IIT-JEE 2010 Paper 1 Offline | Definite Integration Question 44 | Mathematics

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IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

-1

The value of $$\int\limits_0^1 {{{{x^4}{{\left( {1 - x} \right)}^4}} \over {1 + {x^2}}}dx} $$ is (are)

$${{22} \over 7} - \pi $$

$${2 \over {105}}$$

$$0$$

$${{71} \over {15}} - {{3\pi } \over 2}$$

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

Let $$g\left( x \right) = \int\limits_0^{{e^x}} {{{f'\left( t \right)} \over {1 + {t^2}}}} \,dt.$$

Which of the following is true?

$$g'(x)$$ is positive on $$\left( { - \infty ,0} \right)$$ and negative on $$\left( {0,\infty } \right)$$

$$g'(x)$$ is negative on $$\left( { - \infty ,0} \right)$$ and positive on $$\left( {0,\infty } \right)$$

$$g'(x)$$ changes sign on both $$\left( { - \infty ,0} \right)$$ and $$\left( {0,\infty } \right)$$

$$g'(x)$$ does not change sign on $$\left( { - \infty ,0} \right)$$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-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 $$g(0)=0$$

$$\int\limits_{ - 1}^1 {g'\left( x \right)dx = } $$

$$2g(-1)$$

$$0$$

$$-2g(1)$$

$$2g(1)$$

IIT-JEE 2007 Paper 1 Offline

MCQ (Single Correct Answer)

-1

$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\int\limits_2^{{{\sec }^2}x} {f(t)\,dt} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$ equal

$${8 \over \pi }f(2)$$

$${2 \over \pi }f(2)$$

$${2 \over \pi }f\left( {{1 \over 2}} \right)$$

$$4f(2)$$

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