot \sin 119^{\circ}} $
Using the identity for the product of sines:
$ \sin(x-y) = \sin x \cdot \cos y - \cos x \cdot \sin y $
we can transform each term:
$ \sin 1^{\circ} \cdot \alpha = \frac{\sin(61^{\circ} - 60^{\circ})}{\sin 60^{\circ} \cdot \sin 61^{\circ}} + \frac{\sin(63^{\circ} - 62^{\circ})}{\sin 62^{\circ} \cdot \sin 63^{\circ}} + \dots + \frac{\sin(119^{\circ} - 118^{\circ})}{\sin 118^{\circ} \cdot \sin 119^{\circ}} $
This simplifies to:
$ \sin 1^{\circ} \cdot \alpha = \cot 60^{\circ} - \cot 61^{\circ} + \cot 62^{\circ} - \cot 63^{\circ} + \cdots + \cot 118^{\circ} - \cot 119^{\circ} $
This forms a telescoping series, simplifying to:
$ \sin 1^{\circ} \cdot \alpha = \cot 60^{\circ} $
Therefore:
$ \alpha = \frac{\cot 60^{\circ}}{\sin 1^{\circ}} = \frac{1/\sqrt{3}}{\sin 1^{\circ}} = \frac{\operatorname{cosec} 1^{\circ}}{\sqrt{3}} $
Finally, we calculate:
$ \left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2 = \left(\frac{\operatorname{cosec} 1^{\circ}}{\frac{\operatorname{cosec} 1^{\circ}}{\sqrt{3}}}\right)^2 = 3 $
2023
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2023 Paper 2 Online
$
\text { Then the inradius of the triangle } A B C \text { is }
$ :
Show Answer
Correct Answer: 0.25
Explanation:
In radius $ r=\frac{\Delta}{S}=\left[\frac{a}{2 R(\sin A+\sin B+\sin C)}\right)$
2022
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2022 Paper 2 Online
Let $\alpha$ and $\beta$ be real numbers such that $-\frac{\pi}{4}<\beta<0<\alpha<\frac{\pi}{4}$.
Show Answer
Correct Answer: 1
Explanation:
Given, $\sin (\alpha + \beta ) = {1 \over 3}$
and $\cos (\alpha - \beta ) = {2 \over 3}$
Let, $E = {{\sin \alpha } \over {\cos \beta }} + {{\cos \beta } \over {\sin \alpha }} + {{\cos \alpha } \over {\sin \beta }} + {{\sin \beta } \over {\cos \alpha }}$
$ = {{\sin \alpha } \over {\cos \beta }} + {{\cos \alpha } \over {\sin \beta }} + {{\cos \beta } \over {\sin \alpha }} + {{\sin \beta } \over {\cos \alpha }}$
$ = {{\sin \alpha \sin \beta + \cos \alpha \cos \beta } \over {\sin \beta \cos \beta }} + {{\cos \alpha \cos \beta + \sin \alpha \sin \beta } \over {\sin \alpha \cos \alpha }}$
$ = {{\cos (\alpha - \beta )} \over {\sin \beta \cos \beta }} + {{\cos (\alpha - \beta )} \over {\sin \alpha \cos \alpha }}$
$ = \cos (\alpha - \beta )\left[ {{2 \over {2\sin \beta \cos \beta }} + {2 \over {2\sin \alpha \cos \alpha }}} \right]$
$ = {2 \over 3}\left[ {{2 \over {\sin 2\beta }} + {2 \over {\sin 2\alpha }}} \right]$
$ = {4 \over 3}\left[ {{1 \over {\sin 2\beta }} + {1 \over {\sin 2\alpha }}} \right]$
$ = {4 \over 3}\left[ {{{\sin 2\alpha + \sin 2\beta } \over {\sin 2\alpha \sin 2\beta }}} \right]$
$ = {{4 \times 2} \over 3}\left[ {{{2\sin \left( {{{2\alpha + 2\beta } \over 2}} \right)\cos \left( {{{2\alpha - 2\beta } \over 2}} \right)} \over {2\sin 2\alpha \sin 2\beta }}} \right]$
$ = {{16} \over 3}\left[ {{{\sin (\alpha + \beta )\cos (\alpha - \beta )} \over {\cos (2\alpha - 2\beta ) - cos(2\alpha + 2\beta )}}} \right]$
$ = {{16} \over 3}\left[ {{{{1 \over 3} \times {2 \over 3}} \over {(2{{\cos }^2}(\alpha - \beta ) - 1) - (1 - 2{{\sin }^2}(\alpha + \beta ))}}} \right]$
$ = {{32} \over {27}}\left[ {{1 \over {2 \times {4 \over 9} - 2 + 2 \times {1 \over 9}}}} \right]$
$ = {{32} \over {27}}\left[ {{9 \over {8 - 18 + 2}}} \right]$
$ = {{32} \over {27}}\left[ {{9 \over { - 8}}} \right]$
$ = - {4 \over 3}$
$\therefore$ ${E^2} = {{16} \over 9} = 1.77$
$ \Rightarrow [{E^2}] = 1$
2018
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2018 Paper 2 Offline
Let f : R $ \to $ R be a differentiable function with f(0) = 1 and satisfying the equation f(x + y) = f(x) f'(y) + f'(x) f(y) for all x, y$ \in $ R.e (f(4)) is ...........
