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:
- (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)
