Question

Let $f\left(x\right)=a{x}^3+5x^2-bx+1.$ if $f\left(x\right)$ leaves remainder $5$, when divided by $2x+1$, and $f^{\prime }\left(x\right)$ is divisible by $3x-1$, then

• A. $a=26,b=10$
• B. $a=24,b=12$
• C. $a=26,b=12$
• D. none of these

Answer 1

The correct option is C $a=26,b=12$

$f\left(x\right)=a{x}^3+5x^2-bx+1$
When divided by $2x+1$ gives remainder $5$
$\Rightarrow f(-\frac{1}{2})=a{(-\frac{1}{2})}^3+5{(-\frac{1}{2})}^2-b(-\frac{1}{2})+1=5$ $\Rightarrow -\frac{a}{8}+\frac{5}{4}+\frac{b}{2}+1=5$

$\Rightarrow -a+10+4b+8=40$

$\Rightarrow a-4b=-22$ ...(1)

Differentiating $f\left(x\right)$ w.r.t $x$ we get
$f^{\prime }\left(x\right)=3a{x}^2+10x-b$
As $3x-1$ is factor of $f^{\prime }\left(x\right)$ gives
$f^{\prime }\left(\frac{1}{3}\right)=3a{\left(\frac{1}{3}\right)}^2+10\left(\frac{1}{3}\right)-b=0\Rightarrow \frac{a}{3}+\frac{10}{3}-b=0$
$\Rightarrow a-3b=-10$ ...(2)

Solving (1) and (2), we get
$a=26,b=12$
Hence, option 'C' is correct.