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  3. NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.2

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.2

September 06, 2022

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-5 (Continuity And Differentiability)Exercise 5.2 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.2

Exercise 5.2

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

Question 1. Sin(x2 + 5)

Solution: 

y = sin(x2 + 5)

\frac{dy}{dx} = \frac{d(sin(x^2+5)}{dx}

= cos(x2 + 5) × \frac{d(x^2+5)}{dx}

= cos(x2 + 5) × (2x)

dy/dx = 2xcos(x2 + 5)

Question 2. cos(sin x)

Solution:

y = cos(sin x)

\frac{dy}{dx} = \frac{d \ cos(sin x)}{dx}

 = -sin(sin x) × \frac{d \ sinx}{dx}

= -sin(sin x)cos x  

Question 3. sin(ax + b)

Solution:

y = sin(ax + b)

\frac{dy}{dx}=\frac{d}{dx}sin(ax+b)

=cos(ax+b)\frac{d}{dy}(ax+b)

= a cos(ax + b)   

 Question 4. Sec(tan(√x)

Solution:

y = sec(tan√x)

\frac{dy}{dx} = \frac{d \ (sec(tan√x)) }{dx} 

= sec(tan √x) × tan(√x) × \frac{d \ (tan√x) }{dx}

= sec (tan √x) × tan (tan √x) × sec2√x × \frac{d \ (√x)}{dx} 

= sec(tan√x)tan(tan√x)(sec2√x)1/(2√x)

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

Question 5. \frac{sin(ax+b)}{cos(cx+d)}

Solution:

y = \frac{sin(ax+b)}{cos(cx+d)}

\frac{dy}{dx} = \frac{d (\frac{sin(ax+b)}{cos(cx+d)})}{dx}

=\frac{cos(cx+d)\frac{d}{dx}sin(ax+b)-sin(ax+b)\frac{d}{dx}cos(cx+d)}{(cos(cx+d))}

=\frac{cos(cx+d)cos(ax+b)(a)+sin(ax+b)sin(cx+d)(c)}{cos^2(cx+b)}

Question 6. cos x3.sin2(x5)

Solution:

y = cos x3.sin2(x5)

\frac{dy}{dx} = cosx^3\frac{d(sin^2(x^5)}{dx}+sin^2(x^5)\frac{d(cosx^3)}{dx}

= cosx^3.2sin(x^5)\frac{d(sin(x^5)}{dx}+sin^2(x^5)(-sinx^3)\frac{d(x^3)}{dx}

= 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))

\frac{dy}{dx} = \frac{\ d(2\sqrt{cos(x2})}{dx}

= 2\frac{\ d(\sqrt{cos(x^2})}{dx}

= 2\frac{1}{2\sqrt {cotx^2}}.\frac{d(cotx^2)}{dx}

= \frac{1}{\sqrt {cot(x^2)}}.(-cosec^2(x^2)).\frac{d(x^2)}{dx}

= \frac{1}{\sqrt {cot(x^2)}}.(-cosec^2(x^2)).2x

=\frac{-2xcosec^2x^2}{\sqrt{cot(x^2)}}

= \frac{-2x}{sin^2(x^2).\sqrt{\frac{cos(x^2)}{sin(x^2)}}}

= \frac{-2x}{sin(x^2) \times sin(x^2)\sqrt{\frac{cos(x^2)}{sin(x^2)}}}

= \frac{-2x}{sin(x^2) \times \sqrt{sin(x^2) \times \frac{cos(x^2)}{sin(x^2)}}}

= \frac{-2x}{sin(x^2) \times \sqrt{cos(x^2) \times {sin(x^2)}}}

= \frac{-2x}{sin(x^2) \times \sqrt{\frac{2}{2} \times cos(x^2) \times {sin(x^2)}}}

= \frac{-2\sqrt{2}x}{sin(x^2) \times \sqrt{2 \times cos(x^2) \times {sin(x^2)}}}

= \frac{-2\sqrt{2}x}{sin(x^2) \times \sqrt{sin(2x^2)}}

Question 8. cos (√x)

Solution:

y = cos (√x)

dy/dx = -sin√x\frac{d\sqrt{x}}{dx}

=-sin\sqrt{x}\frac{1}{2}(x)^\frac{-1}{2}

=\frac{-sin\sqrt{x}}{2\sqrt{x}}

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

Solution:

RHD=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}                                                            

= \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}                                                      

= \lim_{h\to 0}\frac{((1+h)-1)-(1-1)}{h} 

= \lim_{h\to 0}\frac{h-0}{h} 

=\lim_{h\to 0}(1)                                                      

= +1                                                                                                            

 LHD=\lim_{h\to 0}\frac{f(x)-f(x-h)}{h} 

= \lim_{h\to 0}\frac{(1)-f(1-h)}{h}

\lim_{h\to 0}\frac{(1-1)-(-(1-h)-1)}{h}

= \lim_{h\to 0}\frac{0-h}{h}

= \lim_{h\to 0}(-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) = \lim_{h\to0} \frac{f(x - h)-f(x)}{-h}

= \lim_{h\to0} \frac{f(1 - h)-1}{-h}

=\lim_{h\to0} \frac{0-1}{-h}

= ∞

RHS:

f'(1) = \lim_{h\to0} \frac{f(x + h) - f(x)}{h}

= \lim_{h\to0} \frac{f(1 + h) - f(1)}{h}

= \lim_{h\to0} \frac{1-1}{h}

= \lim_{h\to0} \frac{0}{h}

= 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. 

Chapter-5 (Continuity And Differentiability)

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