IIT-JEE 2007 | Differentiation Question 1 | Mathematics

IIT-JEE 2007

MCQ (Single Correct Answer)

-0.75

$${{{d^2}x} \over {d{y^2}}}$$ equals

$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$

$$ - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$

$$\left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$

$$ - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$

IIT-JEE 2007

MCQ (Single Correct Answer)

-0.75

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 $$t$$.

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 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, ..., $$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = $$

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

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

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

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

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