Question

32:

32. The minimum value of the expression 3 sec²x-4sec x +1 is.

$\left(\mathrm{A}\right)-\frac{1}{3}\left(\mathrm{B}\right)\frac{1}{3}\left(\mathrm{C}\right)\frac{2}{3}\left(\mathrm{D}\right)0$

Answer 1

Dear Student,
$$\begin{gathered} let, \\ f\left(x\right)=3se{c}^{2}x-4secx+1---\left(1\right) \\ differentiatingbothsidesw.r.tx; \\ f\text{'}\left(x\right)=6secx.secx.\tan x-4sec.\tan x \\ f\text{'}\left(x\right)=2\tan x\left(3se{c}^{2}x-2secx\right) \\ put,f\text{'}\left(x\right)=0 \\ 2\tan x\left(3se{c}^{2}x-2secx\right)=0 \\ 3se{c}^{2}x-2secx=0 \\ secx\left(3secx-2\right)=0 \\ secx\ne 0\left(Doesnotcomeinitsrange\right) \\ 3secx-2=0\to secx\ne \frac{2}{3}\left(doesnotcomeintherangeofsecx\right) \\ and,\tan x=0\to x=0 \\ now,substitutethevalueofx=0,in\left(1\right)togetminimumvalueoftheexpression; \\ So, \\ f\left(0\right)=3se{c}^{2}0-4sec0+1=3-4+1=4-4=0 \end{gathered}$$

Regards.