JEE MAIN - Mathematics (2022 - 27th June Evening Shift - No. 4)
Let f be a differentiable function in $$\left( {0,{\pi \over 2}} \right)$$. If $$\int\limits_{\cos x}^1 {{t^2}\,f(t)dt = {{\sin }^3}x + \cos x} $$, then $${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right)$$ is equal to
$$6 - 9\sqrt 2 $$
$$6 - {9 \over {\sqrt 2 }}$$
$${9 \over 2} - 6\sqrt 2 $$
$${9 \over {\sqrt 2 }} - 6$$
Explanation
$$\int\limits_{\cos x}^1 {{t^2}f(t)dt = {{\sin }^3}x + \cos x} $$
$$ \Rightarrow \sin x{\cos ^2}x\,f(\cos x) = 3{\sin ^2}x\cos x - \sin x$$
$$ \Rightarrow f(\cos x) = 3\tan x - {\sec ^2}x$$
$$ \Rightarrow f'(\cos x).\,( - \sin x) = 3{\sec ^2}x - 2{\sec ^2}x\tan x$$
_27th_June_Evening_Shift_en_4_1.png)
Put $$\cos x = {1 \over {\sqrt 3 }}$$,
$$\therefore$$ $$f'\left( {{1 \over {\sqrt 3 }}} \right)\left( { - {{\sqrt 2 } \over {\sqrt 3 }}} \right) = 9 - 6\sqrt 2 $$
$${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right) = 6 - {9 \over {\sqrt 2 }}$$
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