Problem 59 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Let \(f(x)=\frac{1}{\pi}\left(\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x\right)+\frac{(x+1)}{x^{2}+2 x+10}\) such that the maximum value of \(f(x)\) is \(m\), then find the value of \((104 m-90)\).

Step-by-Step Solution

Answer

By now substituting \(m = \frac{13}{6}\) into the final equation to find the value of \(104m - 90\), we get the answer as \(22\)

Step 1 - Examination of terms

Start by evaluating the inverse trigonometric functions and the function \(\frac{x + 1}{x^2 + 2x + 10}\). Note that the maximum ranges of \(\sin^{-1}x\), \(\cos^{-1}x\), and \(\tan^{-1}x\) are \(\frac{\pi}{2}\), \(\pi\), and \(\frac{\pi}{2}\), respectively. Thus, the maximum value of \(\frac{1}{\pi}(\sin^{-1}x + \cos^{-1}x + \tan^{-1}x)\) is 2. To find the maximum value of the function \(\frac{x + 1}{x^2 + 2x + 10}\), you need to consider its derivative.

Step 2 - Derivation

Take derivative of \(g(x) = \frac{x + 1}{x^2 + 2x + 10}\) using quotient rule. It's given by \(g'(x)=\frac{(x^2+2x+10)- {(x+1)(2x+2)}}{(x^2+2x+10)^2}\).

Step 3 - Solving the derivative

Put \(g'(x) = 0\), we get \((x^2+2x+10)- {(x+1)(2x+2)}=0\). After simplifying, we get that \(x=2\).

Step 4 - Evaluate g(2)

Now, we substiute x=2 into the function g(x), we get \(g(2)=\frac{1}{6}\), and this is also the maximum value for this function by the second derivative test.

Step 5 - Total Maximum Value

Now add the two maximum values together to get the maximum value for \(f(x)\). Hence, \(f_{max} = 2+ \frac{1}{6} = \frac{13}{6}\), therefore \(m = f_{max}\).