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 $tan x=t\Rightarrow t\in \left[- 1 , 1\right]$ $\because x\in \left[- \frac{\pi }{4} , \frac{\pi }{4}\right]$
$\therefore f\left(t\right)=2t^{3}-6t^{2}+2$
$\Rightarrow f^{'}\left(t\right)=2\left[3 t^{2} - 6 t\right]$
$=6t\left(t - 2\right)$

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