Answer:
Option D
Explanation:
(P) Plan
(i) A polymer satisfying the given conditions is taken
(ii) The other conditions are also applied and the number of polynomial is taken out
Let f(x) =ax2+bx+c
f(0)= 0 $\Rightarrow$ c=0
Now, $\int_{0}^{1} f(x) dx=1$
$\Rightarrow$ $ \left(\frac{ax^{3}}{3}+\frac{bx^{2}}{2}\right)_0^1=1$
$\Rightarrow$ $ \frac{a}{3}+\frac{b}{2}=1$
$\Rightarrow$ $ 2a+3b=6$
As a,b are non-negative integers
So, a=0, b=2, or a=3, b=0
So, f(x)=2x or f(x)=3x2
(Q) Plan such type of questions are converted into only sine or cosine expression and then the number of points of maximum in given interval are obtained
f(x) = sin(x2)+ cos (x2)
= $\sqrt{2}\left[\frac{1}{\sqrt{2}}\cos (x^{2})+\frac{1}{\sqrt{2}}\sin(x^{2})\right]$
= $\sqrt{2}\left[\cos x^{2}\cos \frac{\pi}{4}+\sin\frac{\pi}{4}\sin(x^{2})\right]$
=$\sqrt{2}\cos (x^{2}-\frac{\pi}{4})$
For maximum value
$x^{2}-\frac{\pi}{4}=2n\pi$
$\Rightarrow$ $ x^{2}=2n\pi+\frac{\pi}{4}$
$\Rightarrow$ $ x=\pm \sqrt{\frac{\pi}{4}},for n=0$
$ 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) $\int_{-a}^{a} f(x) dx=\int_{-a}^{a} f(-x) dx$
(ii) $\int_{-a}^{a} f(x) dx=2\int_{0}^{a} f(x) dx$
f(-x) =f(x) , i.e, f is an even function
$I= \int_{-2}^{2} \frac{3x^{2}}{1+e^{x}}dx$
$I= \int_{-2}^{2} \frac{3x^{2}}{1+e^{-x}}dx$
$2I= \int_{-2}^{2} \left(\frac{3x^{2}}{1+e^{x}}+\frac{3x^{2}(e^{x})}{e^{x}+1}\right) dx$
$2I= \int_{-2}^{2}3x^{2}dx$
$\Rightarrow $ $2I=2 \int_{0}^{2}3x^{2}dx$
$I=[x^{3}]_0^2=8$
(S) Plan $\int_{-a}^{a} f(x) dx=0$
if f(-x)= -f(x)
i.e, f(x) is an odd function
Let f(x)= cos 2x $\log\left(\frac{1+x}{1-x}\right)$
f(-x)= cos 2x $\log\left(\frac{1-x}{1+x}\right)$
=-f(x)
Hence, f(x) is an odd function.
So, $\int_{-1/2}^{1/2} f(x) dx=0$
P:ii, Q:iii, R:i, S:iv
