1

### JEE Main 2019 (Online) 9th January Morning Slot

For any $\theta \in \left( {{\pi \over 4},{\pi \over 2}} \right)$, the expression

$3{(\cos \theta - \sin \theta )^4}$$+ 6{(\sin \theta + \cos \theta )^2} + 4{\sin ^6}\theta equals : A 13 – 4 cos2\theta + 6sin2\theta cos2\theta B 13 – 4 cos6\theta C 13 – 4 cos2\theta + 6cos2\theta D 13 – 4 cos4\theta + 2sin2\theta cos2\theta ## Explanation Given, 3(sin\theta - cos\theta )4 + 6(sin\theta + cos\theta )2 + 4sin6\theta = 3[(sin\theta - cos\theta )2]2 + 6 (sin2\theta + cos2\theta + 2sin\theta cos\theta ) + 4sin6\theta = 3[sin2\theta + cos2\theta -2sin\theta cos\theta ]2 + 6(1 + sin2\theta ) + 4sin6\theta = 3(1 - sin2\theta )2 + 6(1 + sin2\theta ) + 4sin6\theta = 3 (1 - 2 sin2\theta + sin22\theta ) + 6 + 6sin2\theta + 4sin6\theta = 3 - 6sin2\theta + 3sin22\theta + 6 + 6sin2\theta + 4sin6\theta = 9 + 3sin22\theta + 4 sin6\theta = 9 + 3(2sin\theta cos\theta )2 + 4(1 - cos2\theta )3 = 9 + 12sin2\theta cos2\theta + 4 (1 - cos6\theta - 3cos2\theta + 3cos4\theta ) = 13 + 12 (1 - cos2\theta - 4cos6\theta - 12cos\theta + 12 cos4\theta = 13 + 12 cos2\theta - 12 cos4\theta - 4cos6\theta - 12 cos2\theta + 12 cos4\theta = 13 - 4 cos6\theta 2 MCQ (Single Correct Answer) ### JEE Main 2019 (Online) 9th January Evening Slot If 0 \le x < {\pi \over 2}, then the number of values of x for which sin x - sin 2x + sin 3x = 0, is : A 3 B 1 C 4 D 2 ## Explanation sin x - sin 2x + sin 3x = 0 x \in \left[ {0,{\pi \over 2}} \right) \Rightarrow (sin3x + sinx) - sin2x = 0 \Rightarrow 2sin2x.cos2x - sin2x = 0 \Rightarrow sin2x (2cosx - 1) = 0 sin 2x = 0 x = 0 and cos x = {1 \over 2} and x = {\pi \over 3} two solutions 3 MCQ (Single Correct Answer) ### JEE Main 2019 (Online) 10th January Morning Slot The sum of all values of \theta \in$$\left( {0,{\pi \over 2}} \right)$ satisfying
sin2 2$\theta$ + cos4 2$\theta$ = ${3 \over 4}$ is -
A
${{5\pi } \over 4}$
B
${\pi \over 2}$
C
$\pi$
D
${{3\pi } \over 8}$

## Explanation

sin22$\theta$ + cos42$\theta$ = ${3 \over 4}, $$\theta \in \left( {0,{\pi \over 2}} \right) \Rightarrow 1 - cos22\theta + cos42\theta = {3 \over 4} \Rightarrow 4cos2\theta - 4cos22\theta + 1 = 0 \Rightarrow (2cos22\theta - 1)2 = 0 \Rightarrow cos22\theta = {1 \over 2} = cos2{{\pi \over 4}} \Rightarrow 2\theta = n\pi \pm {\pi \over 4}, n \in I \Rightarrow \theta = {{n\pi } \over 2} \pm {\pi \over 8} \Rightarrow \theta = {\pi \over 8},{\pi \over 2} - {\pi \over 8} Sum of solutions {\pi \over 2} 4 MCQ (Single Correct Answer) ### JEE Main 2019 (Online) 10th January Evening Slot The value of \cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}} is - A {1 \over {256}} B {1 \over {2}} C {1 \over {1024}} D {1 \over {512}} ## Explanation Given \cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}} Let {\pi \over {{2^{10}}}}\, = \,\theta \therefore {\pi \over {{2^9}}}\, = \,2\theta {\pi \over {{2^8}}}\, = \,{2^2}\theta {\pi \over {{2^7}}}\, = \,{2^3}\theta . . {\pi \over {{2^2}}}\, = \,{2^8}\theta So given term becomes, \cos {2^8}\theta .\cos {2^7}\theta .....\cos \theta$$.\sin {\pi \over {{2^{10}}}}$

= $(\cos \theta .\cos 2\theta ......\cos {2^8}\theta )\sin {\pi \over {{2^{10}}}}$

= ${{\sin {2^9}\theta } \over {{2^9}\sin \theta }}.\sin {\pi \over {{2^{10}}}}$

= ${{\sin {2^9}\left( {{\pi \over {{2^{10}}}}} \right)} \over {{2^9}\sin {\pi \over {{2^{10}}}}}}.\sin {\pi \over {{2^{10}}}}$

= ${{\sin \left( {{\pi \over 2}} \right)} \over {{2^9}}}$

= ${1 \over {{2^9}}}$ = ${1 \over {512}}$

Note :
$(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )$ = ${{\sin {2^n}\theta } \over {{2^n}\sin \theta }}$

NEET