Show Answer
Correct Answer: 2
Explanation:
Given, C = A
2015
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2015 Paper 1 Offline
The number of distinct solutions of the equation
Show Answer
Correct Answer: 8
Explanation:
Given: $\frac{5}{4} \cos ^2 2 x+\cos ^4 x+\sin ^4 x+\cos ^6 x+$ $\sin ^6 x=2$
2011
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2011 Paper 1 Offline
The positive integer value of $n\, > \,3$ satisfying the equation ${1 \over {\sin \left( {{\pi \over n}} \right)}} = {1 \over {\sin \left( {{{2\pi } \over n}} \right)}} + {1 \over {\sin \left( {{{3\pi } \over n}} \right)}}$ is
Show Answer
Correct Answer: 7
Explanation:
We have,
${1 \over {\sin (\pi /n)}} - {1 \over {\sin (3\pi /n)}} = {1 \over {\sin (2\pi /n)}}$
$ \Rightarrow {{\sin (3\pi /n) - \sin (\pi /n)} \over {\sin (\pi /n)\sin (3\pi /n)}} = {1 \over {\sin (2\pi /n)}}{{(2\sin (\pi /n)\cos (2\pi /n))\sin (2\pi /n)} \over {\sin (\pi /n)\sin (3\pi /n)}} = 1$
$ \Rightarrow \sin {{4\pi } \over n} = \sin {{3\pi } \over n} \Rightarrow {{4\pi } \over n} + {{3\pi } \over n} = \pi \Rightarrow n = 7$
2010
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2010 Paper 1 Offline
The number of all possible values of $\theta $ where $0 < \theta < \pi ,$ for which the system of equations
$$\left( {y + z} \right)\cos {\mkern 1mu} 3\theta = \left( {xyz} \right){\mkern 1mu} \sin 3\theta $$
$$x\sin 3\theta = {{2\cos 3\theta } \over y} + {{2\sin 3\theta } \over z}$$
$$\left( {xyz} \right){\mkern 1mu} \sin 3\theta = \left( {y + 2z} \right){\mkern 1mu} \cos 3\theta + y{\mkern 1mu} sin3\theta $$
have a solution $\left( {{x_0},{y_0},{z_0}} \right)$ with ${y_0}{z_0}{\mkern 1mu} \ne {\mkern 1mu} 0,$ is
Show Answer
Correct Answer: 3
Explanation:
View the equation in xyz, y and t.
We have,
$(xyz)\sin 3\theta - y\cos 3\theta - z\cos 3\theta = 0$
$(xyz)\sin 3\theta - 2y\sin 3\theta - 2z\cos 3\theta = 0$
$(xyz)\sin 3\theta - y(\cos 3\theta + \sin 3\theta ) - 2z\cos 3\theta = 0$
$xyz \ne 0$
Hence, the equation has non-trivial solution which gives
$\left| {\matrix{
{\sin 3\theta } & { - \cos 3\theta } & { - \cos 3\theta } \cr
{\sin 3\theta } & { - 2\sin 3\theta } & { - 2\cos 3\theta } \cr
{\sin 3\theta } & { - (\cos 3\theta + \sin 3\theta )} & { - 2\cos 3\theta } \cr
} } \right| = 0$
$ \Rightarrow \sin 3\theta \cos 3\theta (\sin 3\theta - \cos 3\theta ) = 0$
$ \Rightarrow \sin 3\theta = 0$ then $xyz = 0$ (not possible)
$\cos 3\theta = 0$ not possible
$\sin 3\theta = \cos 3\theta \Rightarrow \tan 3\theta = 1$
$3\theta = n\pi + {\pi \over 4},n \in z$
$\theta = {{n\pi } \over 3} + {\pi \over {12}}$ ; $\theta = {\pi \over {12}},{{5\pi } \over {12}},{{9\pi } \over {12}}$
Thus there are 3 solutions.
