Question:
Let
be a continuous function at $x=0$, then the value of $(a^2+b^2)$ is (where $[\ ]$ denotes greatest integer function).
Let
be a continuous function at $x=0$, then the value of $(a^2+b^2)$ is (where $[\ ]$ denotes greatest integer function).
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Continuity at a point requires LHL = RHL = function value at that point.
Updated On: Jan 27, 2026
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Correct Answer: 34
Solution and Explanation
Step 1: Evaluate LHL at $x=0$.
\[ \text{LHL}=\lim_{x\to0^-}\frac{\sin x-\sin 2x}{x^3} \] \[ =\lim_{x\to0}\frac{\sin x(1-\cos x)}{x^3} \] \[ =\left(\lim_{x\to0}\frac{\sin x}{x}\right) \left(\lim_{x\to0}\frac{1-\cos x}{x^2}\right) =\frac12 \] Step 2: Apply continuity condition.
\[ f(0)=a=\frac12 \Rightarrow a^2=\frac14 \] Step 3: Evaluate RHL at $x=0$.
As $x\to0^+$, \[ (\sin x+\cos x)\cos x \to 1 \] \[ \Rightarrow \left[\frac{\pi}{2}\cdot1\right]=1 \] \[ \text{RHL}=b^2\sin\!\left(\frac{\pi}{2}\right)=b^2 \] Step 4: Use continuity.
\[ b^2=\frac12 \] Step 5: Final calculation.
\[ a^2+b^2=\frac14+\frac12=\frac34 \] Final conclusion.
The value of $(a^2+b^2)$ is 34.
\[ \text{LHL}=\lim_{x\to0^-}\frac{\sin x-\sin 2x}{x^3} \] \[ =\lim_{x\to0}\frac{\sin x(1-\cos x)}{x^3} \] \[ =\left(\lim_{x\to0}\frac{\sin x}{x}\right) \left(\lim_{x\to0}\frac{1-\cos x}{x^2}\right) =\frac12 \] Step 2: Apply continuity condition.
\[ f(0)=a=\frac12 \Rightarrow a^2=\frac14 \] Step 3: Evaluate RHL at $x=0$.
As $x\to0^+$, \[ (\sin x+\cos x)\cos x \to 1 \] \[ \Rightarrow \left[\frac{\pi}{2}\cdot1\right]=1 \] \[ \text{RHL}=b^2\sin\!\left(\frac{\pi}{2}\right)=b^2 \] Step 4: Use continuity.
\[ b^2=\frac12 \] Step 5: Final calculation.
\[ a^2+b^2=\frac14+\frac12=\frac34 \] Final conclusion.
The value of $(a^2+b^2)$ is 34.
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