CBSE Class 12 Maths Revision Notes Chapter 5

Mathematics can be easy or difficult, depending on how well a student understands the concepts. It is essential to have a strong foundation so that it becomes easier to grasp knowledge about high-level topics. This subject cannot be mastered with memorisation. To excel in Mathematics, students need to practice regularly. The Class 12 Mathematics chapter 5 notes- Continuity and Differentiability give a clear understanding of different concepts, such as the algebra of continuous functions. It also includes differentiability of different functions such as inverse trigonometric, implicit, composite and more. In this chapter, students will learn about second-order derivatives and the Mean Value theorem.

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List Of The Topics Covered In Class 12 Mathematics Chapter 5 Notes

The list of topics covered in Extramarks class 12 Mathematics notes chapter 5 Continuity and Differentiability include the following.

Introduction
Continuity and Algebra of Continuous Function
Differentiability
Derivatives of Composite, Implicit and Inverse Trigonometric Function
Logarithmic and Exponential Function
Logarithmic Differentiation
Parametric Form Differentiation
Second-Order Derivative
Mean Value Theorem and Rolle’s Theorem

Key Topics Covered In Class 12 Mathematics Chapter 5 Notes

The key topics covered under Class 12 Mathematics Chapter 5 notes include

Definition of Continuity at a point:

The continuity of a function (f) at any point x = c, where c belongs to the domain of the function, is given as, Left-hand limit of f(x = c) = Right-hand limit of f(x = c) = Value of f(x=c) , i.e.,

If x = c, then LHL = RHL = f(c).

Therefore, xa–f(x) = xa+f(x) = f (x = c) = f(c)

REMEMBER:

To find the solution for LHL of f(x) at x=0, substitute x = a – h and for RHL, substitute x = a + h.
If xa–f(x) and xa+f(x) exists but LHL RHL, then the function is not continuous.
If xa–f(x) and xa+f(x) exists but LHL = RHL f(c), then the function is discontinuous.
At least one limit of the f does not exist; then, the function is discontinuous.

Definition of Continuity in an Interval (INTERMEDIATE VALUE THEOREM):

The continuity of a function (f) in an interval [a, b] where a<b then, if the function is continuous at all points, including the end-points in the given interval.

Continuity at point ‘a’ is given as,

xaf(x) = f (a)

Continuity at point ‘b’ is given as,

xbf(x) = f (b)

Algebra of Continuous Functions

If f(x) and g(x) be two real functions. If they are continuous at any point c R, then

f(x) + g(x) is continuous at c.
f(x) – g(x) is also continuous at c.
f(x) . g(x) is continuous at c.
f(x) / g(x) is continuous at c, g(c) 0.
The product c.f(x) is continuous, where c is a constant.
Consider a composite function fog(x), if g (x) is continuous at x and f(x) is continuous at g(c), then fog(x) is continuous at point c.
If f(x) is continuous, then f(x) is also continuous.

Definition and formula of differentiability:

If f(x) is a real function. It is said to be differentiable at a point c in its domain if the left-hand derivative is equal to the right-hand derivative of the function at point c, i.e.

LHD f(x=c) = RHD f(x=c).

The derivative of a function is given ash0f(c+h) – f(c)h

Therefore, RHD f’(c) = h0f(c+h) – f(c)h and LHD f’(c) = h0f(c-h) – f(c)-h

Types of Discontinuity:

REMOVABLE DISCONTINUITY:

If xaf(x) exists but is not equal to f(c), for c R, then function f(x) has removable discontinuity. In such a case, we can redefine the function and make it continuous.

1. Missing point discontinuity:

In this type of discontinuity, xaf(x) exists but f(c) does not exist.

2. Isolated point discontinuity:

In this type of discontinuity, xaf(x) exists and f(c) exist, but xaf(x) f(c).

NON-REMOVABLE DISCONTINUITY:

In this case, xaf(x) does not exist, and so the function cannot be redefined to make it continuous.

1. Finite Discontinuity:

This type of discontinuity exists when the LHL and RHL limits do not exist, but their values are both finite, yet LHL RHL.

2. Infinite Discontinuity:

This type of discontinuity exists when either LHL or RHL or both limits tend toward infinity.

3. Oscillatory Discontinuity:

This type of discontinuity exists when limits oscillate between two finite values.

Relation between continuity and differentiability:

Every f(x) is differentiable; then it is also continuous. But, if f(x) is continuous, then it is not differentiable.
A function that is not continuous (discontinuous) will not be differentiable.
If f is function it is differentiable a a point x = c, then f is also continuous at x = c.

REMEMBER:

Differentiable Continuous

Not Differentiable Not Continuous

Not Continuous Not Differentiable

Derivatives of inverse functions:

If f and g are two real-valued functions such that y = f(x) and x= g(y) then

g(f(x)) = x ddx g(f(x)) = g’(f(x)). f’(x) = 1

Using this result, we can say that dydx . dxdy= 1 or dydx = 1dxdy or dxdy = 1dydx where dxdy0

Rules of Differentiation:

1. Sum and difference rule:

The derivative of y= f(x) g(x) is given as

dydx= ddyf(x) ddy g(x)

2. Product Rule:

The derivative of y= f(x). g(x) can be remembered with the help of the u-v rule. It is defined as

ddxu.v = (ddxu) v + (ddxv) u i.e.,

dydx= (ddyf(x)) g(x) + (ddyg(x)) f(x)

3. Quotient Rule:

The derivative of y= f(x) / g(x) can be remembered with the help of the the u-v rule. It is defined as

ddx(uv) = u (ddxv) + v (ddx u)v2 i.e.,

ddx(f(x)g(x)) = (ddxg(x)) f(x) + (ddx f(x)) g(x)g(x)2

4. Chain Rule:

i) for two functions:

Let y = f(u), u= f(x) and if dduy and ddxu exists then,

We can write ddxy = dduy . ddxu

ii) for three functions:

Let y = f(u), u= f(w), w= f(x) and if dduy, ddwu and ddxw exists then,

We can write ddxy = dduy . ddxu. ddxw

1. Differentiation in Parametric Form:

If x and y are expressed in the form x = f(t), y = g(t), where t is a parameter. It is given as

ddxy = ddtyddtx, provided ddtx 0.

2. Logarithmic Differentiation:

Consider y = f(x)g(x). This function can be treated in the following way:

Taking loge on both sides, we get loge y = g(x) loge f(x). The derivation can further be calculated by using the chain rule, given as

ddxy= f(x)g(x) g(x)f(x)f'(x) + g'(x) log f(x)

Second-Order Derivative:

It is denoted by y2 or y’’.

