IIT-JEE 2010 Paper 2 Offline

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MCQ (Single Correct Answer)

Let $$f$$ be a real-valued function defined on the interval $$(-1, 1)$$ such that
$${e^{ - x}}f\left( x \right) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,\,dt,} $$ for all $$x \in \left( { - 1,1} \right)$$,
and let $${f^{ - 1}}$$ be the inverse function of $$f$$. Then $$\left( {{f^{ - 1}}} \right)'\left( 2 \right)$$ is equal to

$$1$$

$${{1 \over 3}}$$

$${{1 \over 2}}$$

$${{1 \over e}}$$

Explanation

We have,

$${e^{ - x}}f(x) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,dt\,x \in ( - 1,1)} $$

On differentiating w.r.t. x, we get

$${e^{ - x}}(f'(x) - f(x)) = \sqrt {{x^4} + 1} $$

$$ \Rightarrow f'(x) = f(x) + \sqrt {{x^4} + 1} \,{e^x}$$

$$\because$$ $${f^{ - 1}}$$ is the inverse of f

$$\therefore$$ $${f^{ - 1}}(f(x)) = x$$

$$ \Rightarrow {f^{ - 1'}}(f(x))f'(x) = 1$$

$$ \Rightarrow {f^{ - 1'}}(f(x)) = {1 \over {f'(x)}}$$

$$ \Rightarrow {f^{ - 1'}}(f(x)) = {1 \over {f(x) + \sqrt {{x^4} + 1} \,{e^x}}}$$

At $$x = 0$$, $$f(x) = 2$$

$${f^{ - 1'}}(2) = {1 \over {2 + 1}} = {1 \over 3}$$

IIT-JEE 2009

MCQ (Single Correct Answer)

let $$f$$be a non-negative function defined on the interval
$$\left[ {0,1} \right]$$. If $$\int\limits_0^x {\sqrt {1 - {{\left( {f'\left( t \right)} \right)}^2}} dt = } \int\limits_0^x {f\left( t \right)dt,\,\,\,0 \le x \le 1} ,$$
$$f\left( 0 \right) = 0,$$ then

$$f\left( {{1 \over 2}} \right) < {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) > {1 \over 3}$$

$$f\left( {{1 \over 2}} \right) > {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) > {1 \over 3}$$

$$f\left( {{1 \over 2}} \right) < {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$

$$f\left( {{1 \over 2}} \right) > {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$

IIT-JEE 2008

MCQ (Single Correct Answer)

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

MCQ (Single Correct Answer)

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?

$$f(x)$$ is decreasing on $$(-1,1)$$ and has a local minimum at $$x=1$$

$$f(x)$$ is increasing on $$(-1,1)$$ and has a local minimum at $$x=1$$

$$f(x)$$ is increasing on $$(-1,1)$$ but has neither a local maximum nor a local minimum at $$x=1$$

$$f(x)$$ is decreasing on $$(-1,1)$$ but has neither a local maximum nor a local minimum at $$x=1$$