Question

If $a^2+b^2$ - ab - a - b + 1 $\le$ 0, a, b $ϵ$ R and $f\left(x\right)=(1+sec2x)(1+sec{2}^{2}x)$ ___ $(1+sec{2}^{n}x)$, then the value of $f\left(\frac{\pi }{2^n-1}\right)$ where n $ϵ$ N, is

• A. a - b
• B. a + b
• C. ab
• D. 12 ab

Answer 1

The correct option is C ab

$(1-a{)}^2+(1-b{)}^2+(a-b{)}^2\le 0\Rightarrow a=1,b=1f(x)=\frac{tan\left(2^{n}x\right)}{tanx})\therefore f\left(\frac{\pi }{2^n-1}\right)=1=ab$