2010
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2010 Paper 1 Offline
The number of values of $\theta $ in the interval, $\left( { - {\pi \over 2},\,{\pi \over 2}} \right)$ such
that$\,\theta \ne {{n\pi } \over 5}$ for $n = 0,\, \pm 1,\, \pm 2$ and $\tan \,\theta = \cot \,5\theta \,$ as well as $\sin \,2\theta = \cos \,4 \theta $ is
Show Answer
Correct Answer: 3
Explanation:
Given, $\tan \theta = \cot 5\theta $
$ \Rightarrow \tan \theta = \tan \left( {{\pi \over 2} - 5\theta } \right)$
$ \Rightarrow {\pi \over 2} - 5\theta = n\pi + \theta $
$ \Rightarrow 6\theta = {\pi \over 2} - {{n\pi } \over 6}$
$ \Rightarrow \theta = {\pi \over {12}} - {{n\pi } \over 6}$
Also $\cos 4\theta = \sin 2\theta = \cos \left( {{\pi \over 2} - 2\theta } \right)$
$ \Rightarrow 4\theta = 2n\pi \, \pm \,\left( {{\pi \over 2} - 2\theta } \right)$
Taking positive
$6\theta = 2n\pi + {\pi \over 2} \Rightarrow \theta = {{n\pi } \over 3} + {\pi \over {12}}$
Taking negative
$2\theta = 2n\pi - {\pi \over 2} \Rightarrow \theta = n\pi - {\pi \over 4}$
Above values of $\theta$ suggests that there are only 3 common solutions.
2010
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2010 Paper 1 Offline
The maximum value of the expression ${1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}$ is
Show Answer
Correct Answer: 2
Explanation:
Let
$f(\theta ) = {1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}$
Again let
$g(\theta ) = {\sin ^2}\theta + 3\sin \theta \cos \theta + 5{\cos ^2}\theta $
$ = {{1 - \cos 2\theta } \over 2} + 5\left( {{{1 + \cos 2\theta } \over 2}} \right) + {3 \over 2}\sin 2\theta $
$ = 3 + 2\cos 2\theta + {3 \over 2}\sin 2\theta $
$\therefore$ $g{(\theta )_{\min }} = 3 - \sqrt {4 + {9 \over 4}} = 3 - {5 \over 2} = {1 \over 2}$
$\therefore$ $f(\theta ) = {1 \over {g{{(\theta )}_{\min }}}} = 2$
2010
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2010 Paper 2 Offline
Two parallel chords of a circle of radius 2 are at a distance $\sqrt 3 + 1$ apart. If the chords subtend at the center , angles of ${\pi \over k}$ and ${{2\pi } \over k},$ where$k > 0,$ then the value of $\left[ k \right]$ is
[Note :[k] denotes the largest integer less than or equal to k ]
Show Answer
Correct Answer: 3
Explanation:
Let $\theta = {\pi \over {2k}}$
$\cos \theta = {x \over 2}$
$ \Rightarrow \cos 2\theta = {{\sqrt 3 + 1 - x} \over 2}$
$ \Rightarrow 2{\cos ^2}\theta - 1 = {{\sqrt 3 + 1 - x} \over 2}$
$ \Rightarrow 2\left( {{{{x^2}} \over 4}} \right) - 1 = {{\sqrt 3 + 1 - x} \over 2}$
$ \Rightarrow {x^2} + x - 3 - \sqrt 3 = 0$
$ \Rightarrow x = {{ - 1 \pm \sqrt {1 + 12 + 4\sqrt 3 } } \over 2}$
$ = {{ - 1 \pm \sqrt {13 + 4\sqrt 3 } } \over 2}$
$ = {{ - 1 + 2\sqrt 3 + 1} \over 2} = \sqrt 3 $
$\therefore$ $\cos \theta = {{\sqrt 3 } \over 2} \Rightarrow \theta = {\pi \over 6}$
$\therefore$ Required angle $ = {\pi \over k} = 2\theta = {\pi \over 3}$
$ \Rightarrow k = 3$
2022
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2022 Paper 2 Online
Let $P Q R S$ be a quadrilateral in a plane, where
D.
