Question

Number of integral values of $a$ for which the function $f(x)=\left(\frac{4a-7}{3}\right)x^3+(a-3)x^2+x+5$ is monotonic for every $x\in R$, is

Answer 1

Answer: 7

We have $f(x)=\left(\frac{4a-7}{3}\right)x^3+(a-3)x^2+x+5$
$\Rightarrow f^{\prime }(x)=(4a-7)x^2+2(a-3)x+1$
Now, for $f(x)$ to be monotonic,
$f^{\prime }(x)\ge 0\forall x\in R$
or $f^{\prime }(x)\le 0\forall x\in R.$
$\Rightarrow D\le 0\Rightarrow (a-3{)}^2-(4a-7)\le 0$
$\Rightarrow a^2-10a+16\le 0$
$\Rightarrow (a-2)(a-8)\le 0\Rightarrow a\in [2,8]$
Also, $4a-7\ne 0\Rightarrow a\ne \frac{7}{4}$
So, $a\in [2,8]-\left\{\frac{7}{4}\right\}$.
Hence, number of integral values of a equals $7$.