Question

$$\begin{array}{cc}\text{List I}& \text{List II}\\ \text{(A) If f satisfies}|f\left(u\right)-f\left(v\right)|\le |u-v|\text{for}u\&v\text{in}[a,b]& \\ \text{, then}\text{maximum possible value of}\left|\underset{2}{\overset{4}{\int }}f\left(x\right)dx-f\left(2\right)dx\right|\text{is}& \text{(P)}1\\ \text{(B)}\text{Let}f\left(z\right)\text{being a complex function defined}& \\ \text{as}f\left(z\right)=\frac{az+b}{cz+d}\text{, where}a,b,c,d& \\ \text{are non-zero real numbers. If}f\left(z_1\right)=f\left(z_2\right)& \\ \text{for all}{z}_1\ne z_2\text{and}b,a,c& \\ \text{are in G.P., then the value of}\frac{a}{d}\text{is}& \text{(Q)}2\\ \text{(C) Evaluate:}\underset{x\to 5}{lim}\frac{1-\cos (x^2-9x+20)}{(x-5{)}^2}& \text{(R)}\frac{1}{2}\\ \text{(D) If}\text{The number of values of (a) that satisfying}& \\ \underset{x\to -a}{lim}\frac{x^5+a^5}{x+a}=5\text{is}& \text{(S)}4\\ & \text{(T)}5\\ & \text{(U)}3\end{array}$$

Which of the following is CORRECT option ?

• A. $\left(B\right)\to \left(S\right)$
• B. $\left(C\right)\to \left(T\right)$
• C. $\left(D\right)\to \left(R\right)$
• D. $\left(A\right)\to \left(Q\right)$

Answer 1

The correct option is D $\left(A\right)\to \left(Q\right)$

$\left(A\right)$
$\left|\underset{2}{\overset{4}{\int }}f\left(x\right)dx-f\left(2\right)dx\right|\le \underset{2}{\overset{4}{\int }}|f\left(x\right)-f\left(2\right)|dx\Rightarrow \left|\underset{2}{\overset{4}{\int }}f\left(x\right)dx-f\left(2\right)dx\right|\le \underset{2}{\overset{4}{\int }}(x-2)dx\Rightarrow \left|\underset{2}{\overset{4}{\int }}f\left(x\right)dx-f\left(2\right)dx\right|\le {\left(\frac{(x-2{)}^2}{2}\right)}_2^4\Rightarrow \left|\underset{2}{\overset{4}{\int }}f\left(x\right)dx-f\left(2\right)dx\right|\le 2$
$\left(A\right)\to \left(Q\right)$

$\left(B\right)$
$f\left(z_1\right)=f\left(z_2\right)$
$\Rightarrow \frac{a{z}_1+b}{c{z}_1+d}=\frac{a{z}_2+b}{c{z}_2+d}$
$\Rightarrow ac{z}_1{z}_2+bc{z}_2+ad{z}_1+bd=ac{z}_1{z}_2+bc{z}_1+ad{z}_2+bd$
$\Rightarrow bc(z_2-z_1)=ad(z_2-z_1)$
$\Rightarrow bc=ad(\because z_1\ne z_2)...\left(1\right)$
$\because b,a,c$ are in G.P.
$\therefore bc=a^2...\left(2\right)$
From eqn(1) and (2),
$a^2=ad$
$\therefore \frac{a}{d}=1(\because a,b,c,d\ne 0)$
$\left(B\right)\to \left(P\right)$

$\left(C\right)$
$\underset{x\to 5}{lim}\frac{1-\cos (x^2-9x+20)}{(x-5{)}^2}$
Using L'Hospital's rule,
$=\underset{x\to 5}{lim}\frac{\sin (x^2-9x+20)\times (2x-9)}{2(x-5)}$
$=\underset{x\to 5}{lim}\frac{\sin (x^2-9x+20)}{(x-5)(x-4)}\times \frac{(x-4)(2x-9)}{2}$
$=\frac{1}{2}$
$\left(C\right)\to \left(R\right)$

$\left(D\right)$
$\underset{x\to -a}{lim}\frac{x^5+a^5}{x+a}=5$
$\underset{x\to -a}{lim}\frac{a^5-(-x^5)}{a-(-x)}=5$
$\Rightarrow 5a^4=5$
$\Rightarrow a=\pm 1$
$\Rightarrow$ Two values of $a$
$\left(D\right)\to \left(Q\right)$