$(2 \sqrt{2}, 3 \sqrt{2})$
Show Answer
Correct Answer: A,B
Explanation:
Applying sine rule in $\triangle P Q R$,
$
\begin{aligned}
& \frac{\alpha}{\sin 30^{\circ}}=\frac{1}{\sin 80^{\circ}} \\\\
\Rightarrow & \alpha=\frac{1}{2 \sin 80^{\circ}}
\end{aligned}
$
Applying sine rule in $\triangle P R S$,
$
\begin{aligned}
& \frac{\beta}{\sin 40^{\circ}}=\frac{1}{\sin \theta} \\\\
\Rightarrow \beta \sin \theta=& \sin 40^{\circ}
\end{aligned}
$
From (i) and (ii),
$
4 \alpha \beta \sin \theta=4 \cdot \frac{1}{2 \sin 80^{\circ}} \cdot \sin 40^{\circ}=\frac{1}{\cos 40^{\circ}}
$
Using $\cos 0^{\circ}=1, \cos 30^{\circ}=\frac{\sqrt{3}}{2}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}$ and $\cos 60^{\circ}=\frac{1}{2}$
Only Options (A) and (B) are correct.
2019
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2019 Paper 2 Offline
For non-negative integers n, let$-1$ x takes values in [0, $\pi $], which of the following options is/are correct?
A.
If $\alpha $ = tan(cos$-$1 f(6)), then $\alpha $2 + 2$\alpha $ $-$1 = 0
B.
$f(4) = {{\sqrt 3 } \over 2}$
C.
sin(7 cos$-$1 f(5)) = 0
D.
$\mathop {\lim }\limits_{n \to \infty } \,f(n) = {1 \over 2}$
Show Answer
Correct Answer: A,B,C
Explanation:
It is given, that for non-negative integers 'n',
2018
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2018 Paper 1 Offline
In a $\Delta $PQR = 30$^\circ $ and the sides PQ and QR have lengths 10$\sqrt 3 $ and 10, respectively. Then, which of the following statement(s) is(are) TRUE?
A.
$\angle QPR = 45^\circ $
B.
The area of the $\Delta PQR$ is $25\sqrt 3 $ and $\angle QRP = 120^\circ $
C.
The radius of the incircle of the $\Delta PQR$ is $10\sqrt 3 $ $-$ 15
D.
The area of the circumcircle of the $\Delta PQR$ is 100$\pi $
Show Answer
Correct Answer: B,C,D
Explanation:
We have,
In $\Delta $PQR
$\angle PQR = 30^\circ $
$PQ = 10\sqrt 3 $
$QR = 10$
By cosine rule
$\cos 30^\circ = {{P{Q^2} + Q{R^2} - P{R^2}} \over {2PQ.QR}}$
$ \Rightarrow {{\sqrt 3 } \over 2} = {{300 + 100 - P{R^2}} \over {200\sqrt 3 }}$
$ \Rightarrow 300 = 300 + 100 - P{R^2}$
$ \Rightarrow PR = 10$
Since, $PR = QR = 10$
$ \therefore $ $\angle QPR = 30^\circ $ and $\angle QRP = 120^\circ $
Area of $\Delta PQR = {1 \over 2}PQ.QR.\sin 30^\circ $
$ = {1 \over 2} \times 10\sqrt 3 \times 10 \times {1 \over 2} = 25\sqrt 3 $
Radius of incircle of
$\Delta PQR = {{Area\,of\,\Delta PQR} \over {Semi - perimetre\,of\,\Delta PQR}}$
i.e. $r = {\Delta \over s} = {{25\sqrt 3 } \over {{{10\sqrt 3 + 10 + 10} \over 2}}} = {{25\sqrt 3 } \over {5(\sqrt 3 + 2)}}$
$ \Rightarrow r = 5\sqrt 3 (2 - \sqrt 3 ) = 10\sqrt 3 - 15$
and radius of circumcircle
$(R) = {{abc} \over {4\Delta }} = {{10\sqrt 3 \times 10 \times 10} \over {4 \times 25\sqrt 3 }} = 10$
$ \therefore $ Area of circumcircle of
$\Delta PQR = \pi {R^2} = 100\pi $
Hence, option (b), (c) and (d) are correct answer.
2017
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2017 Paper 2 Offline
Let $\alpha $ and $\beta $ be non zero real numbers such that $2(\cos \beta - \cos \alpha ) + \cos \alpha \cos \beta = 1$. Then which of the following is/are true?
A.
