Continuity and Differentiability: 1 mark Questions (CBSE Class 12)
12 different 1 mark questions on Continuity and Differentiability from CBSE Class 12 Maths board exams 2026, newest first.
If e−x+e−y=2, then dxdy is
- (A)ex−y
- (B)ey−x
- (C)−ex−y
- (D)−ey−x
Show answer & solution
Answer: (D) −ey−x
- Differentiating: −e−x−e−ydxdy=0.
- dxdy=−e−ye−x=−ey−x.
If ex+y=3x, then dxdy is
- (A)ex+y3
- (B)ex+y1
- (C)ex+y1−ex+y
- (D)ex+y3−ex+y
Show answer & solution
Answer: (D) ex+y3−ex+y
- Differentiating: ex+y(1+dxdy)=3.
- dxdy=ex+y3−1=ex+y3−ex+y.
If x+y=xy, then dxdy is
- (A)x−1y
- (B)x−11
- (C)x−1y−1
- (D)x−11−y
Show answer & solution
Answer: (D) x−11−y
- Differentiating: 1+dxdy=y+xdxdy.
- dxdy(1−x)=y−1⇒dxdy=1−xy−1=x−11−y.
The value of k for which the function f(x)={x2sinx1,k(x+1),x=0x=0 is a continuous function, is :
- (A)41
- (B)2
- (C)21
- (D)0
Show answer & solution
Answer: (D) 0
- ∣x2sinx1∣≤x2→0, so limx→0f(x)=0.
- f(0)=k(0+1)=k.
- Continuity at 0 needs k=0.
If sin−1x=y, then dxdy is :
- (A)cos−1x
- (B)cosy
- (C)1−x21
- (D)secy
Show answer & solution
Answer: (D) secy
- x=siny, so dydx=cosy.
- dxdy=cosy1=secy.
If tan−1x=y, then dxdy is equal to :
- (A)(sec−1x)2
- (B)sec2y
- (C)1+x21
- (D)cos2y
Show answer & solution
Answer: (D) cos2y
- x=tany, so dydx=sec2y.
- dxdy=sec2y1=cos2y.
If 2cos−1x=y, then dxdy is :
- (A)−2sin−1x
- (B)−21sin2y
- (C)1−2x2−1
- (D)−2cosec2y
Show answer & solution
Answer: (D) −2cosec2y
- x=cos2y, so dydx=−21sin2y.
- dxdy=sin2y−2=−2cosec2y.
- (Equivalently 1−x2−2.)
Differential of eex with respect to x is :
- (A)logx
- (B)eex
- (C)exeex
- (D)(ex)2
Show answer & solution
Answer: (C) exeex
- By the chain rule, dxdeex=eex⋅dxd(ex).
- =eex⋅ex.
If f(x)={xsinx+cosx,k,x=0x=0 is continuous at x = 0, then the value of k is :
- (A)0
- (B)−2
- (C)−1
- (D)2
Show answer & solution
Answer: (D) 2
- limx→0(xsinx+cosx)=1+1=2.
- For continuity at x=0, k=f(0)=2.
The greatest integer function, f(x)=[x], 0<x<3 is not differentiable at how many points ?
- (A)At only one point
- (B)At only two points
- (C)At no point
- (D)At three points
Show answer & solution
Answer: (B) At only two points
- [x] jumps at every integer, so it is discontinuous (hence not differentiable) at integers.
- In 0<x<3 the integers are x=1 and x=2.
- Elsewhere f is constant on each piece and differentiable. So it fails at exactly two points.
If f(x)={x+1x2−4x−5,k,x=−1x=−1
is continuous at x=−1, then the value of k is :
- (A)Any real value
- (B)6
- (C)−1
- (D)−6
Show answer & solution
Answer: (D) −6
- For x=−1, x+1x2−4x−5=x+1(x−5)(x+1)=x−5.
- limx→−1f(x)=−1−5=−6.
- Continuity at x=−1 needs k=−6.
Derivative of cos−1(2sinx+cosx), −4π<x<4π with respect to x is :
- (A)−1
- (B)1
- (C)4π
- (D)−4π
Show answer & solution
Answer: (A) −1
- Let u=2sinx+cosx, so u′=2cosx−sinx.
- 1−u2=1−21+sin2x=2(cosx−sinx)2, and cosx−sinx>0 for −4π<x<4π.
- dxdcos−1u=−1−u2u′=−(cosx−sinx)/2(cosx−sinx)/2=−1.
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