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### JEE Advanced 2014 Paper 2 Offline

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
A
$${e^2} - 1$$
B
$${e^4} - 1$$
C
$$e - 1$$
D
$${e^4}$$
2

### IIT-JEE 2008

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

A
Statement - 1 is True, Statement - 2 is True; Statement - 2 is a correct explanation for Statement - 1
B
Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1
C
Statement - 1 is True, Statement -2 is False
D
Statement - 1 is False, Statement -2 is True
3

### IIT-JEE 2008

Let $$g\left( x \right) = \log f\left( x \right)$$ where $$f(x)$$ is twice differentible positive function on $$\left( {0,\infty } \right)$$ such that $$f(x+1)=x f(x)$$. Then, for $$N=1, 2, 3, ..........$$
$$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) =$$
A
$$- 4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N - 1} \right)}^2}}}} \right\}$$
B
$$4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N - 1} \right)}^2}}}} \right\}$$
C
$$- 4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N + 1} \right)}^2}}}} \right\}$$
D
$$4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N + 1} \right)}^2}}}} \right\}$$
4

### IIT-JEE 2007

Let $$\,\,\,$$$$f\left( x \right) = 2 + \cos x$$ for all real $$X$$.

STATEMENT - 1: for eachreal $$t$$, there exists a point $$c$$ in $$\left[ {t,t + \pi } \right]$$ such that $$f'\left( c \right) = 0$$ because
STATEMENT - 2: $$f\left( t \right) = f\left( {t + 2\pi } \right)$$ for each real $$t$$.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
C
Statement-1 is True, Statement-2 is False
D
Statement-1 is False, Statement-2 is True.

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