$\sqrt 3 \tan \left( {{\alpha \over 2}} \right) - \tan \left( {{\beta \over 2}} \right) = 2$
B.
$\tan \left( {{\alpha \over 2}} \right) - \sqrt 3 \tan \left( {{\beta \over 2}} \right) = 0$
C.
$\tan \left( {{\alpha \over 2}} \right) + \sqrt 3 \tan \left( {{\beta \over 2}} \right) = 0$
D.
$\sqrt 3 \tan \left( {{\alpha \over 2}} \right) + \tan \left( {{\beta \over 2}} \right) = 2$
Show Answer
Correct Answer: B,C
Explanation:
$2(\cos \beta - \cos \alpha ) + \cos \alpha \cos \beta = 1$
2013
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2013 Paper 1 Offline
Let $f\left( x \right) = x\sin \,\pi x,\,x > 0.$ Then for all natural numbers $n,\,f'\left( x \right)$ vanishes at
A.
A unique point in the interval $\left( {n,\,n + {1 \over 2}} \right)$
B.
A unique point in the interval $\left( {n + {1 \over 2},n + 1} \right)$
C.
A unique point in the interval $\left( {n,\,n + 1} \right)$
D.
Two points in the interval $\left( {n,\,n + 1} \right)$
Show Answer
Correct Answer: B,C
Explanation:
Given, ,$f(x)=x \cdot \sin \pi x, x>0$
$\Rightarrow \quad f^{\prime}(x)=1 \cdot \sin \pi x+x \cdot \pi \cos \pi x$
Apply $f^{\prime}(x)=0$
$\begin{aligned}
& \Rightarrow \sin \pi x+\pi x \cos \pi x=0 \\
& \Rightarrow \sin \pi x=-\pi x \cos \pi x \\
& \Rightarrow \frac{\sin \pi x}{\cos \pi x}=-\pi x \\
& \Rightarrow \tan \pi x=-\pi x
\end{aligned}$
Draw the graph of $y=\tan \pi x$ and $y=-\pi x$ for $x>0$
It is clear that $y=\tan \pi x$ and $y=-\pi x$ intersect at a unique point if $\frac{1}{2} < x < 1$ as $\frac{3}{2} < x < 2$ as $\frac{5}{2} < x < 3$ or.....
So, $y=\tan \pi x$ and $y=-\pi x$ intersect at a unique point is $x \in\left(n+\frac{1}{2}, n+1\right)$ as $(n, n+1)$.
Hints:
The number of point of intersection of graphs of $y=f(x)$ and $y=g(x)$ is equal to the number of solution of $f(x)=g(x)$
2012
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2012 Paper 1 Offline
Let $\theta ,\,\varphi \, \in \,\left[ {0,2\pi } \right]$ be such that
$2\cos \theta \left( {1 - \sin \,\varphi } \right) = {\sin ^2}\theta \,\,\left( {\tan {\theta \over 2} + \cot {\theta \over 2}} \right)\cos \varphi - 1,\,\tan \left( {2\pi - \theta } \right) > 0$ and $ - 1 < \sin \theta \, < - {{\sqrt 3 } \over 2},$
then $\varphi $ cannot satisfy
A.
$0 < \varphi < {\pi \over 2}$
B.
${\pi \over 2} < \varphi < {{4\pi } \over 3}$
C.
${{4\pi } \over 3} < \varphi < {{3\pi } \over 2}$
D.
