Question

Match the entries of col. I with those of col. II.
$\begin{array}{cc}Column-I& Column-II\\ \left(a\right)f\left(x\right)=\frac{1-x+x^2}{1+x-x^2}\text{on}[0,1]& \left(p\right)Greatestvalueoff=1\\ \left(b\right)f\left(x\right)=2tanx-ta{n}^{2}x\text{on}[0,\frac{\pi }{2}]& \left(q\right)Leastvalueoff=\frac{3}{5}\\ \left(c\right)f\left(x\right)=\frac{2}{\pi }(sin2x-x)\text{on}[-\frac{\pi }{2},\frac{\pi }{2}]& \left(r\right)Leastvalueoff=-1\\ \left(d\right)f\left(x\right)=\frac{1}{2},(x^3-3x^2+6x-2)\text{on}(-1,1)& \left(s\right)Leastvalueoff=-6\end{array}$

• A. $\left(a\right)\to \left(p\right),\left(b\right)\to \left(q\right),\left(c\right)\to \left(r\right),\left(d\right)\to \left(s\right)$
• B. $\left(a\right)\to (p,q),\left(b\right)\to \left(p\right),\left(c\right)\to (p,r),\left(d\right)\to (p,s)$
• C. $\left(a\right)\to \left(q\right),\left(b\right)\to \left(p\right),\left(c\right)\to \left(p\right),\left(d\right)\to \left(s\right)$
• D. $\left(a\right)\to \left(s\right),\left(b\right)\to \left(q\right),\left(c\right)\to (p,r),\left(d\right)\to (r,s)$

Answer 1

The correct option is B

$\left(a\right)\to (p,q),\left(b\right)\to \left(p\right),\left(c\right)\to (p,r),\left(d\right)\to (p,s)$

$\left(a\right)\to (p,q),\left(b\right)\to \left(p\right),\left(c\right)\to (p,r),\left(d\right)\to (p,s)$
$\left(a\right)f^{\prime }\left(x\right)=\frac{2(2x-1)}{(1+x-x^2{)}^2}$ by quotient formula
$f^{\prime }\left(x\right)=0\Rightarrow x=\frac{1}{2}f\left(0\right)=1,f\left(1\right)=1butf\left(\frac{1}{2}\right)=\frac{3}{5}$
$\therefore$ Greatest value =1 $\Rightarrow$ (p), Least value $=\frac{3}{5}\Rightarrow \left(q\right)$
$\left(b\right)f^{\prime }\left(x\right)=2se{c}^{2}x(1-tanx)=0$
$\Rightarrow tanx=1$ or $x=\frac{\pi }{4}andf\left(\frac{\pi }{4}\right)=2-1=1$
$\therefore \left(b\right)\to \left(p\right).\left(c\right)f^{\prime }\left(x\right)=\frac{2}{\pi }(2cos2x-1)=0\therefore cos2x=\frac{1}{2}=cos\frac{\pi }{3}\therefore x=\frac{\pi }{6}f(-\frac{c}{2})=1,f\left(\frac{\pi }{2}\right)=-1andf\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{\pi }-\frac{1}{3}<1$
$\therefore$ greatest value =1 and least value =-1
$\therefore \left(C\right)\to (p,r)\left(d\right)f^{\prime }\left(x\right)=\frac{3}{2}(x^2-2x+2)=\frac{3}{2}\left[\right(x-1{)}^2+1]\ne 0$
But f(-1)=-6,f(1)=1
$\therefore$ Least value is -6 and greatest value is 1.
$\therefore \left(d\right)\to (p,s)$