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

 Chapter 5 of the Class 12 NCERT Mathematics textbook focuses on the concepts of continuity and differentiability which are fundamental in understanding calculus. This chapter helps students grasp how functions behave concerning their limits and derivatives. Exercise 5.2 is designed to test and reinforce these concepts through the various problems.

Continuity and Differentiability

The Continuity of a function at a point means that the function's value is consistent and predictable as the input approaches that point. Formally, a function f(x) is continuous at x=a if lim_x→af(x)=f(a). The Differentiability, on the other hand, refers to the function's ability to have a derivative at a point which implies that the function must be continuous there. If a function is differentiable at a point it must also be continuous at that point but the converse is not always true.

Differentiate the function with respect to x in Questions 1 to 8

Question 1. Sin(x² + 5)

Solution: 

y = sin(x² + 5)

 = 

= cos(x² + 5) × 

= cos(x² + 5) × (2x)

dy/dx = 2xcos(x² + 5)

Question 2. cos(sin x)

Solution:

y = cos(sin x)

 = 

 = -sin(sin x) × 

= -sin(sin x)cos x  

Question 3. sin(ax + b)

Solution:

y = sin(ax + b)

= a cos(ax + b)  

 Question 4. Sec(tan(√x)

Solution:

y = sec(tan√x)

 = 

= sec(tan √x) × tan(√x) × 

= sec (tan √x) × tan (tan √x) × sec2√x × 

= sec(tan√x)tan(tan√x)(sec2√x)1/(2√x)

= 1/(2√x) × sec(tan√x)tan(tan√x)(sec2√x)

Question 5. 

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. 2√(cos(x²))

Solution:

y = 2√(cos(x²))

 = 

= 2

= 

= 

= 

=

= 

= 

= 

= 

= 

= 

= 

Question 8. cos (√x)

Solution:

y = cos (√x)

dy/dx = -sin√x

=

=

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

Solution:

= 

= 

=

= +1                                                                                                            

= 

= 

= 

= -1   

LHD ≠ 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) = 

= 

=

= ∞

RHS:

f'(1) = 

= 

= 

= 

= 0

LHS ≠ 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. 

Read More:

• Differentiability of a Function | Class 12 Maths
• Limits, Continuity and Differentiability

Conclusion

Understanding continuity and differentiability is crucial for the analyzing functions in calculus. Exercise 5.2 offers a practical approach to the applying these concepts helping students solidify their knowledge. Mastery of these topics prepares students for the more advanced calculus concepts and real-world applications.