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1

IIT-JEE 2011 Paper 1 Offline

Let the straight line $$x=b$$ divide the area enclosed by
$$y = {\left( {1 - x} \right)^2},y = 0,$$ and $$x=0$$ into two parts $${R_1}\left( {0 \le x \le b} \right)$$ and
$${R_2}\left( {b \le x \le 1} \right)$$ such that $${R_1} - {R_2} = {1 \over 4}.$$ Then $$b$$ equals
A
$${3 \over 4}$$
B
$${ 1\over 2}$$
C
$${1 \over 3}$$
D
$${1 \over 4}$$

Explanation

We can write the integral

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

$$\Rightarrow \left. {{{{{(x - 1)}^3}} \over 3}} \right|_0^b - \left. {{{{{(x - 1)}^3}} \over 3}} \right|_b^1 = {1 \over 4}$$

$$\Rightarrow {{{{(b - 1)}^3}} \over 3} + {1 \over 3} - \left( {0 - {{{{(b - 1)}^3}} \over 3}} \right) = {1 \over 4}$$

$$\Rightarrow {{2{{(b - 1)}^3}} \over 3} = - {1 \over {12}} \Rightarrow {(b - 1)^3} = - {1 \over 8} \Rightarrow b = {1 \over 2}$$

2

IIT-JEE 2011 Paper 1 Offline

The value of $$\,\int\limits_{\sqrt {\ell n2} }^{\sqrt {\ell n3} } {{{x\sin {x^2}} \over {\sin {x^2} + \sin \left( {\ell n6 - {x^2}} \right)}}\,dx}$$ is
A
$${1 \over 4}\,\ell n{3 \over 2}$$
B
$$\,{1 \over 2}\,\ell n{3 \over 2}$$
C
$$\ell n{3 \over 2}$$
D
$$\,\,{1 \over 6}\,\ell n{3 \over 2}$$

Explanation

$${x^2} = t \Rightarrow 2x\,dx = dt$$

$$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin t} \over {\sin t + \sin (\ln 6 - t)}}dt}$$ and $$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin (\ln 6 - t)} \over {\sin (\ln 6 - t) + \sin t}}dt}$$

$$2I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {1dt \Rightarrow I = {1 \over 4}\ln {3 \over 2}}$$.

3

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 function$$f'(x)$$ is

A
increasing in $$\left( { - t, - {1 \over 4}} \right)$$ and decreasing in $$\left( { - {1 \over 4},t} \right)$$
B
decreasing in $$\left( { - t, - {1 \over 4}} \right)$$ and increasing in $$\left( { - {1 \over 4},t} \right)$$
C
increasing in $$(-t, t)$$
D
decreasing in $$(-t, t)$$

Explanation

4

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

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

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

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

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

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