Continuity and Differentiability
Continuity at x=0 — GIF + Limit
nta_pyq_2026_jan
Grade 12
Question:
Let $[t]$ denote the greatest integer less than or equal to $t$. If the function $f(x) = \begin{cases} b^2\sin\!\left(\dfrac{\pi}{2}\!\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\right) & x < 0 \\[6pt] \dfrac{\sin x - \tfrac{1}{2}\sin 2x}{x^3} & x > 0 \\[4pt] a & x = 0 \end{cases}$ is continuous at $x=0$, then $a^2+b^2$ is equal to
$\dfrac{5}{8}$
$\dfrac{1}{2}$
$\dfrac{9}{16}$
$\dfrac{3}{4}$
Step-by-Step Solution
Key Concept: For $x>0$: $\dfrac{\sin x-\frac{1}{2}\sin 2x}{x^3}=\dfrac{\sin x(1-\cos x)}{x^3}$. $\lim_{x\to0^+}\dfrac{\sin x}{x}\cdot\dfrac{1-\cos x}{x^2}=1\cdot\dfrac{1}{2}=\dfrac{1}{2}$. So $a=\dfrac{1}{2}$.
$a=1/2$, $b^2=1/2$. $a^2+b^2=3/4$.
Correct Answer: 4