Question

Column-I	Column-II
(A) Let $f_n\left(x\right)=e^x\bullet e^{x^2}\bullet e^{x^3}\dots .e^{x^n},nϵN\text{and}g\left(x\right)={lim}_{n\to \infty }{f}_n\left(x\right)\text{then}{g}^{\prime }\left(\frac{1}{2}\right)=\lambda \cdot e$, then $\lambda$ is greater than	(p) 0
(B) a, b are distinct real numbers satisfying $|a–1\neg |+|b–1|=|a|+|b|=|a+1|+|b+1|$. If the minimum value of $|a–b|$ is $\mu ,\text{then}\mu$ is greater than	(q) 1
(C) Let $nϵN$. If the value of c prescribed in Rolle’s theorem for the function $f\left(x\right)=2x(x–3{)}^n\text{on}[0,3]\text{is}\frac{3}{4}$, then n is equal to	(r) 2
(D) If $x_1,x_2$ are abscissae of two points on the curve $f\left(x\right)=x–x^2$ in the interval [0, 1], then the maximum value of expression$(x_1+x_2)-(x_1^2+x_2^2)$is less than	(s) 3
	(t) 4

• A. A(P, Q, R, S); B(P, Q); C(S); D(Q, R, S, T)
• B. A(P, Q, R, S); B(P, Q); C(S); D(Q, R, T, S)

Answer 1

The correct option is A A(P, Q, R, S); B(P, Q); C(S); D(Q, R, S, T)

(A) : $f_n\left(x\right)=e^{x+x^2+x^3+\dots ..+x^n}$
$g\left(x\right)={lim}_{x\to \infty }{f}_n\left(x\right)=e^{\frac{x}{1-x}}$
$\therefore g^{\prime }\left(x\right)=e^{\frac{x}{1-x}\cdot \frac{1}{{(1-x)}^2}}$
$\Rightarrow g^{\prime }\left(\frac{1}{2}\right)=4\cdot e$
$\therefore \lambda =4$
(B) : Let $a<b\text{and}f\left(x\right)=|x–a|+|x–b|,\forall xϵR$
So $f\left(x\right)$ is decreasing in $(–\infty ,a),\text{constant in}[a,b]$ and increasing in $(b,\infty ),$
We have, $f\left(0\right)=f\left(1\right)=f(–1)$
$\Rightarrow \{-1,0,1\}ϵ[a,b]$
$\therefore |a–b{|}_{min}=2$
$\therefore \mu =2$
(C) : $\because f\left(x\right)=2x(x–3{)}^n,nϵN$
$f\left(x\right)$ is differentiable and continuous everywhere.
$\therefore f\left(x\right)=2(x–3{)}^n+2nx(x–3{)}^{n–1}$
$\because f^{\prime }\left(\frac{3}{4}\right)=0$
$\Rightarrow 2{(\frac{3}{4}-3)}^n+2n\cdot \frac{3}{4}{(\frac{3}{4}-3)}^{n-1}=0$
$\therefore n=3$
(D) Let $x_1,y_1\text{and}(x_2,y_2)$ are two points.
$\therefore y_1+y_2=(x_1+x_2)-(x_1^2+x_2^2)$
When $x_2\to \frac{1}{2}$
$y_1=\frac{1}{4}$
$y_2={\left(\frac{1}{4}\right)}^{-}$
$\Rightarrow y_1+y_2={\left(\frac{1}{2}\right)}^{-}$