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NCERT Solutions for Exercise 5.2 Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solutions for Exercise 5.2 Class 12 Maths Chapter 5 - Continuity and Differentiability

Edited By Komal Miglani | Updated on Apr 23, 2025 11:00 PM IST | #CBSE Class 12th

If we can imagine a function as a road, then continuity is that the road has no gaps or breaks. While differentiability means that the road is smooth enough to ride a bike comfortably. Together, they make a perfect road for calculus to travel on. In exercise 5.2 of the chapter Continuity and Differentiability, we will look beyond the concepts of continuity and dive deep into the world of differentiability. We will learn if a function is properly smooth and consistent, so that we can differentiate that function. This article on the NCERT Solutions for Exercise 5.2 Class 12 Maths Chapter 5 - Continuity and Differentiability provides clear and step-by-step solutions for the exercise problems, so that students can clear their doubts and improve their understanding about differentiability. For syllabus, notes, and PDF, refer to this link: NCERT.

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Class 12 Maths Chapter 5 Exercise 5.2 Solutions: Download PDF

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Continuity and Differentiability Exercise: 5.2

Question:1.Differentiate the functions with respect to x in

sin(x2+5)

Answer:

Given function is
f(x)=sin(x2+5)
when we differentiate it w.r.t. x.
Lets take t=x2+5 . then,
f(t)=sint
df(t)dx=df(t)dt.dtdx (By chain rule)
df(t)dt=d(sint)dt=cost=cos(x2+5)
dtdx=d(x2+5)dx=2x
Now,
df(t)dx=df(t)dt.dtdx=cos(x2+5).2x
Therefore, the answer is 2xcos(x2+5)

Question:2. Differentiate the functions with respect to x in

cos(sinx)

Answer:

Given function is
f(x)=cos(sinx)
Lets take t=sinx then,
f(t)=cost
df(t)dx=df(t)dt.dtdx ( By chain rule)
df(t)dt=d(cost)dt=sint=sin(sinx)
dtdx=d(sinx)dt=cosx
Now,
df(t)dx=df(t)dt.dtdx=sin(sinx).cosx
Therefore, the answer is sin(sinx).cosx

Question:3. Differentiate the functions with respect to x in

sin(ax+b)

Answer:

Given function is
f(x)=sin(ax+b)
when we differentiate it w.r.t. x.
Lets take t=ax+b . then,
f(t)=sint
df(t)dx=df(t)dt.dtdx (By chain rule)
df(t)dt=d(sint)dt=cost=cos(ax+b)
dtdx=d(ax+b)dx=a
Now,
df(t)dx=df(t)dt.dtdx=cos(ax+b).a
Therefore, the answer is acos(ax+b)

Question:4. Differentiate the functions with respect to x in

sec(tan(x))

Answer:

Given function is
f(x)=sec(tan(x))
when we differentiate it w.r.t. x.
Lets take t=x . then,
f(t)=sec(tant)
take tant=k. then,
f(k)=seck
df(k)dx=df(k)dk.dkdt.dtdx (By chain rule)
df(k)dk=d(seck)dk=secktank=sec(tanx)tan(tanx)
(k=tant and t=x)
df(t)dt=d(tant)dt=sec2t=sec2(x)      (t=x)
dtdx=d(x)dx=12x
Now,
df(k)dx=df(k)dk.dkdt.dtdx=sec(tanx)tan(tanx).sec2(x).12x
Therefore, the answer is sec(tanx).tan(tanx).sec2(x)2x

Question:5. Differentiate the functions with respect to x in

sin(ax+b)cos(cx+d)

Answer:

