Question

Let $[t]$ denote the greatest integer less than or equal to $t$.
If the function
\[ f(x)= \begin{cases} b^2 \sin\!\left[\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\right], & x < 0 \\ \dfrac{\sin x-\dfrac{1}{2}\sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases} \] is continuous at $x=0$, then $a^2+b^2$ is equal to

• A. $\dfrac{3}{4}$
• B. $\dfrac{1}{2}$
• C. $\dfrac{5}{8}$
• D. $\dfrac{9}{16}$

Hint:

For piecewise functions with limits, always compute LHL and RHL separately before applying continuity.

Answer 1

The Correct Option is C

To determine the value of \(a^2 + b^2\) such that the function \(f(x)\) is continuous at \(x = 0\), we need to ensure that the left-hand limit, right-hand limit, and the function value at \(x = 0\) are equal. Given the function:

\(f(x)= \begin{cases} b^2 \sin\!\left[\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\right], & x<0 \\ \dfrac{\sin x-\dfrac{1}{2}\sin 2x}{x^3}, & x>0 \\ a, & x=0 \end{cases}\)

1. Check the left-hand limit (\(LHL\)) as \(x \to 0^-\:\)
   • For \(x \lt 0\), the value inside the greatest integer function becomes: \(\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\).
   • As \(x \to 0^-\:\), (cos x + sin x) → 1, thus inside integral brackets becomes zero, i.e., the
   • This implies \(\sin(0) = 0\).
   • Thus, \(\lim_{{x \to 0^-}} f(x) = b^2 \cdot 0 = 0\).
2. Check the right-hand limit (\(RHL\)) as \(x \to 0^+\:\)
   • For \(x \gt 0\), consider \(f(x) = \dfrac{\sin x - \dfrac{1}{2} \sin 2x}{x^3}\).
   • We apply L'Hôpital's Rule because it is indeterminate: \(\frac{0}{0}\).
   • Derivative first time: \(\dfrac{\cos x - \cos 2x}{3x^2}\). L'Hôpital's Rule applied again.
   • Second derivative: \(\dfrac{-\sin x + 2\sin 2x}{6x}\). Again apply L'Hôpital's Rule.
   • Third derivative gives: \(\dfrac{-\cos x + 4 \cos 2x}{6}\).
   • Evaluating at \(x=0\) results in:
     • \(-1 + 4(1) = 3\)
     • \(\lim_{{x \to 0^+}} f(x)= 6 \cdot \dfrac{1}{6} = \dfrac{1}{4}\).
3. Ensuring continuity at \(x = 0\) (\(a = f(0)\)):
   • Since the function is continuous at \(x = 0\), we must have \(\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^+}} f(x) = a\).
   • This means \(0 = \dfrac{1}{4} = a\).
   • From above, \(a^2 = \dfrac{1}{16} + b^2 = \dfrac{b^2}{4}\).
4. Substituting:

Therefore, the value of \(a^2 + b^2\) is \(\dfrac{5}{8}\).