Indefinite Integration 1 Question 64

65. Match List I with List II and select the correct answer using codes given below the lists.

(2014)

List I List II
P. The number of polynomials $f(x)$ with
non-negative integer coefficients of degree
$\leq 2$, satisfying $f(0)=0$ and $\int _0^{1} f(x) d x=1$, is
(i) 8
Q. The number of points in the interval
$[-\sqrt{13}, \sqrt{13}]$ at which $f(x)=\sin \left(x^{2}\right)+\cos \left(x^{2}\right)$
attains its maximum value, is
(ii) 2
R. $\int _{-2}^{2} \frac{3 x^{2}}{1+e^{x}} d x$ equals (iii) 4
S. $\frac{\int _{-1 / 2}^{1 / 2} \cos 2 x \log \frac{1+x}{1-x} d x}{\int _0^{1 / 2} \cos 2 x \log \frac{1+x}{1-x} d x}$ equals (iv) 0

Codes $\begin{array}{llll}P & Q & R & S\end{array}$ $\begin{array}{llll}P & Q & R & S\end{array}$

(a) (iii) (ii) (iv) (i)

(b) (ii) (iii) (iv) (i)

(c) (iii) (ii) (i) (iv)

(d) (ii) (iii) (i) (iv)

Analytical & Descriptive Questions

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Solution:

  1. (P) PLAN (i) A polynomial satisfying the given conditions is taken.

(ii) The other conditions are also applied and the number of polynomial is taken out.

Let $\quad f(x)=a x^{2}+b x+c$

$$ f(0)=0 \Rightarrow c=0 $$

Now, $\quad \int _0^{1} f(x) d x=1$

$$ \begin{aligned} & \Rightarrow \quad \frac{a x^{3}}{3}+{\frac{b x^{2}}{2}} _0^{1}=1 \Rightarrow \frac{\alpha}{3}+\frac{\beta}{2}=1 \\ & \Rightarrow \quad 2 a+3 b=6 \end{aligned} $$

As $a, b$ are non-negative integers.

So, $\quad a=0, b=2$ or $a=3, b=0$

$$ \therefore \quad f(x)=2 x \text { or } f(x)=3 x^{2} $$

(Q) PLAN Such type of questions are converted into only sine or cosine expression and then the number of points of maxima in given interval are obtained.

$$ \begin{aligned} f(x) & =\sin \left(x^{2}\right)+\cos \left(x^{2}\right) \\ & =\sqrt{2} \frac{1}{\sqrt{2}} \cos \left(x^{2}\right)+\frac{1}{\sqrt{2}} \sin \left(x^{2}\right) \\ & =\sqrt{2} \cos x^{2} \cos \frac{\pi}{4}+\sin \frac{\pi}{4} \sin \left(x^{2}\right) \\ & =\sqrt{2} \cos x^{2}-\frac{\pi}{4} \end{aligned} $$

For maximum value, $x^{2}-\frac{\pi}{4}=2 n \pi \Rightarrow x^{2}=2 n \pi+\frac{\pi}{4}$

$\Rightarrow \quad x= \pm \sqrt{\frac{\pi}{4}}$, for $n=0 \Rightarrow x= \pm \sqrt{\frac{9 \pi}{4}}$, for $n=1$

So, $f(x)$ attains maximum at 4 points in $[-\sqrt{13}, \sqrt{13}]$.

(R) PLAN

(i) $\int _{-a}^{a} f(x) d x=\int _{-a}^{a} f(-x) d x$

(ii) $\int _{-a}^{a} f(x) d x=2 \int _0^{a} f(x) d x$, if $f(-x)=f(x)$, i.e. $f$ is an even function.

$$ I=\int _{-2}^{2} \frac{3 x^{2}}{1+e^{x}} d x $$

and $\quad I=\int _{-2}^{2} \frac{3 x^{2}}{1+e^{-x}} d x$

$\Rightarrow \quad 2 I=\int _{-2}^{2} \frac{3 x^{2}}{1+e^{x}}+\frac{3 x^{2}\left(e^{x}\right)}{e^{x}+1} d x$

$$ 2 I=\int _{-2}^{2} 3 x^{2} d x \Rightarrow 2 I=2 \int _0^{2} 3 x^{2} d x $$

$$ I=\left[x^{3}\right] _0^{2}=8 $$

(S) PLAN

$$ f(x) d x=0 $$

If $f(-x)=-f(x)$, i.e. $f(x)$ is an odd function.

Let

$$ f(x)=\cos 2 x \log \frac{1+x}{1-x} $$

$$ f(-x)=\cos 2 x \log \frac{1-x}{1+x}=-f(x) $$

Hence, $f(x)$ is an odd function.

So, $\quad \int _{-1 / 2}^{1 / 2} f(x) d x=0$

$(P) \rightarrow$ (ii); (Q) $\rightarrow$ (iii); (R) $\rightarrow$ (i); (S) $\rightarrow$ (iv)



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