1
JEE Advanced 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-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
$$F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt} $$ 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
2
IIT-JEE 2008 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-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, ..., $$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = $$
3
IIT-JEE 2008 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Consider the function $$f:\left( { - \infty ,\infty } \right) \to \left( { - \infty ,\infty } \right)$$ defined by
$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$
$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$
Which of the following is true?
4
IIT-JEE 2008 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1
Consider the functions defined implicitly by the equation $$y^3-3y+x=0$$ on various intervals in the real line. If $$x\in(-\infty,-2)\cup(2,\infty)$$, the equation implicitly defines a unique real valued differentiable function $$y=f(x)$$. If $$x\in(-2,2)$$, the equation implicitly defines a unique real valued differentiable function $$y=g(x)$$ satisfying $$g(0)=0$$
If $$f\left( { - 10\sqrt 2 } \right) = 2\sqrt 2 ,$$ then $$f''\left( { - 10\sqrt 2 } \right) = $$
Questions Asked from Differentiation (MCQ (Single Correct Answer))
Number in Brackets after Paper Indicates No. of Questions
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