The derivative of the first-order derivative of a function is called the second-order derivative.

d2dx2y= ddx(ddxy)

Derivative of infinite series:

If y = f(x) +f(x) +f(x) +…. ∞

Then y can be treated as y= f(x) + y

By differentiating, we get, (2y-1) ddxy = f’(x)

Rolle’s Theorem:

If function f:[a, b] R is continuous of [a, b] and differentiable on (a, b) and f(a) = f(b), then at least one number c in the interval (a, b) such that f’(c) = 0, a,b,c R.

Lagrange’s Mean Value Theorem:

It is an expansion of Rolle’s Theorem. It states that if function f:[a, b] R is continuous of [a, b] and differentiable on (a, b) and f(a) = f(b), then at least one number c in the interval (a, b) such that

f’(c) = f(b) – f(a)b – a, where a,b,c R.

Useful substitutions for finding derivatives:

Expression:	Substitution:
1)a2 + x2	x = a tan or x = a cot
2)a2 – x2	x = a sin or x = a cos
3) x2 – a2	x = a sec or x = a cosec
4) a – xa + x or a + xa – x	x = a cos 2
5) a2 – x2a2 + x2 or a2 + x2a2 – x2	x2= a2 cos 2

Derivatives of Trigonometric functions:

Expression:	Derivatives:	Expression:	Derivatives:
ddxsin x	cos x	ddxsec x	sec x . tan x
ddxcos x	– sin x	ddxcosec x	– cosec x . cot x
ddxtan x	sec2x	ddxcot x	– cosec2x

Derivatives of Inverse Trigonometric functions:

ddxsin-1x	11 – x2	ddxcosec-1x	1x x2 – 1
ddxcos-1x	-11 – x2	ddxsec-1x	-1x x2 – 1
ddxtan-1x	11 + x2	ddxcot-1x	-11 + x2

Derivatives of standard functions:

ddxxn	n.xn-1	ddx (constant)	0
ddxex	ex	ddxLog x	1x, where x 0
ddxax	ax log a, where a > 0

List of Continuous Functions:

Function f(x)	Interval in which f is continuous
Constant c	(-∞, ∞)
xn, n is an integer, n 0	(-∞, ∞)
x-n, x is a positive integer	(-∞, ∞) – {0}
x-a	(-∞, ∞)
P (x) = a0xn+ a1xn-1+ a2xn-2+…..+ an	(-∞, ∞)
p(x)q(x), where p(x) and q(x) are polynomial in x	(-∞, ∞), provided q(x) 0
sin x	(-∞, ∞)
cos x	(-∞, ∞)
tan x	(-∞, ∞) – {(2n +1)2: n I
cot x	(-∞, ∞) – {n: n I}
sec x	(-∞, ∞) – {(2n +1)2: n I
cosex x	(-∞, ∞) – {n : n I}
ex	(-∞, ∞)
Log x	(0, ∞)

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Class 12 Mathematics chapter 5 notes Exercises & Answer Solutions.

Chapter 5 Mathematics class 12 notes begin with a recall of all concepts students learnt in Class 11. To revise the previse topics, students can refer to CBSE Solutions for class 11. With the help of the notes, students can learn in detail about Continuity and Differentiation. Topics such as second-order derivatives, Rolle’s theorem, and MVT theorem are explained in detail in the Class 12 Mathematics chapter 5 notes. Students will also gain knowledge about Derivatives of standard functions and Continuous Functions.

To help students gain knowledge of the chapter, Extramarks provides detailed and well-structured Class 12 Mathematics chapter 5 notes. All important definitions, formulas, properties, and theorems are included in the notes. With the help of the CBSE solutions, it will become easier for students to tackle the most difficult questions in no time. The step-by-step solutions in the Class 12 Mathematics chapter 5 notes help students to understand each step and how to solve the problems ahead. It empowers students to get closer to pursuing their dream careers.

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Q.1

$Differentiate : y = \sqrt{logx + \sqrt{logx + \sqrt{logx + . \dots}}}$

Ans

$\begin{array}{l} y = \sqrt{logx + \sqrt{logx + \sqrt{logx + . \dots}}} \\ Squaring both sides \\ y^{2} = \log + y \\ Differentiating w . r . t . x \\ \frac{d}{dx} (y^{2}) = \frac{d}{dx} (logx + y) \\ \Rightarrow 2 y \frac{dy}{dx} = \frac{1}{x} + \frac{dy}{dx} \\ \Rightarrow 2 y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{x} \\ \Rightarrow \frac{dy}{dx} = \frac{1}{x (2 y - 1)} \end{array}$

Q.2

$If x = a (θ - sinθ), y = a (1 + cosθ), find \frac{d^{2} y}{{dx}^{2}} at θ = \frac{π}{2} .$

Ans

$\begin{array}{l} x = a (θ - sinθ) \Rightarrow \frac{dx}{dθ} = a (1 - cosθ) \\ y = a (1 + cosθ) \Rightarrow \frac{dx}{dθ} = - asinθ \\ \Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}} \frac{- asinθ}{a (1 - cosθ)} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} = - \cot \frac{θ}{2} \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = \frac{1}{2} {cosec}^{2} \frac{θ}{2} . \frac{dθ}{dx} \\ = {cosec}^{2} \frac{θ}{2} . \frac{dθ}{dx} = \frac{1}{2} {cosec}^{2} \frac{θ}{2} . \frac{1}{a (1 - cosθ)} \\ \Rightarrow {\frac{d^{2} y}{{dx}^{2}}|}_{θ = \frac{π}{2}} = \frac{1}{2} {cosec}^{2} \frac{π}{4} . \frac{1}{a (1 - \cos \frac{π}{2})} \\ = \frac{1}{2} {(\sqrt{2})}^{2} \frac{1}{a (1 - 0)} = \frac{2}{2 a} = \frac{1}{a} \end{array}$

Q.3

$Differentiate \tan^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}] w . r . t \tan^{- 1} x .$

