Discussion Forum

1)

Match the following

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