Question

If $f\left(x\right)=x^3+3(a-7)x^2+3(a^2-9)x-2$ where $a>0$, has +ive point of maximum then a varies over an interval of length

• A. $\frac{8}{7}$
• B. $\frac{6}{7}$
• C. $\frac{4}{7}$
• D. $\frac{3}{7}$

Answer 1

The correct option is A $\frac{8}{7}$

$f^{\prime }x=0\Rightarrow x^2+2(a-7)x+(a^2-9)=0$

$\therefore x=\frac{-2(a-7)\pm \sqrt{4{(a-7)}^2-4(a^2-9)}}{2}$

$x=7-a\pm \sqrt{58-14a}$

$58-4a>0\Rightarrow a<\frac{29}{7}$ ...(1)

Also smaller root is $(7-a)-\sqrt{58-14a}=+ive$

$\therefore {(7-a)}^2>58-4a\Rightarrow a^2-9>0$

$(a+3)(a-3)>0\Rightarrow a>3$ as a is +ive ...(2)

$\therefore aϵ(3,\frac{29}{7})$ by (1) and (2)

$\therefore$ Length of interval is $\frac{29}{7}-3=\frac{8}{7}$

Ans: A