Given function is
f(x)=sin(ax+b)cos(cx+d)=g(x)h(x)
We know that,
f(x)=g(x)h(x)g(x)h(x)h2(x)
g(x)=sin(ax+b) and h(x)=cos(cx+d)
Lets take u=(ax+b) and v=(cx+d)
Then,
sin(ax+b)=sinu and cos(cx+d)=cosc
g(x)=d(g(x))dx=d(g(x))du.dudx (By chain rule)
d(g(x))du=d(sinu)du=cosu=cos(ax+b)         (u=ax+b)
dudx=d(ax+b)dx=a
g(x)=acos(ax+b) -(i)
Similarly,
h(x)=d(h(x))dx=d(h(x))dv.dvdx
d(h(x))dv=d(cosv)dv=sinv=sin(cx+d)       (v=(cx+d))
dvdx=d(cx+d)dv=c
h(x)=csin(cx+d) -(ii)
Now, put (i) and (ii) in
f(x)=g(x)h(x)g(x)h(x)h2(x)=acos(ax+b).cos(cx+d)sin(ax+b).(c.sin(cx+d))cos2(cx+d)
=acos(ax+b).cos(cx+d)cos2(cx+d)+sin(ax+b).c.sin(cx+d)cos2(cx+d)
=acos(ax+b).sec(cx+d)+csin(ax+b).tan(cx+d).sec(cx+d)
Therefore, the answer is acos(ax+b).sec(cx+d)+csin(ax+b).tan(cx+d).sec(cx+d)

Question:6. Differentiate the functions with respect to x in

cosx3.sin2(x5)

Answer:

Given function is
f(x)=cosx3.sin2(x5)
Differentitation w.r.t. x is
f(x)=g(x).h(x)+g(x).h(x)
g(x)=cosx3 and h(x)=sin2(x5)
Lets take u=x3 and v=x5
Our functions become,
cosx3=cosu and sin2(x5)=sin2v
Now,
g(x)=d(g(x))dx=d(g(u))du.dudx ( By chain rule)
d(g(u))du=d(cosu)du=sinu=sinx3    (u=x3)
dudx=d(x3)dx=3x2
g(x)=sinx3.3x2 -(i)
Similarly,
h(x)=d(h(x))dx=d(h(v))dv.dvdx
d(h(v))dv=d(sin2v)dv=2sinvcosv=2sinx5cosx5   (v=x5)

dvdx=d(x5)dx=5x4
h(x)=2sinx5cosx5.5x4=10x4sinx5cosx5 -(ii)
Put (i) and (ii) in
f(x)=g(x).h(x)+g(x).h(x)=3x2sinx3.sin2x5+cosx3.10x4sinx5cosx5
Therefore, the answer is 10x4sinx5cosx5.cosx33x2sinx3.sin2x5

Question:7. Differentiate the functions with respect to x in

2cot(x2)

Answer:

Give function is
f(x)=2cot(x2)
Let's take t=x2
f(t)=2cott
Now, take cott=k2
f(k)=2k
Differentiation w.r.t. x
d(f(k))dx=d(f(k))dk.dkdt.dtdx -(By chain rule)
d(f(k))dk=d(2k)dk=2
dkdt=d(cott)dt=12cott.(cosec2t)=cosec2x22cotx2   (t=x2)
dtdx=d(x2)dx=2x
So,
d(f(k))dx=2.cosec2x22cotx2.2x=22xsin2x22sinx2cosx2sin2x2 ( Multiply and divide by 2 and multiply and divide cotx2 by sinx2

(cotx=cosxsinx and cosecx=1sinx)
=22xsinx2sin2x2    (2sinxcosx=sin2x)
There, the answer is 22xsinx2sin2x2

Question:8 Differentiate the functions with respect to x in

cos(x)

Answer:

Let us assume : y = cos(x)

Differentiating y with respect to x, we get :

dydx = d(cos(x))dx

or = sinx.d(x)dx

or = sinx2x

Question:9. Prove that the function f given byf(x)=|x1|,xϵR is not differentiable at x = 1.

Answer:

Given function is

f(x)=|x1|,xR

We know that any function is differentiable when both

limh0f(c+h)f(c)h and limh0+f(c+h)f(c)h are finite and equal.

The required condition for the function to be differentiable at x=1 is

limh0f(1+h)f(1)h=limh0+f(1+h)f(1)h.

Now, the Left-hand limit of the function at x=1 is

limh0f(1+h)f(1)h=limh0|1+h1||11|h=limh0|h|0h

=limh0hh=1(h<0).

