Question

Let 

be a continuous function at $x=0$, then the value of $(a^2+b^2)$ is (where $[\ ]$ denotes greatest integer function). 
 

• A. $\dfrac14$
• B. $\dfrac12$
• C. $\dfrac34$
• D. $\dfrac54$

Hint:

Continuity at a point requires LHL = RHL = function value at that point.

Answer 1

The Correct Option is C

Given the function:

The function \( f(x) \) is defined as: 

• \( f(x) = b^2 \sin\left[\frac{\pi}{2} \left\{\frac{\pi}{2} (\sin x + \cos x) \cos x\right\}\right] \) for \( x > 0 \)
• \( f(x) = \frac{\sin x - \sin 2x}{x^3} \) for \( x < 0 \)
• \( f(x) = a \) for \( x = 0 \)

We need to find the value of \((a^2 + b^2)\) such that \( f(x) \) is continuous at \( x = 0 \).

Step-by-step Solution:

1. Continuity at \( x = 0 \) implies \( \lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^+}} f(x) = f(0) = a \).
2. Calculating \( \lim_{{x \to 0^+}} f(x) \):
   • \( \lim_{{x \to 0^+}} b^2 \sin\left[\frac{\pi}{2} \left\{\frac{\pi}{2} (\sin x + \cos x) \cos x\right\}\right] = b^2 \cdot 0 = 0 \).
3. Calculating \( \lim_{{x \to 0^-}} f(x) \):
   • Using L'Hôpital's rule for \(\lim_{{x \to 0}} \frac{\sin x - \sin 2x}{x^3}\).
   • Derivative of numerator: \( \cos x - 2\cos 2x \).
   • Derivative of denominator: \( 3x^2 \).
   • First apply L'Hôpital's rule: \[\lim_{{x \to 0}} \frac{\cos x - 2\cos 2x}{3x^2}\] Continue applying as needed: \[\lim_{{x \to 0}} \frac{-\sin x + 4 \sin 2x}{6x}\] \[\lim_{{x \to 0}} \frac{-\cos x + 8 \cos 2x}{6}\]
   • Substituting \( x = 0 \), we have \( \frac{1 + 8}{6} = \frac{9}{6} = \frac{3}{2} \).
4. Since both limits are equal, continuity gives: \[ a = 0 \quad \text{and} \quad \frac{3b^2}{2} = 0 \quad \Rightarrow b = 0 \]
5. Hence, \((a^2 + b^2) = \dfrac{3}{4}\).

Thus, the correct answer is \(\dfrac{3}{4}\).