${{3\pi } \over 2} < \varphi < 2\pi $
Show Answer
Correct Answer: C,D,A
Explanation:
Given, $\tan (2 \pi-\theta)>0$
$\begin{aligned}
& \Rightarrow 0<(2 \pi-\theta)<\frac{\pi}{2} \text { or } \pi<(2 \pi-\theta)<\frac{3 \pi}{2} \\
& \Rightarrow \frac{3 \pi}{2}<\theta<2 \pi \text { or } \frac{\pi}{2}<\theta<\pi \quad \text{... (i)}
\end{aligned}$
Also, $-1<\sin \theta<\frac{-\sqrt{3}}{2}$
$\Rightarrow \frac{3 \pi}{2}<\theta<\frac{5 \pi}{3} \quad \text{... (ii)}$
From (i) and (ii),
$\Rightarrow \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\quad \text{... (iii)}$
Now,
$\begin{aligned}
& 2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1 \\
& \Rightarrow 2 \cos \theta(1-\sin \phi) \\
& =\sin ^2 \theta\left(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}+\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}\right) \cos \phi-1 \\
& \Rightarrow 2 \cos \theta(1-\sin \phi) \\
& =\sin ^2 \theta\left(\frac{\sin ^2 \frac{\theta}{2}+\cos ^2 \frac{\theta}{2}}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \cos \phi-1
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad 2 \cos \theta(1-\sin \phi) \\
&=\sin \theta \cdot \sin \theta\left(\frac{1}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \cos \phi-1
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad 2 \cos (1-\sin \phi)=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cdot \cos \phi \cdot \sin \theta}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}-1 \\
& \Rightarrow \quad 2 \cos \theta(1-\sin \phi)=2 \sin \theta \cos \phi-1 \\
& \Rightarrow \quad 2 \cos \theta+1=2 \sin (\theta+\phi) \\
& \text { As } \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right) \quad \text{[from (iii)]}\\
& \Rightarrow 2 \cos \theta+1 \in(1,2) \\
& \Rightarrow 1<2 \sin (\theta+\phi)<2 \\
& \Rightarrow \frac{1}{2}<\sin (\theta+\phi)<1
\end{aligned}$
$\begin{aligned}
& \text { Now, } \theta, \phi \in[0,2 \pi] \\
& \theta+\phi \in[0,4 \pi] \\
& \Rightarrow \theta+\phi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right) \text { or } \theta+\phi \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right) \\
& \Rightarrow \frac{\pi}{6}-\theta<\phi<\frac{5 \pi}{6}-\theta \text { or } \frac{13 \pi}{6}-\theta<\phi<\frac{17 \pi}{6}-\theta \\
& \Rightarrow \phi \in\left(-\frac{3 \pi}{2}, \frac{-2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right) \\
& {\left[\because \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\right] }
\end{aligned}$
2009
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2009 Paper 2 Offline
For $0 < \theta < {\pi \over 2},$ the solution (s) of
$$\sum\limits_{m = 1}^6 {\cos ec\,\left( {\theta + {{\left( {m - 1} \right)\pi } \over 4}} \right)\,\cos ec\,\left( {\theta + {{m\pi } \over 4}} \right) = 4\sqrt 2 } $$ is (are)
Show Answer
Correct Answer: C,D
Explanation:
Given solutions
${1 \over {\sin (\pi /4)}}\left[ {{{\sin (\theta + \pi /4 - \theta )} \over {\sin \theta \,.\,\sin (\theta + \pi /4)}} + {{\sin (\theta + \pi /2 - (\theta + \pi /4))} \over {\sin (\theta + \pi /4)\,.\,(\theta + \pi /2)}}\, + \,...\, + \,{{\sin ((\theta + 3\pi /2) - (\theta + 5\pi /4))} \over {\sin (\theta + 3\pi /2)\,.\,\sin (\theta + 5\pi /4)}}} \right] = 4\sqrt 2 $
$ \Rightarrow \sqrt 2 [\cos \theta - \cot (\theta + \pi /4) + \cot (\theta + \pi /4) - \cot (\theta + \pi /2)\, + \,...\, + \,\cot (\theta + 5\pi /4) - \cot (\theta + 3\pi /2)] = 4\sqrt 2 $
$ \Rightarrow \tan \theta + \cot \theta = 4 \Rightarrow \tan \theta = 2 \pm \sqrt 3 $
$ \Rightarrow \theta = {\pi \over {12}}$ or ${{5\pi } \over {12}}$
2009
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2009 Paper 1 Offline
If ${{{{\sin }^4}x} \over 2} + {{{{\cos }^4}x} \over 3} = {1 \over 5},$ then
A.
${\tan ^2}x = {2 \over 3}$
B.
${{{{\sin }^8}x} \over 8} + {{{{\cos }^8}x} \over {27}} = {1 \over {125}}$
C.
${\tan ^2}x = {1 \over 3}$
D.
${{{{\sin }^8}x} \over 8} + {{{{\cos }^8}x} \over {27}} = {2 \over {125}}$
Show Answer
Correct Answer: A,B
Explanation:
It is given that
${{{{\sin }^4}x} \over 2} + {{{{\cos }^4}x} \over 3} = {1 \over 5}$
$3{\sin ^4}x + 2{(1 - {\sin ^2}x)^2} = {6 \over 5}$
$ \Rightarrow 25{\sin ^4}x - 20{\sin ^2}x + 4 = 0$
$ \Rightarrow {\sin ^2}x = {2 \over 5}$ and ${\cos ^2}x = {3 \over 5}$
Hence, ${\tan ^2}x = {2 \over 3}$
Therefore, ${{{{\sin }^8}x} \over 8} + {{{{\cos }^8}x} \over {27}} = {1 \over {125}}$
1999
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1999
For a positive integer $\,n$, let
A.