Ans

$\begin{array}{l} Let y = \tan^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}], u = \tan^{- 1} x \\ y = \tan^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}} \times [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}]] \\ = \tan^{- 1} [\frac{{(\sqrt{1 + sinx} + \sqrt{1 - sinx})}^{2}}{\sqrt{1 + sinx} - \sqrt{1 + sinx}}] \\ = \tan^{- 1} [\frac{1 + sinx + 1 - sinx + 2 \sqrt{1 - \sin^{2} x}}{2 sinx}] \\ = \tan^{- 1} [\frac{2 + 2 cosx}{2 sinx}] \\ = \tan^{- 1} [\frac{2 \cos^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}] \\ = \tan^{- 1} [\cot \frac{x}{2}] \\ = \tan^{- 1} [\tan (\frac{π}{2} - \frac{x}{2})] \\ = \frac{π}{2} - \frac{x}{2} \\ \frac{dy}{dx} = - \frac{1}{2} . \dots.. (1) \\ Now, u = \tan^{- 1} x \\ \Rightarrow \frac{du}{dx} = \frac{1}{1 + x^{2}} . \dots.. (2) \\ By (1) and (2) \\ \frac{dy}{du} = \frac{\frac{dy}{dx}}{\frac{du}{dx}} = \frac{- \frac{1}{2}}{\frac{1}{1 + x^{2}}} = - \frac{1}{2} (1 + x^{2}) \end{array}$

Q.4

${If y = e}^{{acos}^{- 1} x}, - 1 \leq x \leq 1, show that (1 - x^{2}) \frac{d^{2 y}}{{dx}^{2}} - x \frac{dy}{dx} - a^{2} y = 0$

Ans

$\begin{array}{l} Here, y = e^{a \cos^{- 1}} \\ differentiating y w . r . t x \\ \frac{dy}{dx} = \frac{d}{dx} (e^{{acos}^{- 1} x}) \\ = {e^{{acos}^{- 1} x}}^{} \frac{d}{dx} ({acos}^{- 1} x) \\ = \frac{- ay}{\sqrt{1 - x^{2}}} \\ \sqrt{1 - x^{2}} \frac{dy}{dx} = - ay \\ again differentiating w . r . tx \\ \Rightarrow \sqrt{1 - x^{2}} \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} \frac{d}{dx} (\sqrt{1 - x^{2}}) = - a \frac{dy}{dx} \\ \Rightarrow \sqrt{1 - x^{2}} \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} . [\frac{1}{2 \sqrt{1 - x^{2}}} . (- 2 x)] = - a \frac{dy}{dx} \\ \Rightarrow (\sqrt{1 - x^{2}}) \frac{d^{2} y}{{dx}^{2}} - . (\frac{x}{\sqrt{1 - x^{2}}}) \frac{dy}{dx} = - a (\frac{- ay}{\sqrt{1 - x^{2}}}) \\ \Rightarrow (\sqrt{1 - x^{2}}) \frac{d^{2} y}{{dx}^{2}} - x \frac{dy}{dx} - a^{2} y = 0 \end{array}$

Q.5

$If \sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y) then prove that \frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}} .}$

Ans

$\begin{array}{l} Here, \sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y) \\ Let x = sinθ \Rightarrow θ = \sin^{- 1} x \\ y = sinÏ• \Rightarrow Ï• = \sin^{- 1} y \\ \Rightarrow \sqrt{1 - \sin^{2} θ} + \sqrt{1 - \sin^{2} Ï•} = a (sinθ - sinÏ•) \\ \Rightarrow cosθ + cosÏ• = a (sinθ - sinÏ•) \\ \Rightarrow \frac{cosθ + cosÏ•}{sinθ - sinÏ•} = a \\ \Rightarrow \frac{2 \cos (\frac{θ + Ï•}{2}) \cos (\frac{θ - Ï•}{2})}{2 \cos (\frac{θ + Ï•}{2}) \sin (\frac{θ - Ï•}{2})} = a \\ \Rightarrow \cot (\frac{θ - Ï•}{2}) = a \\ \Rightarrow θ - Ï• = 2 \cot^{- 1} a \\ \Rightarrow \sin^{- 1} x - \sin^{- 1} y = 2 \cot^{- 1} a \\ differentiating w . r . t . x \\ \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - y^{2}}} \frac{dy}{dx} = 0 \\ \Rightarrow \frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} \end{array}$

Q.6

$\begin{array}{l} Using mathematical induction prove that \\ \frac{d}{dx} (x^{n}) = {nx}^{n - 1}, \forall n \in I^{+} \end{array}$

Ans

$\begin{array}{l} Let p (n) : \frac{d}{dx} (x^{n}) = {nx}^{n - 1} \\ At n = 1 \\ P (1) : \frac{d}{dx} (x^{1}) = 1 = 1 x^{1 - 1} = x^{0} \\ ∴ P (1) is true \\ Suppose p (k) is true \\ P (k) : \frac{d}{dx} (x^{k}) = {kx}^{k - 1} \\ We have to prove P (k + 1) is true \\ i . e . P (k) : \frac{d}{dx} (x^{k + 1}) = (k + 1) x^{k} \\ P (k + 1) : \frac{d}{dx} (x^{k + 1}) = \frac{d}{dx} (x^{k} . x) \\ = x^{k} \frac{d}{dx} (x) + x \frac{d}{dx} (x^{k}) \\ = x^{k} . 1 + {kxx}^{k - 1} \\ = x^{k} + {kxx}^{k - 1} \\ = x^{k} + {kx}^{k} \\ = (k + 1) x^{k} \\ ∴ P (k + 1) is true \\ Hence by mathematical induction \\ P (n) : \frac{d}{dx} (x^{n}) = {nx}^{n - 1} is true \end{array}$

Q.7

$\begin{array}{l} If {(x - a)}^{2} + {(y - b)}^{2} = c^{2} for some c > 0, \\ Prove that {\frac{\{1 + {(\frac{dy}{dx})}^{2}\}}{\frac{d^{2} y}{{dx}^{2}}}}^{\frac{3}{2}} is a constant, independent of a and b . \end{array}$

