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1

### IIT-JEE 2010 Paper 2 Offline

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 area bounded by the curve $$y=f(x)$$ and the lines $$x=0,$$ $$y=0$$ and $$x=t,$$ lies in the interval

A
$$\left( {{3 \over 4},3} \right)$$
B
$$\left( {{{21} \over {64}},{{11} \over {16}}} \right)$$
C
$$\left( {9,10} \right)$$
D
$$\left( {0,{{21} \over {64}}} \right)$$

## Explanation

$$\int\limits_0^{1/2} {f(x)dx < \int\limits_0^t {f(x)dx < \int\limits_0^{3/4} {f(x)dx} } }$$

Now, $$\int\limits_{}^{} {f(x)dx}$$

$$= \int\limits_{}^{} {(1 + 2x + 3{x^2} + 4{x^3})dx}$$

$$= x + {x^2} + {x^3} + {x^4}$$

$$\Rightarrow \int\limits_0^{1/2} {f(x)dx = {{15} \over {16}} > {3 \over 4}}$$

$$\int\limits_0^{3/4} {f(x)dx = {{530} \over {256}} < 3}$$

2

### IIT-JEE 2010 Paper 2 Offline

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 $$\left( { - {3 \over 4},{{ - 1} \over 2}} \right)$$.

3

### IIT-JEE 2010 Paper 1 Offline

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

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

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

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

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

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

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

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

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

4

### IIT-JEE 2010 Paper 1 Offline

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, $$\mathop {\lim }\limits_{x \to 0} {{\log (1 + x)} \over x} = 1$$]

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