Question

Suppose $f(x)$ is a real valued polynomial function of degree 6 satisfying the following condition
(a) ' $f$ ' has minimum value at $x=0$ & 2
(b) ' $f$ ' has maximum value at $x=1$
(c) for all $x,\underset{x\to 0}{\mathrm{Lim}}\frac{1}{x}\ln \left|\begin{array}{ccc}f(x)/x& 1& 0\\ 0& 1/x& 1\\ 1& 0& 1/x\end{array}\right|=2$
Number of solutions of the equation $8f(x)-1=0$ is -

• A. one
• B. two
• C. three
• D. four

Answer 1

Answer: (D)

$\underset{x\to 0}{\mathrm{Lt}}\frac{1}{x}\ln \left|\begin{array}{ccc}\frac{f(x)}{x}& 1& 0\\ 0& \frac{1}{x}& 1\\ 1& 0& \frac{1}{x}\end{array}\right|=2$
$\underset{x\to 0}{\mathrm{Lt}}\frac{1}{x}\ln \left(\frac{f(x)}{x^3}+1\right)=\mathrm{2....}$(1)
for limit to exist
$\underset{x\to 0}{\mathrm{Lt}}\frac{f(x)}{x^3}=0$
$\Rightarrow f(x)=a_0{x}^6+a_1{x}^5+a_2{x}^4$
Also $f^{\prime }(0)=f^{\prime }(2)=f^{\prime }(1)=0$
$f^{\prime }(x)=6a_0{x}^5+5a_1{x}^4+4a_2{x}^3$
$=x^3\left(6a_0{x}^2+5a_{1}x+4a_2\right)$
$f^{\prime }(2)=0$
$\Rightarrow 24a_0+10a_1+4a_2=0$......(2)
$f^{\prime }(1)=0$
$6a_0+5a_1+4a_2=0$.......(3)
Consider $e^n\cdot (1)$
$\ln \left\{\underset{x\to 0}{\mathrm{Lt}}{\left(\frac{f(x)}{x^3}+1\right)}^{\frac{1}{x}}\right\}=2$
$\ln e^{\left(\underset{x\to 0}{\lim }\frac{f(x)}{x^4}\right)}=2\Rightarrow \underset{x\to 0}{\mathrm{Lt}}\frac{a_0{x}^6+a_1{x}^5+a_2{x}^4}{x^4}=2$
$\Rightarrow a_2=2$
Putting $a_2$ in (2) & (3)
$24a_0+10a_1=-8$
$6a_0+5a_1=-8$
on solving this we get
$a_1=-\frac{12}{5},a_0=\frac{2}{3}$
$f(x)=\frac{2}{3}{x}^6-\frac{12}{5}{x}^5+2x^4$
$f^{\prime }(x)=4x^5-12x^4+8x^3=4x^3\left(x^2-3x+2\right)$
$=4x^3(x-2)(x-1)$