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1

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

Consider the polynomial
$$f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}.$$
Let $$s$$ be the sum of all distinct real roots of $$f(x)$$ and let $$t = \left| s \right|.$$

The real numbers lies in the interval

A
$$\left( { - {1 \over 4},0} \right)$$
B
$$\left( { - 11, - {3 \over 4}} \right)$$
C
$$\left( { - {3 \over 4}, - {1 \over 2}} \right)$$
D
$$\left( {0,{1 \over 4}} \right)$$

Explanation

Given, $$f(x) = 4{x^3} + 3{x^2} + 2x + 1$$

$$f'(x) = 2(6{x^2} + 3x + 1)$$

$$D = 9 - 24 < 0$$

Hence, f(x) = 0 has only one real root.

$$f\left( { - {1 \over 2}} \right) = 1 - 1 + {3 \over 4} - {4 \over 8} > 0$$

$$f\left( { - {3 \over 4}} \right) = 1 - {6 \over 4} + {{27} \over {16}} - {{108} \over {64}}$$

$$ = {{64 - 96 + 108 - 108} \over {64}} < 0$$

f(x) changes its sign in $$\left( { - {3 \over 4},{{ - 1} \over 2}} \right)$$,

hence f(x) = 0 has a root in $$\left( { - {3 \over 4},{{ - 1} \over 2}} \right)$$.

2

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)
The value of $$\int\limits_0^1 {{{{x^4}{{\left( {1 - x} \right)}^4}} \over {1 + {x^2}}}dx} $$ is (are)
A
$${{22} \over 7} - \pi $$
B
$${2 \over {105}}$$
C
$$0$$
D
$${{71} \over {15}} - {{3\pi } \over 2}$$

Explanation

$$\int\limits_0^1 {{{{x^4}{{(1 - x)}^4}} \over {1 + {x^2}}}dx = \int\limits_0^1 {{{{x^4}{{\{ (1 + {x^2}) - 2x\} }^2}} \over {1 + {x^2}}}dx} } $$

$$ = \int\limits_0^1 {{x^4}{{{{(1 + {x^2})}^2} - 4x(1 + {x^2}) + 4{x^2}} \over {1 + {x^2}}}dx} $$

$$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + {{4\{ 1 + {x^2} - 1\} } \over {1 + {x^2}}}} \right]dx} $$

$$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + 4 - {4 \over {1 + {x^2}}}} \right]dx} $$

$$ = \int\limits_0^1 {\left[ {{x^6} - 4{x^5} + 5{x^4} - 4{{{x^4} - 1 + 1} \over {1 + {x^2}}}} \right]dx} $$

$$ = \int\limits_0^1 {({x^6} - 4{x^5} + 5{x^4} - 4{x^2} + 4)dx - 4\int\limits_0^1 {{{dx} \over {1 + {x^2}}}} } $$

$$ = \left[ {{{{x^7}} \over 7} - {{2{x^6}} \over 3} + {x^5} - {{4{x^3}} \over 3} + 4x} \right]_0^1 - 4[{\tan ^{ - 1}}x]_0^1$$

$$ = \left[ {{1 \over 7} - {2 \over 3} + 1 - {4 \over 3} + 4} \right] - \pi = {{22} \over 7} - \pi $$

3

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)
The value of $$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^3}}}\int\limits_0^x {{{t\ln \left( {1 + t} \right)} \over {{t^4} + 4}}} dt$$ is
A
$$0$$
B
$${1 \over 12}$$
C
$${1 \over 24}$$
D
$${1 \over 64}$$

Explanation

$$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^3}}}\int_0^x {{{t\log (1 + t)} \over {4 + {t^4}}}dt} $$

Using L' Hospital's rule,

$$ = \mathop {\lim }\limits_{x \to 0} {{{{x\log (1 + x)} \over {4 + {x^4}}}} \over {3{x^2}}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + x)} \over {3x}}\,.\,{1 \over {4 + {x^4}}}$$

$$ = {1 \over 3}\,.\,{1 \over 4} = {1 \over {12}}$$ [using, $$\mathop {\lim }\limits_{x \to 0} {{\log (1 + x)} \over x} = 1$$]

4

IIT-JEE 2010 Paper 2 Offline

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
A
$$1$$
B
$${{1 \over 3}}$$
C
$${{1 \over 2}}$$
D
$${{1 \over e}}$$

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

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

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