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JEE Main 20205 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Morning Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2020 · 5 September · Morning Shift · Mathematics · Continuity and Differentiability

Let f ( x ) = x · [(x)/(2)], for - 10 < x < 10, where [t] denotes...

Let f(x)=x⋅[x2],f ( x ) = x \cdot \left[\dfrac{x}{2}\right], for −10<x<10,- 10 < x < 10, where \left[t\right]denotesthegreatestintegerfunction.Thenthenumberofpointsofdiscontinuityofdenotes the greatest integer function. Then the number of points of discontinuity off\left(x\right) is equal to

Correct answer

0

Step-by-step solution

We know that the greatest integer function is discontinuous at integral points.

−5<x2<5- 5 < \dfrac{x}{2} < 5

⇒[x2]=−5,−4,−3,−2,−1,0,1,2,3,4⇒ \left[\dfrac{x}{2}\right] = - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, are the integer values.

At x=0x = 0, f(0+)=f(0)=f(0−)=0f\left({0}^{+}\right) = f\left(0\right) = f\left({0}^{-}\right) = 0, which means that function is continuous at x=0x = 0.

∵x≠−10⇒x2≠−5∵ x \ne - 10 ⇒ \dfrac{x}{2} \ne - 5

Hence, the function is discontinuous at points −4,−3,−2,−1,1,2,3,4- 4, - 3, - 2, - 1, 1, 2, 3, 4

The number of values is 88 .

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