Ans

$\begin{array}{l} Here, {(x - a)}^{2} + {(y - b)}^{2} = c^{2} \\ differentiating w . r . t . x \\ \frac{d}{dx} [{(x - a)}^{2} + {(y - b)}^{2}] = 0 \\ \Rightarrow 2 (x - a) + 2 (y - b) \frac{dy}{dx} = 0 \\ \Rightarrow 2 (y - b) \frac{dy}{dx} = - 2 (x - a) \\ \Rightarrow \frac{dy}{dx} = - \frac{(x - a)}{(y - b)} . \dots\dots. (1) \\ again differentiating w . r . t . x \\ \frac{d^{2} y}{{dx}^{2}} = - \frac{(y - b) . 1 - (x - a) . \frac{dy}{dx}}{{(y - b)}^{2}} \\ = - \frac{(y - b) + (x - a) . \frac{(x - a)}{(y - b)}}{{(y - b)}^{2}} \\ = - \frac{{(y - b)}^{2} + {(x - a)}^{2}}{{(y - b)}^{3}} \\ = - \frac{c^{2}}{{(y - b)}^{3}} . \dots.. (2) \\ Now, \frac{{\{1 + {(\frac{dy}{dx})}^{2}\}}^{\frac{3}{2}}}{\frac{d^{2} y}{{dx}^{2}}} = \frac{{\{1 + {(\frac{x - a}{y - b})}^{2}\}}^{\frac{3}{2}}}{\frac{c^{2}}{{(y - b)}^{3}}} \\ \frac{{\{1 + {(\frac{x - a}{y - b})}^{2}\}}^{\frac{3}{2}}}{\frac{c^{2}}{{(y - b)}^{3}}} = \frac{{\{\frac{{(y - b)}^{2} + {(x - a)}^{2}}{{(y - b)}^{2}}\}}^{\frac{3}{2}}}{\frac{c^{2}}{{(y - b)}^{3}}} \\ = \frac{{\{\frac{c^{2}}{{(y - b)}^{2}}\}}^{\frac{3}{2}}}{\frac{c^{2}}{{(y - b)}^{3}}} = \frac{\frac{c^{3}}{{(y - b)}^{3}}}{\frac{c^{2}}{{(y - b)}^{3}}} = - \frac{c^{3}}{c^{2}} = - c \end{array}$

Q.8

$If x \sqrt{1 + y} + y \sqrt{1 + x} = 0 for - 1 < x < 1, prove that \frac{dy}{dx} = - \frac{1}{{(1 + x)}^{2}}$

Ans

$\begin{array}{l} Here x \sqrt{1 + y} + y \sqrt{1 + x} = 0 \\ \Rightarrow x \sqrt{1 + y} = - y \sqrt{1 + x} \\ squaring both sides we get \\ x^{2} (1 + y) = y^{2} (1 + x) \\ \Rightarrow x^{2} + x^{2} y = y^{2} + y^{2} x \\ \Rightarrow x^{2} - y^{2} = y^{2} x - x^{2} y \\ \Rightarrow (x - y) (x + y) = - xy (x - y) \\ \Rightarrow x + y = - xy \\ \Rightarrow x =- (xy + y) \\ \Rightarrow \frac{x}{x + 1} = - y \\ differentiating w . r . t x \\ - \frac{dy}{dx} = \frac{(x + 1) . 1 - x . 1}{{(x + 1)}^{2}} = \frac{1}{{(x + 1)}^{2}} \\ - \frac{dy}{dx} = = \frac{1}{{(x + 1)}^{2}} \end{array}$

Q.9 Verify the Mean value theorem for the function: f(x) = log_ex on [1, 2].

Ans

(a) f(x) is continuous on [1, 2].

(b) f‘(x) = 1/x therefore function is differentiable on [1, 2].

Both the two conditions of Mean Value Theorem are true

∴∃ at least one point c ∈ (1, 2)

$\begin{array}{l} s . t . f ‘ (c) = \frac{f (2) - f (1)}{2 - 1} \\ \Rightarrow \frac{1}{c} = \frac{\log_{e} 2 - \log_{e} 1}{2 - 1} = \log_{e} \frac{2}{1} \\ \Rightarrow c = \frac{1}{\log_{e} 2} = \log_{2} e . \end{array}$

Q.10 Verify Rolle’s Theorem for the following functions: f(x) = (x – 1)(x – 2)² on [1, 2].

Ans

(a) f(x) is a polynomial function, therefore the function is continuous on [1, 2].

(b) f‘(x) = 1.(x – 2)² + (x – 1).2(x – 2)
= (x – 2)[x – 2 + 2x – 2]
= (x – 2)(3x – 4)

Thus, function is differentiable on (1, 2)

∴ ∃ at least one point c (1, 2) s.t. f‘(∈ c) = 0

⇒ (c – 2)(3c – 4) = 0

⇒ c = 4/3 or c = 2
but 2 ∉(1, 2), hence this choice is rejected and the value of c is 4/3.

Q.11

$Differentiate \sqrt{\frac{(x - 3 (x^{2} + 4))}{3 x^{2} + 4 x + 5}} w . r . t . x .$

Ans

$\begin{array}{l} Let y = \sqrt{\frac{(x - 3 (x^{2} + 4))}{3 x^{2} + 4 x + 5}} \\ Taking \log both sides we have \\ \log y = \log \sqrt{\frac{(x - 3 (x^{2} + 4))}{3 x^{2} + 4 x + 5}} \\ = \log {(\frac{(x - 3) (x^{2} + 4)}{3 x^{2} + 4 x + 5})}^{\frac{1}{2}} \\ = \frac{1}{2} [\log (x - 3) + \log (x^{2} + 4) - \log (3 x^{2} + 4 x + 5)] \\ differentiating w . r . t . x \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} [\frac{1}{x - 3} + \frac{1}{x^{2} + 4} \frac{d}{dx} (x^{2 + 4}) - \frac{1}{3 x^{2} + 4 x + 5} \frac{d}{dx} (3 x^{2} + 4 x + 5)] \\ \frac{dy}{dx} = \frac{y}{2} [\frac{1}{x - 3} + \frac{2 x}{x^{2} + 4} - \frac{6 x + 4}{3 x^{2} + 4 x + 5}] \\ \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x - 3) (x^{2 + 4})}{3 x^{2} + 4 x + 5} -} [\frac{1}{x - 3} + \frac{2 x}{x^{2} + 4} - \frac{6 x + 4}{3 x^{2} + 4 x + 5}] \end{array}$

Q.12

$If y = {(\tan^{- 1} x)}^{2}, show that {(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2$

Ans

$\begin{array}{l} Here, y = {(\tan^{- 1} x)}^{2} \\ differentiating y w . r . t . x \\ y_{1} = \frac{dy}{dx} = \frac{d}{dx} {(\tan^{- 1} x)}^{2} \\ = 2 (\tan^{- 1} x) \frac{d}{dx} (\tan^{- 1} x) \\ = \frac{2 \tan^{- 1} x}{x^{2} + 1} \\ (x^{2} + 1) y_{1} = 2 \tan^{- 1} x \\ again differentiating w . r . t . x \\ \frac{d}{dx} [(x^{2} + 1) y_{1}] = \frac{d}{dx} (2 \tan^{- 1} x) \\ \Rightarrow (x^{2} + 1) \frac{d}{dx} y_{1} + y_{1} \frac{d}{dx} (x^{2} + 1) = \frac{d}{dx} (2 \tan^{- 1} x) \\ \Rightarrow (x^{2} + 1) y_{2} + y_{1} (2 x) = \frac{2}{x^{2} + 1} [as y_{2} = \frac{d^{2 y}}{{dx}^{2}}] \\ \Rightarrow {(x^{2} + 1)}^{2} y_{2} + y_{1} (x^{2} + 1) (2 x) = 2 \\ \Rightarrow {(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2 . \end{array}$

