If f(x) = 16 (sec^(-1) x)^2 + (cosec^(-1) x)^2. Then the maximum and minimum value of f(x) is

Question

If \(f(x)=16\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2.\) Then the maximum and minimum value of f(x) is

(1) \(\frac{1001 \pi^2}{33}\text{and} \frac{2 \pi^2}{9}\)

(2) \(\frac{1105 \pi^2}{68} \text{and} \frac{4 \pi^2}{17}\)

(3) \(\frac{1117 \pi^2}{59} \text{and} \frac{6 \pi^2}{19}\)

(4) \(\frac{1268 \pi^2}{27} \text{and} \frac{3 \pi^2}{16}\)

Answer 1

Correct option is (2) \(\frac{1105 \pi^2}{68} \text{and} \frac{4 \pi^2}{17}\)  

\(f(x)=\left(4 \sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\)  

\(=\left(4 \sec ^{-1} x+\operatorname{cosec}^{-1} x\right)^2-8 \sec ^{-1} x \operatorname{cosec}^{-1} x\)   

\(=\left(3 \sec ^{-1} x+\frac{\pi}{2}\right)^2-8 \sec ^{-1} x\left[\frac{\pi}{2}-\sec ^{-1} x\right]\)   

\(=9\left(\sec ^{-1} x\right)^2+\frac{\pi^2}{4}+3 \pi \sec ^{-1} x-4 \pi \sec ^{-1} x+ 8\left(\sec ^{-1} x\right)^2 \)   

\(=17\left(\sec ^{-1} x\right)^2=\pi\left(\sec ^{-1} x\right)+\frac{\pi^2}{4} \)   

\(=17\left[\left(\sec ^{-1} x\right)^2-\frac{\pi}{17}\left(\sec ^{-1} x\right)+\frac{\pi^2}{34^2}\right]+\frac{\pi^2}{4}-\frac{17 \pi^2}{34^2} \)     

\(=17\left[\left(\sec ^{-1} x-\frac{\pi}{34}\right)^2\right]+\frac{\pi^2}{4}-\frac{\pi^2}{68}\)        

\(=17\left[\left(\sec ^{-1} x-\frac{\pi}{34}\right)^2\right]+\frac{4 \pi^2}{17}\)    

\(M in=\frac{4 \pi^2}{17}\)   

Max if \(\sec ^{-1} x=x\)  

\(17\left[\left(\pi-\frac{\pi}{34}\right)^2\right]+\frac{4 \pi^2}{17}\)   

\(\frac{1089}{68} \pi^2+\frac{4 \pi^2}{17}=\frac{1105 \pi^2}{68}\)