Question

Let $f(x)=(x-1)(x-2)(x-3)$ $(x-n),n\in N$ and $\int \frac{f(x)f^{\prime \prime }(x)-{\left(f^{\prime }(x)\right)}^2}{f^2(x)}dx=g(x)+C$, where $C$ is arbitrary constant.

Column I		Column II
A	If $f^{\prime }(n)=5040$, then $n$ is divisible by	P	4
B	If $g(x)$ is discontinuous at 9 points, $\forall x\in R$ then $n$ is greater than	Q	6
C	If $g(x)=5$ has 8 solutions, then $n$ may be equal to	R	8
D	If the number of roots of equation $f^{\prime }(x)=0$, be $(n-5{)}^2(n-1),(n>1)$	S	9 then possible values of $n$ is/are

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

Answer 1

Answer: (D)

(A) $\ln (f(x))=\ln (x-1)+\ln (x-2)+\dots \dots +\ln (x-n)$
$f^{\prime }(x)=f(x)\left[\frac{1}{x-1}+\frac{1}{x-2}+\dots \dots +\frac{1}{x-n}\right]$
$f^{\prime }(x)=(x-2)(x-3)\dots .(x-n)+(x-1)(x-3)\dots .(x-n)+\dots ..+(x-1)(x-2)\dots ..(x-(n-1))$
$f^{\prime }(n)=(n-1)(n-2)(n-3)\cdot 3\cdot 2\cdot 1(\text{ all other factors except the last vanishes when }x=n)$
$5040=(n-1)!$
$7!=(n-1)!\Rightarrow n=8$
(B)$g(x)=\frac{f^{\prime }(x)}{f(x)}=\frac{1}{x-1}+\frac{1}{x-2}+\dots \dots \dots \frac{1}{x-n}+C;$
$\text{ hence }{g}^{\prime }(x)<0\text{ (think!) }$
$\Rightarrow g(x)\text{ is discontinuous at }n\text{ points }n=9$
(C) $g(x)=5$ has either $n-1$ or $n$ roots
$g(x)=\frac{f^{\prime }(x)}{f(x)}\text{ or }g(x)=\frac{f^{\prime }(x)}{f(x)}+C$
$\therefore n=8\text{ or }9$
(D) No. of roots $f^{\prime }(x)=0$ is $n-1$ (use L.M.V.T.)
$\Rightarrow (n-1)(n-5{)}^2=n-1\Rightarrow (n-1)(n-6)(n-4)=0$