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Continuity and Differentiability · Question 2

JEE Main 2026 (Online) 6th April Evening ShiftNumericalMedium📈 Topic in 35% papers+4
1

Let f(x)= casesx3+8; & x<0,\ x2-4; & x 0, cases and g(x)= cases(x-8)1/3; & x<0,\ (x+4)1/2; & x 0. cases Then the number of points, where the function g f is discontinuous, is .

Answer

3

Solution

[figure]

g f(x)=g(f(x)); the sign of f(x) decides which branch of g is used.

For x<0: f(x)=x3+8<0 x<-2

For x 0: f(x)=x2-4<0 0 x<2

g(f(x))= casesx; & x<-2\ x3+12; & -2 x<0\ (x2-12 )1/3; & 0 x<2\ x; & x 2 cases

(x<-2: (x3+8-8 )1/3=x; x 2: x2-4+4=x)

Each piece is continuous inside its own interval; check the joining points:

x=-2: LHL =-2, value = 4=2 discontinuous

x=0: LHL = 12, value =(-12)1/3 discontinuous

x=2: LHL =(-8)1/3=-2, value =2 discontinuous

Number of points of discontinuity =3

Answer: 3

JEE Main 2026 (Online) 6th April Evening Shift

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If y= -1 ( 3 x-4 x4 x+3 x )+2 -1 ( x1+ 1-x2 ), then dydx at x= 32 is equal to :

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3

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6

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JEE Main 2026 (Online) 24th January Evening Shift
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JEE Main 2026 (Online) 23rd January Evening Shift

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