JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 13)
If $$\theta $$1
and $$\theta $$2
be respectively the smallest and the largest values of $$\theta $$ in (0, 2$$\pi $$) - {$$\pi $$} which satisfy
the equation,
2cot2$$\theta $$ - $${5 \over {\sin \theta }}$$ + 4 = 0, then
$$\int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } $$ is equal to :
2cot2$$\theta $$ - $${5 \over {\sin \theta }}$$ + 4 = 0, then
$$\int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } $$ is equal to :
$${\pi \over 9}$$
$${{2\pi } \over 3}$$
$${{\pi } \over 3}$$
$${\pi \over 3} + {1 \over 6}$$
Explanation
2cot2$$\theta $$ - $${5 \over {\sin \theta }}$$ + 4 = 0
$$ \Rightarrow $$ $$2{{{{\cos }^2}\theta } \over {{{\sin }^2}\theta }} - {5 \over {\sin \theta }} + 4$$ = 0
$$ \Rightarrow $$ 2sin2 $$\theta $$ – 5sin$$\theta $$ + 2 = 0
$$ \Rightarrow $$ (2sin$$\theta $$ – 1)(sin$$\theta $$ – 2) = 0
$$ \therefore $$ sin$$\theta $$ = $${1 \over 2}$$ [ sin$$\theta $$ = 2 not possible]
$$ \therefore $$ $$\theta $$1 = $${\pi \over 6}$$ and $$\theta $$2 = $${{5\pi } \over 6}$$ as $$\theta $$1 $$<$$ $$\theta $$2
$$ \therefore $$ I = $$\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{\cos }^2}3\theta d\theta } $$
= $$\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{1 + \cos 6\theta } \over 2}d\theta } $$
= $${1 \over 2}\left[ {\theta + {{\sin 6\theta } \over 2}} \right]_{{\pi \over 6}}^{{{5\pi } \over 6}}$$
= $${{\pi \over 3}}$$
$$ \Rightarrow $$ $$2{{{{\cos }^2}\theta } \over {{{\sin }^2}\theta }} - {5 \over {\sin \theta }} + 4$$ = 0
$$ \Rightarrow $$ 2sin2 $$\theta $$ – 5sin$$\theta $$ + 2 = 0
$$ \Rightarrow $$ (2sin$$\theta $$ – 1)(sin$$\theta $$ – 2) = 0
$$ \therefore $$ sin$$\theta $$ = $${1 \over 2}$$ [ sin$$\theta $$ = 2 not possible]
$$ \therefore $$ $$\theta $$1 = $${\pi \over 6}$$ and $$\theta $$2 = $${{5\pi } \over 6}$$ as $$\theta $$1 $$<$$ $$\theta $$2
$$ \therefore $$ I = $$\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{\cos }^2}3\theta d\theta } $$
= $$\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{1 + \cos 6\theta } \over 2}d\theta } $$
= $${1 \over 2}\left[ {\theta + {{\sin 6\theta } \over 2}} \right]_{{\pi \over 6}}^{{{5\pi } \over 6}}$$
= $${{\pi \over 3}}$$
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