The Right-hand limit of the function at x=1 is

limh0+f(1+h)f(1)h=limh0+|1+h1||11|h=limh0+|h|0h

=limh0+hh=1.

Now, it is clear that

R.H.L. at x=1L.H.L. at x=1.
Therefore, function f(x)=|x1| 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.

Answer:

Given function is

f(x)=x,0<x<3

We know that any function is differentiable when both

limh0f(c+h)f(c)h and limh0+f(c+h)f(c)h are finite and equal.

The required condition for the function to be differentiable at x=1 is

limh0f(1+h)f(1)h=limh0+f(1+h)f(1)h.

Now, the Left-hand limit of the function at x=1 is

limh0f(1+h)f(1)h=limh01+h1h=limh001h

=limh01h=(h<01+h<1,1+h=0).

The Right-hand limit of the function at x=1 is

limh0+f(1+h)f(1)h=limh0+1+h1h=limh0+11h

=limh0+0h=0(h>01+h>1,1+h=1).

Now, it is clear that

R.H.L. at x=1L.H.L. at x=1, and L.H.L. is not finite as well.

Therefore, the function f(x)=x is not differentiable at x=1.

Similarly, for x=2,

The required condition for the function to be differentiable at x=2 is

limh0f(2+h)f(2)h=limh0+f(2+h)f(2)h.

Now, the Left-hand limit of the function at x=2 is

limh0f(2+h)f(2)h=limh02+h2h=limh012h

=limh01h=(h<02+h<2,2+h=1).

The Right-hand limit of the function at x=2 is

limh0+f(2+h)f(2)h=limh0+2+h2h=limh0+22h

=limh0+0h=0(h>02+h>2,2+h=2).
Now, it is clear that
R.H.L. at x= 2 L.H.L. at x= 2 and L.H.L. is not finite as well
Therefore, function f(x)=[x] is not differentiable at x = 2


Also Read,

Topics covered in Chapter 5, Continuity and Differentiability: Exercise 5.2

The main topics covered in Chapter 5 of continuity and differentiability, exercises 5.1 are:

  • Differentiability: Differentiability is the property of a function which denotes that the function has a derivative at the given point or interval. Mathematically, we can say that,
    If limh0f(c+h)f(c)h=limh0+f(c+h)f(c)h then f(x) is said to be differentiable at x=c.
  • Related theorems: Some theorems related to differentiability and continuity are-
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If a function f(x) is said to be differentiable at a point x=c, then it will also be continuous at that point.

Conversely, if a function f(x) is continuous at a point x=c, then it might not be Differentiable at that point.

  • Derivatives of composite functions: This refers to finding the derivative of a function that is made up of two or more functions. To solve this type of problem, we can use the chain rule.
  • Chain rule: Let f be a real function which is composed of two functions u and v. Suppose t=u(x), then dfdx=dvdtdtdx.
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NCERT Solutions Subject Wise

Below are some useful links for subject-wise NCERT solutions for class 12.

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Here are some links to subject-wise solutions for the NCERT exemplar class 12.

Frequently Asked Questions (FAQs)

1. What is the definition of the composite function ?

A composite function is obtained from other functions where the output of one function is the input of another function.

2. Give an example of composite function ?

sin (x/2)  is an example of a composite function.

3. What is derivative of sin(x/2) ?

d(sin(x/2)/dx = cos(x/2)/2 

4. What is derivative of cos(x/2) ?

d(cos(x/2)/dx = -sin(x/2)/2

5. what is a limit of a function ?

Limit of function is defined as the value of functions reaches when the limit reaches.

6. Does calculus is so hard ?

Some people consider it hard as it is a new concept included in class 11 maths and the level of maths till class 10 is too easy.

7. Why calculus is so hard ?

As the foundation of your maths is low, you may find it hard to grasp but with more practice, you will grasp the concept easily.

8. Can I get free NCERT Syllabus for Class 12 Maths ?

Yes, You can check here for the NCERT Syllabus for Class 12 Maths. NCERT book exercise questions, NCERT exemplar problems and solutions are helpful to practice questions for the CBSE board exam.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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