JEE Advanced 2014 Paper 1 Offline

MCQ (More than One Correct Answer)

-0

Let $$f:\left( { - {\pi \over 2},{\pi \over 2}} \right) \to R$$ be given by $$f(x) = {[\log (\sec x + \tan x)]^3}$$. Then,

f(x) is an odd function

f(x) is a one-one function

f(x) is an onto function

f(x) is an even function

IIT-JEE 2012 Paper 2 Offline

MCQ (More than One Correct Answer)

-1

Let $$f:( - 1,1) \to R$$ be such that $$f(\cos 4\theta ) = {2 \over {2 - {{\sec }^2}\theta }}$$ for $$\theta \in \left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 4},{\pi \over 2}} \right)$$. Then the value(s) of $$f\left( {{1 \over 3}} \right)$$ is(are)

$$1 - \sqrt {{3 \over 2}} $$

$$1 + \sqrt {{3 \over 2}} $$

$$1 - \sqrt {{2 \over 3}} $$

$$1 + \sqrt {{2 \over 3}} $$

IIT-JEE 2011 Paper 2 Offline

MCQ (More than One Correct Answer)

-1

Let $$f:(0,1) \to R$$ be defined by $$f(x) = {{b - x} \over {1 - bx}}$$, where b is a constant such that $$0 < b < 1$$. Then

f is not invertible on (0, 1).

f $$\ne$$ f^$$-$$1 on (0, 1) and $$f'(b) = {1 \over {f'(0)}}$$.

f = f^$$-$$1 on (0, 1) and $$f'(b) = {1 \over {f'(0)}}$$.

f^$$-$$1 is differentiable on (0, 1).

JEE Advanced Subjects