JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 1)
Let f : R $$\to$$ R be a continuous function. Then $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$ is equal to :
f (2)
2f (2)
2f $$\left( {\sqrt 2 } \right)$$
4f (2)
Explanation
$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$
= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}.{{\left[ {f({{\sec }^2}x).2\sec x.\sec x\tan x} \right]} \over {2x}}$$
= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}f({\sec ^2}x){\sec ^3}x.{{\sin x} \over x}$$
= $${\pi \over 4}f(2).{\left( {\sqrt 2 } \right)^3}.{1 \over {\sqrt 2 }} \times {4 \over \pi }$$
= 2f (2)
= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}.{{\left[ {f({{\sec }^2}x).2\sec x.\sec x\tan x} \right]} \over {2x}}$$
= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}f({\sec ^2}x){\sec ^3}x.{{\sin x} \over x}$$
= $${\pi \over 4}f(2).{\left( {\sqrt 2 } \right)^3}.{1 \over {\sqrt 2 }} \times {4 \over \pi }$$
= 2f (2)
Comments (0)