Q.13

$If x = \sqrt{a^{\sin^{- 1}},} y = \sqrt{a^{\cos^{- 1}}} show that \frac{dy}{dx} = - \frac{y}{x} .$

Ans

$\begin{array}{l} Here, x = \sqrt{a^{\sin^{- 1}},} \Rightarrow x^{2} = a^{\sin - 1} t \\ d i f f e r e n t i a t i n g x w.r.t.t \\ 2x \frac{d x}{d t} = a^{\sin^{- 1}} \log a . \frac{d}{d t} [\sin^{- 1} t] \\ 2x \frac{d x}{d t} = x^{2} . \log a . \frac{1}{\sqrt{1 - t^{2}}} \\ \frac{d x}{d t} = \frac{1}{2} x . \log a . \frac{1}{\sqrt{1 - t^{2}}} ….. (1) \\ Now y = \sqrt{a^{\cos^{- 1} t}} \Rightarrow y^{2} = a^{\cos^{- 1} t} \\ d i f f e r e n t i a t i n g y w . r . t . t \\ 2 y \frac{d y}{d t} = a^{\cos^{- 1} t} \log a \frac{a}{d t} [\cos^{- 1} t] \\ 2 y \frac{d y}{d t} = y^{2} \log a . \frac{1}{\sqrt{1 - t^{2}}} \\ \frac{d y}{d t} = – \frac{1}{2} y \log a . \frac{1}{\sqrt{1 - t^{2}}} ……. (2) \\ As \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- \frac{1}{2} y \log a . \frac{1}{\sqrt{1 - t^{2}}}}{\frac{1}{2} x . \log a . \frac{1}{\sqrt{1 - t^{2}}}} b y (1) a n d (2) \\ = - \frac{y}{x} \end{array}$

Q.14

$Find \frac{dy}{dx} if x = a [cost + logtan \frac{t}{2}, and y = a sint]$

Ans

$\begin{array}{l} Here x = a [cost + logtan \frac{t}{2}] \\ differentiating x w . r . t . t \\ \frac{dx}{dt} = a \frac{d}{dt} [cost + logtan \frac{t}{2}] \\ = a [- sint + \frac{1}{\tan \frac{t}{2}} . \frac{d}{dt} (tam \frac{t}{2})] \\ = a [- sint + \frac{1}{\tan \frac{t}{2}} . \frac{1}{2} \sec^{2} \frac{t}{2}] \\ = a [- sint + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}] \\ = a [sint + \frac{1}{sint}] [\sin 2 x = 2 sinxcosx] \\ = a [\frac{- \sin^{2} t + 1}{sint}] = [\frac{\cos^{2} t}{sint}] . \dots. (1) \\ Now y = a sint \\ differentiating y w . r . t . t \\ \frac{dy}{dx} = a cost . \dots (2) \\ As \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{acost}{\cos^{2} t} (by (1) and (2)) \\ = tant \end{array}$

Q.15

$Find \frac{dy}{dx}, if x^{y} + y^{x} =1.$

Ans

$\begin{array}{l} Here x^{y} + y^{x} =1 \\ \Rightarrow e^{{logx}^{y}} + e^{{logy}^{x}} = 1 \\ diff w . r . t x \\ \Rightarrow \frac{d}{dx} (e^{{logx}^{y}} + e^{{logy}^{x}}) = 0 \\ \Rightarrow e^{{logx}^{y}} \frac{d}{dx} ({logx}^{y}) + e^{{logy}^{x}} \frac{d}{dx} ({logy}^{x}) = 0 \\ \Rightarrow e^{{logx}^{y}} \frac{d}{dx} (y . logx) + e^{{logy}^{x}} \frac{d}{dx} (x . logy) = 0 \\ \Rightarrow x^{y} [y \frac{d}{dx} (logx) + logx \frac{dy}{dx}] + x^{y} [x \frac{d}{dx} (logy) + logy \frac{dy}{dx} (x)] = 0 \\ \Rightarrow x^{y} [\frac{y}{x} + logx \frac{dy}{dx}] + y^{x} [\frac{x}{y} . \frac{dy}{dx} + (logy)] = 0 \\ \Rightarrow x^{y - 1} y + x^{y} logx \frac{dy}{dx} + {xy}^{x - 1}, \frac{dy}{dx} + y^{x} logy = 0 \\ \Rightarrow [x^{y} logx + {xy}^{x - 1}] \frac{dy}{dx} = - [y^{x} logy + x^{y - 1} y] \\ \Rightarrow \frac{dy}{dx} = - \frac{y^{x} logy + x^{y - 1} y}{x^{y} logx + {xy}^{x - 1}} \end{array}$

Q.16

${Differentiate y=x}^{sinx} + {(sinx)}^{cosx} w . r . t . x$

Ans

$\begin{array}{l} y = e^{{logx}^{sinx}} + e^{\log {(sinx)}^{cosx}} [e^{loga} = a] \\ diff y w . r . t . x \\ \frac{dy}{dx} = \frac{d}{dx} [e^{{logx}^{sinx}} + e^{\log {(sinx)}^{cosx}}] \\ = e^{{logx}^{sinx}} \frac{d}{dx} ({logx}^{sinx}) + e^{\log {(sinx)}^{cosx}} \frac{d}{dx} (\log {(sinx)}^{cosx}) \\ = x^{sinx} \frac{d}{dx} (sinx . logx) + {(sinx)}^{cosx} \frac{d}{dx} (cosx . logsinx) \\ = x^{sinx} [sinx \frac{d}{dx} (logx) + logx \frac{d}{dx} (sinx)] + {(sinx)}^{cosx} [cosx \frac{d}{dx} (logsinx) + logsinx \frac{d}{dx} (cosx)] \\ = x^{sinx} [\frac{sinx}{x} cosx . logx] + {(sinx)}^{cosx} [cosx . cotx - sinx . logsinx] \end{array}$

Q.17

$Determine, if f defined by f (x) = \{\begin{cases} x^{2} \sin \frac{1}{x} if x \neq 0 \\ 0 if x = 0 \end{cases} is a continuous function ?$

