Question:
Consider the function
\[
f:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to(-\infty,\infty)
\]
defined by
\[
f(x)=\left(|x|+|x-1|\right)\sin x+[x\sin x]
\]
where \([x\sin x]\) denotes the greatest integer less than or equal to \(x\sin x\).
Let \(\alpha\) be the total number of points in the interval
\[
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
\]
at which \(f\) is NOT continuous, and let \(\beta\) be the total number of points in the interval
\[
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
\]
at which \(f\) is NOT differentiable.
Then the value of
\[
\alpha+\beta
\]
is ________.
Consider the function
\[
f:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to(-\infty,\infty)
\]
defined by
\[
f(x)=\left(|x|+|x-1|\right)\sin x+[x\sin x]
\]
where \([x\sin x]\) denotes the greatest integer less than or equal to \(x\sin x\).
Let \(\alpha\) be the total number of points in the interval
\[
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
\]
at which \(f\) is NOT continuous, and let \(\beta\) be the total number of points in the interval
\[
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
\]
at which \(f\) is NOT differentiable.
Then the value of
\[
\alpha+\beta
\]
is ________.
Show Hint
Greatest integer functions are discontinuous whenever the inside expression crosses an integer value.
Updated On: May 20, 2026
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Verified By Collegedunia
Correct Answer: 3
Solution and Explanation
Step 1: Simplify the expression involving modulus.
For: \[ x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \] we have: \[ x-1<0 \] Hence: \[ |x-1|=1-x \] Thus: \[ |x|+|x-1|= \begin{cases} 1,& x\ge0 1-2x,& x<0 \end{cases} \] Therefore: \[ f(x)= \begin{cases} \sin x+[x\sin x],& x\ge0 (1-2x)\sin x+[x\sin x],& x<0 \end{cases} \]
Step 2: Study the greatest integer term.
Consider: \[ x\sin x \] Since: \[ x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \] we get: \[ 0\le x\sin x<\frac{\pi}{2} \] Also: \[ x\sin x=0 \] only at: \[ x=0 \] Further: \[ x\sin x<1 \] for: \[ |x|<\frac{\pi}{2} \] Hence: \[ [x\sin x]= \begin{cases} 0,& x\sin x 1,& x\sin x\ge1 \end{cases} \] Now solve: \[ x\sin x=1 \] There exists exactly one positive solution: \[ x=a \] in: \[ \left(0,\frac{\pi}{2}\right) \] and exactly one negative solution: \[ x=-a \] Thus discontinuities occur at: \[ x=\pm a \] Hence: \[ \alpha=2 \]
Step 3: Find points of non-differentiability.
The function: \[ [x\sin x] \] is not differentiable at: \[ x=\pm a \] Also modulus term changes form at: \[ x=0 \] Check derivatives at \(0\): For: \[ x>0 \] \[ f(x)=\sin x \] Thus: \[ f'_+(0)=1 \] For: \[ x<0 \] \[ f(x)=(1-2x)\sin x \] Derivative: \[ f'_-(0)=1 \] Thus differentiable at: \[ x=0 \] Therefore only non-differentiable points are: \[ x=\pm a \] Hence: \[ \beta=1 \]
Step 4: Compute \(\alpha+\beta\).
\[ \alpha=2,\qquad \beta=1 \] Thus: \[ \alpha+\beta=3 \]
Step 5: Identify the final answer.
Therefore: \[ \boxed{3} \]
For: \[ x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \] we have: \[ x-1<0 \] Hence: \[ |x-1|=1-x \] Thus: \[ |x|+|x-1|= \begin{cases} 1,& x\ge0 1-2x,& x<0 \end{cases} \] Therefore: \[ f(x)= \begin{cases} \sin x+[x\sin x],& x\ge0 (1-2x)\sin x+[x\sin x],& x<0 \end{cases} \]
Step 2: Study the greatest integer term.
Consider: \[ x\sin x \] Since: \[ x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \] we get: \[ 0\le x\sin x<\frac{\pi}{2} \] Also: \[ x\sin x=0 \] only at: \[ x=0 \] Further: \[ x\sin x<1 \] for: \[ |x|<\frac{\pi}{2} \] Hence: \[ [x\sin x]= \begin{cases} 0,& x\sin x 1,& x\sin x\ge1 \end{cases} \] Now solve: \[ x\sin x=1 \] There exists exactly one positive solution: \[ x=a \] in: \[ \left(0,\frac{\pi}{2}\right) \] and exactly one negative solution: \[ x=-a \] Thus discontinuities occur at: \[ x=\pm a \] Hence: \[ \alpha=2 \]
Step 3: Find points of non-differentiability.
The function: \[ [x\sin x] \] is not differentiable at: \[ x=\pm a \] Also modulus term changes form at: \[ x=0 \] Check derivatives at \(0\): For: \[ x>0 \] \[ f(x)=\sin x \] Thus: \[ f'_+(0)=1 \] For: \[ x<0 \] \[ f(x)=(1-2x)\sin x \] Derivative: \[ f'_-(0)=1 \] Thus differentiable at: \[ x=0 \] Therefore only non-differentiable points are: \[ x=\pm a \] Hence: \[ \beta=1 \]
Step 4: Compute \(\alpha+\beta\).
\[ \alpha=2,\qquad \beta=1 \] Thus: \[ \alpha+\beta=3 \]
Step 5: Identify the final answer.
Therefore: \[ \boxed{3} \]
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