JEE Advanced 2014 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Let $$f:\left[ {0,2} \right] \to R$$ be a function which is continuous on $$\left[ {0,2} \right]$$ and is differentiable on $$(0,2)$$ with $$f(0)=1$$. Let
$$F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt} $$ for $$x \in \left[ {0,2} \right]$$. If $$F'\left( x \right) = f'\left( x \right)$$ for all $$x \in \left[ {0,2} \right]$$, then $$F(2)$$ equals

$${e^2} - 1$$

$${e^4} - 1$$

$$e - 1$$

$${e^4}$$

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Let $$g(x) = \log f(x)$$, where $$f(x)$$ is a twice differentiable positive function on (0, $$\infty$$) such that $$f(x + 1) = xf(x)$$. Then for N = 1, 2, 3, ..., $$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = $$

$$ - 4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N - 1} \right)}^2}}}} \right\}$$

$$4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N - 1} \right)}^2}}}} \right\}$$

$$ - 4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N + 1} \right)}^2}}}} \right\}$$

$$4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N + 1} \right)}^2}}}} \right\}$$

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Consider the function $$f:\left( { - \infty ,\infty } \right) \to \left( { - \infty ,\infty } \right)$$ defined by

$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$

Which of the following is true?

$${\left( {2 + a} \right)^2}f''\left( 1 \right) + {\left( {2 - a} \right)^2}f''\left( { - 1} \right) = 0$$

$${\left( {2 - a} \right)^2}f''\left( 1 \right) - {\left( {2 + a} \right)^2}f''\left( { - 1} \right) = 0$$

$$f'\left( 1 \right)f'\left( { - 1} \right) = {\left( {2 - a} \right)^2}$$

$$f'\left( 1 \right)f'\left( { - 1} \right) = -{\left( {2 + a} \right)^2}$$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Let $$f$$ and $$g$$ be real valued functions defined on interval $$(-1, 1)$$ such that $$g''(x)$$ is continuous, $$g\left( 0 \right) \ne 0.$$ $$g'\left( 0 \right) = 0$$, $$g''\left( 0 \right) \ne 0$$, and $$f\left( x \right) = g\left( x \right)\sin x$$

STATEMENT - 1: $$\mathop {\lim }\limits_{x \to 0} \,\,\left[ {g\left( x \right)\cot x - g\left( 0 \right)\cos ec\,x} \right] = f''\left( 0 \right)$$ and

STATEMENT - 2: $$f'\left( 0 \right) = g\left( 0 \right)$$

Statement - 1 is True, Statement - 2 is True; Statement - 2 is a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1

Statement - 1 is True, Statement -2 is False

Statement - 1 is False, Statement -2 is True

JEE Advanced Subjects