${f_2}\left( {{\pi \over {16}}} \right) = 1$
B.
${f_3}\left( {{\pi \over {32}}} \right) = 1$
C.
${f_4}\left( {{\pi \over {64}}} \right) = 1$
D.
${f_5}\left( {{\pi \over {128}}} \right) = 1$
Show Answer
1988
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1988
The values of $\theta $ lying between $\theta = \theta $ and $\theta = \pi /2$ and satisfying the equation
$\left| {\matrix{
{1 + {{\sin }^2}\theta } & {{{\cos }^2}\theta } & {4\sin 4\theta } \cr
{{{\sin }^2}\theta } & {1 + {{\cos }^2}\theta } & {4\sin 4\theta } \cr
{{{\sin }^2}\theta } & {{{\cos }^2}\theta } & {1 + 4\sin 4\theta } \cr
} } \right| = 0$ are
Show Answer
2005
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2005
Find the range of values of $\,t$ for which $$2\,\sin \,t = {{1 - 2x + 5{x^2}} \over {3{x^2} - 2x - 1}},\,\,\,\,\,t\, \in \,\left[ { - {\pi \over 2},\,{\pi \over 2}} \right].$$
Show Answer
Correct Answer: $$\,\left[ {{{ - \pi } \over 2},\,{{ - \pi } \over {10}}} \right]\, \cup \,\left[ {{{3\pi } \over {10}},\,{\pi \over 2}} \right]$$
2000
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2000
In any triangle $ABC,$ prove that
$$\cot {A \over 2} + \cot {B \over 2} + \cot {C \over 2} = \cot {A \over 2}\cot {B \over 2}\cot {C \over 2}.$$
Show Answer
Correct Answer: Solve it.
1998
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1998
Prove that $\tan \,\alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot 8\alpha = \cot \alpha $
Show Answer
Correct Answer: Solve it.
1997
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1997
Prove that $\sum\limits_{k = 1}^{n - 1} {\left( {n - k} \right)\,\cos \,{{2k\pi } \over n} = - {n \over 2},} $ where $n \ge 3$ is an integer.
Show Answer
Correct Answer: Solve it.
1997
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1997
Prove that the values of the function ${{\sin x\cos 3x} \over {\sin 3x\cos x}}$ do not lie between ${1 \over 3}$ and 3 for any real $x.$
Show Answer
Correct Answer: Solve it.
1996
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1996
Find all values of $\theta $ in the interval $\left( { - {\pi \over 2},{\pi \over 2}} \right)$ satisfying the equation $\left( {1 - \tan \,\theta } \right)\left( {1 + \tan \,\theta } \right)\,\,{\sec ^2}\theta + \,\,{2^{{{\tan }^2}\theta }} = 0.$
Show Answer
Correct Answer: $$ \pm {\pi \over 3}$$
1995
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1995
Find the smallest positive number $p$ for which the equation $\cos \left( {p\,\sin x} \right) = \sin \left( {p\cos x} \right)$ has a solution $x\, \in \,\left[ {0,2\pi } \right]$.
Show Answer
Correct Answer: $${{\pi \sqrt 2 } \over 4}$$
1993
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1993
Determine the smallest positive value of number $x$ (in degrees) for which
$$\tan \left( {x + {{100}^ \circ }} \right) = \tan \left( {x + {{50}^ \circ }} \right)\,\tan \left( x \right)\tan \left( {x - {{50}^ \circ }} \right).$$
Show Answer
Correct Answer: $${30^ \circ }$$
1992
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1992
Show that the value of ${{\tan x} \over {\tan 3x}},$ wherever defined never lies between ${1 \over 3}$ and 3.
Show Answer
Correct Answer: Solve it.
1991
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1991
If $\exp \,\,\,\left\{ {\left( {\left( {{{\sin }^2}x + {{\sin }^4}x + {{\sin }^6}x + \,\,\,..............\infty } \right)\,In\,\,2} \right)} \right\}$ satiesfies the equation ${x^2} - 9x + 8 = 0,$ find the value of ${{\cos x} \over {\cos x + \sin x}},\,0 < x < {\pi \over 2}.$
Show Answer
Correct Answer: $${{\sqrt 3 - 1} \over 2}$$
1990
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1990
$ABC$ is a triangle such that
$$\sin \left( {2A + B} \right) = \sin \left( {C - A} \right) = \, - \sin \left( {B + 2C} \right) = {1 \over 2}.$$
If $A,\,B$ and $C$ are in arithmetic progression, determine the values of $A,\,B$ and $C$.
