Question

Match the following:
$\begin{array}{cccc}& Column-I& & Column-II\\ \left(P\right)& Minimumof\frac{x^4}{y^4}+\frac{y^4}{x^4};& \left(1\right)& \frac{1}{4}\\ & (x,y\ne 0;x,y>0)is& & \\ \left(Q\right)& xϵ(0,1),thefunction{x}^{25}(1-x{)}^{75}& \left(2\right)& 2\\ & takesthemaximumvalueatx& & \\ \left(R\right)& Theminimumvalueof& \left(3\right)& 8^2\times {16}^4\\ & 2^{(x^2-3{)}^3+27}is& & \\ \left(S\right)& Ifx+y=24,(x>0,y>0),then& \left(4\right)& 1\\ & themaximumvalueof{x}^2{y}^{4}is& & \end{array}$

• A. P-2; Q-1; R-3; S-4
• B. P-2; Q-1; R-4; S-3
• C. P-3; Q-1; R-4; S-3
• D. P-3; Q-1; R-3; S-4

Answer 1

The correct option is B P-2; Q-1; R-4; S-3

A) $A.M\ge G.M$
$\frac{x^4}{y^4}+\frac{y^4}{x^4}\ge 2\sqrt{\frac{x^4}{y^4}.\frac{y^4}{x^4}}\ge 2$

B) $y=x^{25}(1-x{)}^{75}$
log y = 25 log x + 75 log (1 - x )
$\frac{1}{y}\frac{dy}{dx}=\frac{25}{x}-\frac{75}{1-x}$
For the maximum and minimum $\frac{dy}{dx}=0$
25 (1 - x) = 75x
1 - x = 3x
$\Rightarrow$ 4x = 1
$x=\frac{1}{4}$

C) At x = 0
$f\left(x\right)=2^{-27+27}=2^{\circ }=1$

D) y = 24 - x
$f\left(x\right)=x^2(24-x{)}^4$
$f^{\prime }\left(x\right)=2x(24-x{)}^4-4x^2(24-x{)}^3$
$f^{\prime }\left(x\right)=0\Rightarrow 2x(2x-x{)}^3(24-x-2x)=0$
$x=0,24,8$
$\therefore x\ne 0,24\therefore x=8$
$y=16$
$\therefore$ maximum value of $x^2{y}^4=8^2\times {16}^4$.