Question

Let $f\left(x\right)=2{\left(tan\right)}^{3}x-6{\left(tan\right)}^{2}x+1+sgn\left(e^x\right),\forall x\in \left[-\frac{\pi }{4},\frac{\pi }{4}\right].$ Then the positive difference between the least value and the local maximum value of the function is (where $sgn(f(x)$ represents the signum function)

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

Answer: (B)

Here, $sgn\left(e^x\right)=1$
Let $tanx=t\Rightarrow t\in \left[-1,1\right]$ $\because x\in \left[-\frac{\pi }{4},\frac{\pi }{4}\right]$
$\therefore f\left(t\right)=2t^3-6t^2+2$
$\Rightarrow f^{{}^{\prime }}\left(t\right)=2\left[3t^2-6t\right]$
$=6t\left(t-2\right)$

$\therefore f\left(-1\right)=2\left[-1-3+1\right]=-6$
$f\left(1\right)=2\left[1-3+1\right]=-2$
$f\left(0\right)=2$
$\therefore$ Least value $=-6$
Local maximum value $=2$
$\therefore$ the required positive difference $=8$