Ans

$\begin{array}{l} Here f (0) = 0, \\ L . H . L . = \lim_{x \to 0^{-}} (x^{2} \sin \frac{1}{x}) . Putting x = 0 - h \\ = \lim_{h \to 0^{-}} {(0 - h)}^{2} \sin \frac{1}{0 - h} = - \lim_{h \to 0^{-}} h^{2} \sin \frac{1}{h} \\ \sin \frac{1}{h} lies between -1 and 1 \\ Thus L . H . L .= - \lim_{h \to 0^{-}} h^{2} \sin \frac{1}{h} = 0 \\ R . H . L . = \lim_{h \to 0^{+}} (x^{2} \sin \frac{1}{x}), Putting x = 0 + h \\ = \lim_{h \to 0^{}} (0 + h^{2}) \sin \frac{1}{0 + h} = \lim_{h \to 0^{}} h^{2} \sin \frac{1}{h} \\ ∴ R . H . L . = \lim_{h \to 0^{+}} h^{2} \sin \frac{1}{h} = 0 \\ R . H . L . = L . H . L = f (o) \\ Hence, f is continuous for all x \in R \end{array}$

Q.18

$\begin{array}{l} Find the values ofa and b so that the function defined by \\ f (x) = \{\begin{cases} 5 if x \leq 2 \\ ax + b if 2< x <10 \\ 21 if x \geq 10 \end{cases} is continous function . \end{array}$

Ans

$\begin{array}{l} Since f is continuous at x = 10, we obtain \\ \lim_{x \to 10^{-}} f (x) = \lim_{x \to 10^{+}} f (x) = f (10) \\ \Rightarrow \lim_{x \to 10^{-}} (ax + b) = \lim_{x \to 10^{+}} f (21) = (21) \\ \Rightarrow 10 a + b = 21 = 21 \\ \Rightarrow 10 a + b = 21 . \dots. (2) \end{array}$

Q.19

$\begin{array}{l} Find the relationship betweena andb so that the function defined by \\ f (x) = \{\begin{cases} ax +1 if x \leq 3 \\ bx +3 if x > 3 \end{cases} is continuous at x = 3. \end{array}$

Ans

$\begin{array}{l} Here f (3) = a (3) + 1 = 3 a + 1 \dots\dots (1) \\ R . H . L . = \lim_{x \to 3^{+}} (bx + 3) . \\ Putting x = 3 + h \\ R . H . L . = \lim_{h \to 0} b (3 + h) + 3 = 3 b + 3 . \dots (2) \\ As f is continuous at x = 3 \\ By (1) and (2) \\ R . H . L = f (3) \\ 3 b + 3 = 3 a + 1 \\ a - b = \frac{2}{3} or a = b + \frac{2}{3} \end{array}$

Q.20

$If y = \cos^{- 1} x, find \frac{d^{2} y}{{dx}^{2}} in terms of y alone .$

Ans

$\begin{array}{l} Here y = \cos^{- 1} x \Rightarrow x = \cos y \\ differentiating w . r . t x \\ \frac{d}{dx} (x) = \frac{d}{dx} (\cos y) \\ 1 = - siny . \frac{dy}{dx} \\ ∴ \frac{dy}{dx} = - cosec y . \dots.. (1) \\ differentiating \frac{dy}{dx} w . r . t . x \\ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (- cosec y) \\ = cosecy . coty . \frac{dy}{dx} \\ = - {cosec}^{2} y . coty . .. by (1) \end{array}$

Q.21

$Find \frac{d^{2} y}{{dx}^{2}} if y = \sin^{- 1} x .$

Ans

$\begin{array}{l} Here y = \sin^{- 1} x \\ differentiating y w . r . t . x \\ \frac{dy}{dx} = \frac{d}{dx} (\sin^{- 1} x) \\ = \frac{1}{\sqrt{1 - x^{2}}} \\ diff \frac{dy}{dx} w . r . t . x \\ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (\frac{1}{\sqrt{1 - x^{2}}}) = \frac{d}{dx} {(1 - x^{2})}^{- \frac{1}{2}} \\ = - \frac{1}{2} {(1 - x^{2})}^{- \frac{1}{2} - 1} \frac{d}{dx} (1 - x^{2}) \\ = - \frac{1}{2} {(1 - x^{2})}^{- \frac{1}{2} - 1} . (- 2 x) \\ = x {(1 - x^{2})}^{- \frac{3}{2}} = \frac{x}{{(1 - x^{2})}^{\frac{3}{2}}} \end{array}$

Q.22

$Find \frac{dy}{dx} if y = \sin (\tan^{- 1} e^{- x}) .$

Ans

$\begin{array}{l} Here y = \sin (\tan^{- 1} e^{- 1}) \\ differentiating y w . r . t x \\ \frac{dy}{dx} = \frac{d}{dx} [\sin (\tan^{- 1} e^{- 1})] \\ = \cos (\tan^{- 1} e^{- 1}) \frac{d}{dx} (\tan^{- 1} e^{- x}) \\ = \cos (\tan^{- 1} e^{- 1}) . \frac{1}{1 + {(e^{- x})}^{2}} \frac{d}{dx} (e^{- x}) \\ = \cos (\tan^{- 1} e^{- x}) . \frac{1}{1 + {(e^{- x})}^{2}} . e^{- x} \frac{d}{dx} (- x) \\ = \cos (\tan^{- 1} e^{- x}) . \frac{1}{1 + {(e^{- x})}^{2}} . e^{- x} . (- 1) \\ = - \frac{e^{- x} \cos (\tan^{- 1} e^{- x})}{1 + e^{- 2 x}} \end{array}$

Q.23

$Find \frac{dy}{dx}, if y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) .$

Ans

$\begin{array}{l} Let x = tanθ \Rightarrow θ = \tan^{- 1} x \\ y = \tan^{- 1} (\frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) \\ = \tan^{- 1} (\tan 3 θ) \\ = 3 θ = 3 \tan^{- 1} x \\ Now differentiating y w . r . t . x \\ \frac{dy}{dx} = \frac{3}{1 + x^{2}} . \end{array}$

Q.24

$\begin{array}{l} If f (x) = \{\begin{cases} \frac{kcosx}{π - 2 x} if x \neq \frac{π}{2}, find the value of k, \\ 3 if x = \frac{π}{2}, \end{cases} \\ if f is continuous at x = \frac{π}{2} . \end{array}$

