Let f(x)= cases 13, & x 2\ b(1- x)( -2x)2, & x> 2 cases. If f is continuous at x= 2, then the value of 03b-6 |x2+2x-3 |dx is :
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Previous year questions with solutions · 2025 – 2026
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 13 | 19 | 0.68 | 2.74% ▲ 62.5% vs last year |
| 2025 | 8 | 19 | 0.42 | 1.68% |
Let f(x)= cases 13, & x 2\ b(1- x)( -2x)2, & x> 2 cases. If f is continuous at x= 2, then the value of 03b-6 |x2+2x-3 |dx is :
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 .
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The number of points, at which the function f(x)= \6x, 2+3x2 \+|x-1| |x2- 14 |, x (- , ), is not differentiable, is .
Let f(x)= casesex-1, & x<0\ x2-5x+6, & x 0 cases and g(x)=f(|x|)+|f(x)|. If the number of points where g is not continuous and is not differentiable are and respectively, then + is equal to .
The number of points in the interval [2, 4], at which the function f(x)= [x2-x- 12 ], where [ ] denotes the greatest integer function, is discontinuous, is .
Let f(x) = 0 ( x - x ( 2 ) (x - 1)1 + x ( 2 )(x - 1) ), x R. Consider the following two statements : (I) f(x) is discontinuous at x = 1. (II) f(x) is continuous at x = -1. Then,
Let , R be such that the function f(x) = cases 2 (x2 - 2 ) + 2 x, & x < 1 \ ( + 3)x + ( - ), & x 1 cases be differentiable at all x R. Then 34( + ) is equal to
Let [t] denote the greatest integer less than or equal to t. If the function f(x) = cases b2 ( 2 [ 2( x + x) x ] ), & x < 0 \ x - 12 2xx3, & x > 0 \ a, & x = 0 cases is continuous at x = 0, then a2 + b2 is equal to :
Let f(x) = cases ax2 + 2ax + 34x2 + 4x - 3 & , x - 32, 12 \ b & , x = - 32, 12 cases be continuous at x = - 32. If f f(x) = 75, then x is equal to:
If f(x) = cases a|x| + x2 - 2( |x|)( |x|)x & , x 0 \ b & , x = 0 cases is continuous at x = 0, then a + b is equal to :
Let [ ] denote the greatest integer function, and let f(x) = \ 2x, x2 \. Let S = \x (-2, 2) : the function g(x) = |x| [x2 ] is discontinuous at x\. Then x S f(x) equals
The number of points of discontinuity of the function f(x) = [ x22 ] - [ x], x [0, 4], where [ ] denotes the greatest integer function, is .
If the function f(x) = ( x) - ( x) x - x is continuous at x = 0, then f(0) is equal to .
Let m and n be the number of points at which the function f(x) = \x, x3, x5, , x21 \, x R, is not differentiable and not continuous, respectively. Then m + n is equal to .
Let f(x) = cases (1 + ax)1/x, & x < 0 \ 1 + b, & x = 0 \ (x+4)1/2 - 2(x+c)1/3 - 2, & x > 0 cases be continuous at x = 0. Then ea bc is equal to:
Let the function f(x)= (x2-1 ) |x2-ax+2 |+ |x| be not differentiable at the two points x= =2 and x= . Then the distance of the point ( , ) from the line 12x+5y+10=0 is equal to :
Let f(x) = cases 3x, & x < 0 \ \1 + x + [x], x + 2[x]\, & 0 x 2 \ 5, & x > 2 cases where [.] denotes greatest integer function. If and are the number of points, where f is not continuous and is not differentiable, respectively, then + equals .
Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function f(x)=[x]+|x-2|, -2<x<3, is not continuous and not differentiable. Then m+n is equal to :
If the function f(x)= cases 2x \ (k1+1)x+ (k2-1)x \, & x<0\\[2mm] 4, & x=0\\[2mm] 2x e ( 2+k1x2+k2x ), & x>0 cases is continuous at x=0, then k12+k22 is equal to :