Question

If \[ \frac{dy}{dx} = (x^3-x) -(1-3x^2)\tan x +(x^3-x)\tan^2x \] and \[ y(1)=0, \] then \[ \frac{64}{\pi}y\!\left(\frac{\pi}{4}\right) = ? \]

• A. \(1\)
• B. \(\pi^2+16\)
• C. \(\pi^2-16\)
• D. \(16\pi^2\)

Hint:

When a derivative contains \(\tan^2x\), rewrite it using \[ \sec^2x=1+\tan^2x \] to identify a product-rule pattern.

Answer 1

The Correct Option is C

Concept: Observe the structure \[ (x^3-x)\tan^2x \] and \[ (1-3x^2)\tan x. \] This suggests differentiating an expression involving \[ (x^3-x)\tan x. \]

Step 1: Recognize the derivative pattern.
Consider \[ F(x)=(x^3-x)\tan x. \] Differentiating, \[ F'(x) = (3x^2-1)\tan x +(x^3-x)\sec^2x. \] Using \[ \sec^2x=1+\tan^2x, \] \[ F'(x) = (3x^2-1)\tan x +(x^3-x) +(x^3-x)\tan^2x. \] Since \[ (3x^2-1)\tan x = -(1-3x^2)\tan x, \] we obtain \[ F'(x) = (x^3-x) -(1-3x^2)\tan x +(x^3-x)\tan^2x. \] Thus \[ \frac{dy}{dx}=F'(x). \]

Step 2: Integrate.
Hence \[ y=(x^3-x)\tan x+C. \]

Step 3: Use the condition \(y(1)=0\).
\[ 0=(1-1)\tan1+C. \] \[ C=0. \] Therefore \[ y=(x^3-x)\tan x. \]

Step 4: Evaluate at \(x=\frac{\pi}{4}\).
Since \[ \tan\frac{\pi}{4}=1, \] \[ y\!\left(\frac{\pi}{4}\right) = \left(\frac{\pi^3}{64}-\frac{\pi}{4}\right). \] \[ = \frac{\pi(\pi^2-16)}{64}. \]

Step 5: Compute the required quantity.
\[ \frac{64}{\pi} y\!\left(\frac{\pi}{4}\right) = \frac{64}{\pi} \cdot \frac{\pi(\pi^2-16)}{64}. \] \[ =\pi^2-16. \] Hence \[ \boxed{\pi^2-16}. \]