Show Answer
Correct Answer: $${45^ \circ },\,\,\,{60^ \circ },\,\,\,{75^ \circ }$$
1984
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1984
Find the values of $x \in \left( { - \pi , + \pi } \right)$ which satisfy the equation ${g^{(1 + \left| {\cos x} \right| + \left| {{{\cos }^2}x} \right| + \left| {{{\cos }^3}x} \right| + ...)}} = {4^3}$
Show Answer
Correct Answer: $$ \pm {\pi \over 3}, \pm {{2\pi } \over 3}$$
1983
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1983
Show that $$16\cos \left( {{{2\pi } \over {15}}} \right)\cos \left( {{{4\pi } \over {15}}} \right)\cos \left( {{{8\pi } \over {15}}} \right)\cos \left( {{{16\pi } \over {15}}} \right) = 1$$
Show Answer
Correct Answer: Solve by yourself
1983
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1983
Find all solutions of $4{\cos ^2}\,x\sin x - 2{\sin ^2}x = 3\sin x$
Show Answer
Correct Answer: Solve by yourself
1982
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1982
Without using tables, prove that $\left( {\sin \,{{12}^ \circ }} \right)\left( {\sin \,{{48}^ \circ }} \right)\left( {\sin \,{{54}^ \circ }} \right) = {1 \over 8}.$
Show Answer
Correct Answer: Solve it.
1980
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1980
Given $A = \left\{ {x:{\pi \over 6} \le x \le {\pi \over 3}} \right\}$ and
Show Answer
Correct Answer: $$\left[ {{1 \over 2} - {\pi \over 3}\left( {1 + {\pi \over 3}} \right),\,{{\sqrt 3 } \over 2} - {\pi \over 6}\left( {1 + {\pi \over 6}} \right)} \right]$$
1980
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1980
Given $\alpha + \beta - \gamma = \pi ,$ prove that
Show Answer
Correct Answer: Solve it.
1980
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1980
For all $\theta $ in $\left[ {0,\,\pi /2} \right]$ show that, $\cos \left( {\sin \theta } \right) \ge \,\sin \,\left( {\cos \theta } \right).$
Show Answer
Correct Answer: Solve it.
1979
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1979
(a) Draw the graph of $y = {1 \over {\sqrt 2 }}\left( {cinx + \cos x} \right)$ from $x = - {\pi \over 2}$ to $x = {\pi \over 2}$.
(b) If $\cos \left( {\alpha + \beta } \right) = {4 \over 5},\,\,\sin \,\left( {\alpha - \beta } \right) = \,{5 \over {13}},$ and $\alpha ,\,\beta $ lies between 0 and ${\pi \over 4}$, find tan2$\alpha $.
Show Answer
Correct Answer: $${56 \over {33}}$$
1978
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1978
If $\tan \alpha = {m \over {m + 1}}\,$ and $\tan \beta = {2 \over {2m + 1}},$ find the possible values of $\left( {\alpha + \beta } \right).$
Show Answer
Correct Answer: $$n\pi + {\pi \over 4}$$
1997
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1997
The real roots of the equation $\,{\cos ^7}x + {\sin ^4}x = 1$ in the interval $\left( { - \pi ,\pi } \right)$ are ...., ...., and ______.
Show Answer
Correct Answer: $$ - {\pi \over 2},{\pi \over 2},0$$
1996
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1996
General value of $\theta $ satisfying the equation ${\tan ^2}\theta + \sec \,2\,\theta = 1$ is _________.
Show Answer
Correct Answer: $$n\pi ,n\pi \pm {\pi \over 3}$$
1993
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1993
If $K = \sin \left( {\pi /18} \right)\sin \left( {5\pi /18} \right)\sin \left( {7\pi /18} \right),$ then the numerical value of K is ______.
Show Answer
Correct Answer: $${\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 8$}}$$
1993
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1993
If $A > 0,B > 0\,$ and $A + B = \pi /3,$ then the maximum value of tan A tan B is _______.
Show Answer
Correct Answer: $${\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 3$}}$$
1991
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1991
The value of
Show Answer
Correct Answer: $${\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle {64}$}}$$