Ans

$\begin{array}{l} Here, f \frac{π}{2} = 3 \\ Left hand limit of f at x = \frac{π}{2} is \\ \lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{kcosx}{π - 2 x} \\ Putting x - \frac{π}{2} - h \\ \lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{kcosx}{π - 2 x} = \lim_{x \to 0} \frac{kcos (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} = \lim_{x \to 0} \frac{ksin h}{2 h} = \frac{k}{2} . \\ As f is continuous \\ ∴ \frac{k}{2} = 3 or k = 6 . \end{array}$

Q.25

$Examine the continuity of f (x) = x -5$

Ans

$\begin{array}{l} f (x) = |x - 5| or f (x) = \{\begin{cases} (x - 5) if x \geq 5 \\ (x - 5) if x < 5 \end{cases} \\ This function f is defined at all points of the real line . \\ Let c be a point on a real line . \\ Then, c < 5 or c =0 or c > 5 \\ Case I : c < 5 \\ Then, f (c) = 5- c \\ \lim_{x \to c} f (x) = \lim_{x \to c} f (5 - x) = 5- c = f (c) \\ Therefore, f is continuous at all real numbers less than 5. \\ Case II : c = 5 \\ Then, f (c) = f (5) = (5 - 5) = 0 \\ \lim_{x \to 5} f (x) =0 = \lim_{x \to 5} f (5 - x) = (5 - 5) = 0 \\ \lim_{x \to 5} f (x) =0 = \lim_{x \to 5} (x - 5) = (5 - 5) = 0 \\ ∴ f is continuous at x = 5 \\ Case III : c > 5 \\ Then, f (c) = c - 5 \\ \lim_{x \to c} f (x) = \lim_{x \to 5} (x - 5) = c - 5 = f (c) \\ Therefore, f is continuous at all real numbers greater than 5. \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \end{array}$

Q.26

$Differentiate : y = \sqrt{logx + \sqrt{logx + \sqrt{logx + . \dots}}}$

Ans

$\begin{array}{l} y = \sqrt{logx + \sqrt{logx + \sqrt{logx + . \dots}}} \\ y = \sqrt{logx + y} \\ Squaring both sides \\ y^{2} = logx + y \\ Differentiating w . r . t . x \\ \frac{d}{dx} (y^{2}) = \frac{d}{dx} (logx + y) \\ \Rightarrow 2 y \frac{dy}{dx} = \frac{1}{x} + \frac{dy}{dx} \\ \Rightarrow 2 y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{x} \\ \Rightarrow (2 y - 1) \frac{dy}{dx} = \frac{1}{x} \\ \Rightarrow \frac{dy}{dx} = \frac{1}{x (2 y - 1)} \end{array}$

Q.27

$Differentiate y = \frac{x}{x +1} with respect to x .$

Ans

$\begin{array}{l} y = \frac{x}{x +1} \\ \frac{dy}{dx} = \frac{d}{dx} (\frac{x}{x + 1}) \\ \frac{dy}{dx} = \frac{(x + 1) \frac{d}{dx} (x) - x \frac{d}{dx} (x + 1)}{{(x + 1)}^{2}} \\ = \frac{x + 1 - x}{{(x + 1)}^{2}} = \frac{\frac{1}{{(x + 1)}^{2}}}{} \end{array}$

Q.28

$Find the points where the constant function f (x) = k is continuous .$

Ans

$\begin{array}{l} The function is defined at all real numbers and by definition, its \\ value at any real number equalsk . Letcbe any real number . Then \\ \lim_{x \to c} f (x) = \lim_{x \to c} k = k \\ Since, f (c) = k = \lim_{x \to c} f (x) for any real number c, \\ the function f is continuous at every real number . \end{array}$

Q.29

$\begin{array}{l} Examine the continuity of functionfgiven by \\ f (x) = 2 x + 5 at x = 1. \end{array}$

Ans

$\begin{array}{l} Function is defined on x = 1 at its value is 7 \\ \lim_{x \to 1} f (x) = \lim_{x \to 1} (2 x + 5) = 2 (1) + 5 = 7 \\ \lim_{x \to 1} f (x) = f (1) = 7 \\ Hence, f is continuous at x = 1 \end{array}$

Q.30

$\begin{array}{l} Prove that the identity function on real numbers \\ given byf (x) = x is continuous at every real number . \end{array}$

Ans

$\begin{array}{l} Here, f (c) = cfor every real numberc . Also, \\ \lim_{x \to c} f (x) = c = f (c) and hence the function is \\ continuous at every real number . \end{array}$

Q.31 If f and g be two real functions continuous at a real number c, then f + g is continuous at x = c.

Ans

$\begin{array}{l} \lim_{x \to c} (f + g) (x) = \lim_{x \to c} [f (x) + g (x)] \\ = \lim_{x \to c} f (x) + \lim_{x \to c} g (x) \\ = f (c) + g (c) \\ = (f + g) (c) \end{array}$

Q.32

$\begin{array}{l} Find the point of discontinuity of f where f is defined by \\ f (x) = \{\begin{cases} 2 x + 3, if x \leq 2 \\ 2 x - 3 if x > 2 \end{cases} \end{array}$

Ans

$\begin{array}{l} f (2) = 2 \times 2 +3 = 7 \\ R . H . L = f (2 + h) = \lim_{h \to 0} (2 + h) = (2 \times 2 - 3) = 1 \\ f (2) is not equal to its R . H . L \\ hence, function is discontinus at x = 2 \end{array}$

Q.33 Prove that every rational function is continuous.

Ans

Every rational function is defined by

f(x) = p(x)/q(x), q(x) ≠ 0

Where p and q are the polynomial functions. The domain of f is all real numbers except points at which q is zero. Since polynomial functions are continuous, f is continuous.

Q.34

$Differentiate \cos (sinx) with respect to x .$

Ans

$\begin{array}{l} Let y = \cos (sinx) \\ \Rightarrow \frac{dy}{dx} = \frac{d}{dx} \cos (sinx) \\ = - \sin (sinx) \frac{d}{dx} (sinx) \\ = - \sin (sinx) cosx \end{array}$

Q.35

$Differentiate y = e^{- 2 x} with respect to x .$

Ans

$\begin{array}{l} y = e^{- 2 x} \\ \frac{dy}{dx} = \frac{d}{dx} (e^{- 2 x}) \\ \frac{dy}{dx} = e^{- 2 x} \frac{d}{dx} = (- 2 x) \\ = e^{- 2 x} \times - 2 \\ = - 2 e^{- 2 x} \end{array}$

Q.36 Differentiate y = sin (xy) with respect to x.

