# Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.2

Last Updated : 03 Mar, 2021

### Question 1. Sin(x2 + 5)

Solution:

y = sin(x2 + 5)

=

= cos(x2 + 5) Ã—

= cos(x2 + 5) Ã— (2x)

dy/dx = 2xcos(x2 + 5)

### Question 2. cos(sin x)

Solution:

y = cos(sin x)

=

= -sin(sin x) Ã—

= -sin(sin x)cos x

### Question 3. sin(ax + b)

Solution:

y = sin(ax + b)

= a cos(ax + b)

### Question 4. Sec(tan(âˆšx)

Solution:

y = sec(tanâˆšx)

=

= sec(tan âˆšx) Ã— tan(âˆšx) Ã—

= sec (tan âˆšx) Ã— tan (tan âˆšx) Ã— sec2âˆšx Ã—

= sec(tanâˆšx)tan(tanâˆšx)(sec2âˆšx)1/(2âˆšx)

= 1/(2âˆšx) Ã— sec(tanâˆšx)tan(tanâˆšx)(sec2âˆšx)

Solution:

y =

=

### Question 6. cos x3.sin2(x5)

Solution:

y = cos x3.sin2(x5)

= cos x3.2sin(x5) .cos(x5(5x4)(5x4) – sin2(x5).sin x3.3x2

= 10x4 cos x3sin(x5)cos(x5) – 3x2 sin2(x5)sin x3

### Question 7. 2âˆš(cos(x2))

Solution:

y = 2âˆš(cos(x2))

=

= 2

=

Solution:

y = cos (âˆšx)

dy/dx = -sinâˆšx

=

=

### Question 9. Prove that the function f given by f(x) = |x – 1|, x âˆˆ R is not differentiable at x = 1.

Solution:

=

= +1

= -1

LHD â‰  RHD

Hence, f(x) is not differentiable at x = 1

### Question 10. Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.

Solution:

Given: f(x) = [x], 0 < x < 3

LHS:

f'(1) =

=

= âˆž

RHS:

f'(1) =

= 0

LHS â‰  RHS

So, the given f(x) = [x] is not differentiable at x = 1.

Similarly, the given f(x) = [x] is not differentiable at x = 2.

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