Question

If \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots \infty}}} \), then the value of \( \frac{d^2y}{dx^2} \) at \( (\pi,1) \) is:

• A. \( 2 \)
• B. \( -2 \)
• C. \( -\frac{1}{2} \)
• D. \( \frac{1}{2} \)

Hint:

For nested functions, express them as a self-contained equation and use implicit differentiation.

Answer 1

The Correct Option is B

Step 1: Establishing the Functional Equation Since the function follows a repeating pattern, we define: \[ y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots}}} \] Since the same function appears infinitely, we square both sides: \[ y^2 = \sin x + y. \] Rearrange: \[ y^2 - y - \sin x = 0. \] 
Step 2: Differentiate Implicitly Differentiating both sides with respect to \( x \): \[ 2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}. \] Rearranging: \[ \frac{dy}{dx} (2y - 1) = \cos x. \] \[ \frac{dy}{dx} = \frac{\cos x}{2y - 1}. \] 
Step 3: Compute Second Derivative Differentiating again and substituting \( x = \pi \), \( y = 1 \), we obtain: \[ \frac{d^2y}{dx^2} = -2. \] Thus, the correct answer is: \(Option (2): -2.\)

Answer 2

Given \(y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots \infty}}}\), you can denote \(y = \sqrt{\sin x + y}\) since the expression continues to infinity.
Squaring both sides: \(y^2 = \sin x + y\).
Rearrange to get: \(y^2 - y - \sin x = 0\). (Equation 1)
The given point is \((\pi,1)\), substitute to verify: \(1^2 - 1 - \sin \pi = 0\) which holds true since \(\sin \pi = 0\).
To find \(\frac{d^2y}{dx^2}\), differentiate Equation 1 w.r.t \(x\): \(\frac{d}{dx}(y^2 - y - \sin x) = 0\).
Using implicit differentiation: \(2y \cdot \frac{dy}{dx} - \frac{dy}{dx} - \cos x = 0\).
This simplifies to: \((2y-1)\frac{dy}{dx} = \cos x\). (Equation 2)
At \((\pi,1)\), \(y=1\) and \(\cos \pi = -1\), substitute these values: \((2\cdot 1-1)\frac{dy}{dx} = -1\) leads to \(\frac{dy}{dx} = -1\).
Differentiate Equation 2 again w.r.t \(x\): \(\frac{d}{dx}((2y-1)\frac{dy}{dx}) = \frac{d}{dx}(\cos x)\).
Using the product rule: \((2 \cdot \frac{dy}{dx} \cdot \frac{dy}{dx} + (2y-1) \cdot \frac{d^2y}{dx^2}) = -\sin x\).
At \((\pi,1)\), \(\sin \pi=0\), substitute \(\frac{dy}{dx} = -1\):
\((2(-1)^2 + (2 \cdot 1 - 1) \cdot \frac{d^2y}{dx^2}) = 0\) simplifies to \(2 + \frac{d^2y}{dx^2} = 0\) hence \(\frac{d^2y}{dx^2} = -2\).
The value of \(\frac{d^2y}{dx^2}\) at \((\pi,1)\) is \(-2\).