Question

For the Function \(f(x)=\tan x-4x,\ x\in\left(0,\frac{\pi}{2}\right)\), which of the following statements are correct?
A. \(f(x)\) is increasing in \(\left(0,\frac{\pi}{3}\right)\)
B. \(f(x)\) is increasing in \(\left(\frac{\pi}{3},\frac{\pi}{2}\right)\)
C. \(f(x)\) is decreasing in \(\left(0,\frac{\pi}{3}\right)\)
D. \(f(x)\) is decreasing in \(\left(\frac{\pi}{3},\frac{\pi}{2}\right)\)
Choose the correct answer from the options given below:

• A. A and B only
• B. A and D only
• C. B and C only
• D. B and D only

Hint:

Find \(f'(x)=\sec^2 x-4\). It is zero at \(x=\frac{\pi}{3}\), negative before it and positive after it.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A function is increasing where its derivative is positive and decreasing where its derivative is negative. So we find \(f'(x)\) and study its sign on \((0,\frac{\pi}{2})\).

Step 2: Find the derivative:
\[ f(x)=\tan x-4x \Rightarrow f'(x)=\sec^2 x-4 \]

Step 3: Find the turning point:
Put \(f'(x)=0\). Then \(\sec^2 x=4\), so \(\sec x=2\) (secant is positive here). This gives \(\cos x=\frac{1}{2}\), so \(x=\frac{\pi}{3}\).

Step 4: Sign of f'(x) on \((0,\frac{\pi}{3})\):
In this interval \(\cos x\) is larger than \(\frac{1}{2}\), so \(\sec x<2\) and \(\sec^2 x<4\). Hence \(f'(x)<0\) and \(f\) is decreasing. So C is true and A is false.

Step 5: Sign of f'(x) on \((\frac{\pi}{3},\frac{\pi}{2})\):
Here \(\cos x\) is smaller than \(\frac{1}{2}\), so \(\sec x>2\) and \(\sec^2 x>4\). Hence \(f'(x)>0\) and \(f\) is increasing. So B is true and D is false.

Step 6: Match with the options:
Correct statements are B and C. Option 1 (A and B) has the false statement A. Option 2 (A and D) has two false statements. Option 4 (B and D) has the false statement D. Option 3 (B and C only) is correct.

Final Answer:
\(f\) decreases on \((0,\frac{\pi}{3})\) and increases on \((\frac{\pi}{3},\frac{\pi}{2})\), so B and C are correct, which is option 3. \[ \boxed{\text{B and C only}} \]