Ans

$\begin{array}{l} y = \sin (xy) \\ \frac{dy}{dx} = \frac{d}{dx} [\sin (xy)] \\ \frac{dy}{dx} = \cos (xy) \frac{d}{dx} (xy) \\ = \cos (xy) (x \frac{dy}{dx} + y) \\ = \cos (xy) \frac{dy}{dx} + ycos (xy) \\ \frac{dy}{dx} = \frac{y \cos (xy)}{1 - xcos (xy)} \end{array}$

Q.37

$If y = 5cos x-3sinx, prove that \frac{d^{2} y}{{dx}^{2}} + y = 0 .$

Ans

$\begin{array}{l} y = 5 \cos x - 3 sinx \\ \Rightarrow \frac{dy}{dx} = 5 \frac{d}{dx} cosx - 3 \frac{d}{dx} sinx \\ = - 5 sinx - 3 \cos x \\ \Rightarrow \frac{d^{2} y}{d^{2} x} = - 5 \frac{d}{dx} sinx - 3 \frac{d}{dx} cosx \\ = - 5 cosx + 3 sinx = - (5 cosx - 3 sinx) = - y \\ \Rightarrow \frac{d^{2} y}{d^{2} x} + y = 0 \end{array}$

Q.38

$\begin{array}{l} Discuss the continuity of the function f defined \\ by f (x) = \frac{1}{x}, x \neq 0 . \end{array}$

Ans

$\begin{array}{l} Let c be any non - zero real number . \\ We have, \\ \lim_{x \to c} f (x) = \lim_{x \to c} (\frac{1}{x}) = (\frac{1}{c}) \\ Also, for c \neq 0, f (c) = \frac{1}{c} \\ \lim_{x \to c} f (x) = f (c) \\ \Rightarrow f is continuous . \end{array}$

Q.39

$\begin{array}{l} Discuss the continuity of the function f defined \\ by f (x) = \frac{1}{x}, at x = 0. \end{array}$

Ans

$\begin{array}{l} We have, \\ LHL = \lim_{x \to \bar{0}} f (x) = \lim_{x \to \bar{0}} (\frac{1}{x}) = (\frac{1}{- 0}) = - \infty \\ RHL = \lim_{x \to \bar{\overset{+}{0}}} f (x) = \lim_{x \to \bar{\bar{\overset{+}{0}}}} (\frac{1}{x}) = (\frac{1}{+ 0}) = + \infty \\ LHL \neq RHL = f (0) \\ \Rightarrow f is discontinuous at x = 0. \end{array}$

Q.40

$\begin{array}{l} Discuss the continuity of the function f defined \\ by f (x) = |x|, at x = 0. \end{array}$

Ans

$\begin{array}{l} We have, \\ LHL = \lim_{x \to \bar{0}} f (x) = \lim_{x \to \bar{0}} |x| = |- 0| = 0 \\ RHL = \lim_{x \to \bar{\overset{+}{0}}} f (x) = \lim_{x \to \overset{+}{0}} |x| = |+ 0| = 0 \\ LHL = RHL = f (0) \\ \Rightarrow f is continuous at x = 0. \end{array}$

Q.41

$\begin{array}{l} Discuss the differentiability of the function f defined \\ by f (x) = |x| at x =0. \end{array}$

Ans

$\begin{array}{l} LHD = \lim_{h \to 0} \frac{f (a - h) - f (a)}{- h} = \lim_{h \to \bar{0}} \frac{f (0 - h) - f (0)}{- h} \\ = \lim_{h \to \bar{0}} \frac{|- h| - |0|}{- h} = \lim_{h \to \bar{0}} \frac{|h|}{- h} = - 1 \\ RHD = \lim_{h \to 0} \frac{f (a + h) - f (a)}{- h} = \lim_{h \to \bar{0}} \frac{f (0 + h) - f (0)}{- h} \\ = \lim_{h \to \bar{0}} \frac{|h| - |0|}{- h} = \lim_{h \to \bar{0}} \frac{|h|}{- h} = 1 \\ LHD \neq RHD \\ \Rightarrow Function f (x) = |x| is not differentiable at x = 0. \end{array}$

Q.42

$\begin{array}{l} Discuss the continuity of the function f defined \\ by f (x) = 5 x +3 at x =1. \end{array}$

Ans

$\begin{array}{l} We have, \\ LHL = \lim_{x \to \bar{1}} f (x) \\ = \lim_{x \to \bar{1}} f (5 x + 3) \\ = (5 \times 1 + 3) = 8 \\ RHL = \lim_{x \overset{Ë™}{\to} \overset{+}{1}} f (x) \\ = \lim_{x \overset{Ë™}{\to} \overset{+}{1}} f (5 x + 3) \\ = (5 \times 1 + 3) = 8 \\ f (1) = (5 \times 1 + 3) = 8 \\ LHL = RHL = f (1) \\ \Rightarrow f is continuous at x = 1 \end{array}$

Q.43

$\begin{array}{l} Find the set of all points where the function \\ f (x) = x |x| is differentiable . \end{array}$

Ans

$\begin{array}{l} Since domain of the function, f (x) = x |x| is the set of all \\ real numbers, therefore, it is differentiable at the set of all \\ real numbers, that is, (- \infty, \infty) . \end{array}$

Q.44

$\begin{array}{l} Let f be a continuous function on satisfying \\ f (x + y) = f (x) + f (y) \forall x, y \in R and f (1) = 5, \\ then evaluate \lim_{x \to 4} f (x) . \end{array}$

Ans

$\begin{array}{l} \lim_{x \to 4} f (x) = f (4) \\ = f (2 + 2) \\ = f (2) + f (2) [f (x + y) = f (x) + f (y)] \\ = f (1 + 1) + f (1 + 1) \\ = f (1) + f (1) + f (1) + f (1) \\ = 5 + 5 + 5 + 5 [f (1) = 5] \\ = 20 \end{array}$

Q.45

$\begin{array}{l} Let f be a continuous function on satisfying \\ f (x) f (y) = f (x) + f (y) + f (xy) - 2 \forall x, y \in R and f (2) = 5, \\ then evaluate \lim_{x \to 4} f (x) . \end{array}$

Ans

$\begin{array}{l} \lim_{x \to 4} f (x) = f (4) \\ = f (2 \times 2) [f (xy) = f (x) f (y) + 2 - f (x) - f (y)] \\ = f (2) f (2) + 2 - f (2) - f (2) \\ = 5 \times 5 + 2 - 5 - 5 \\ = 25 + 2 - 10 [f (2) - 5] \\ = 17 \end{array}$

Q.46

$If f (x) = \log_{7} x, then f ‘ (x) .$

Ans

$\begin{array}{l} We have, \\ f (x) = \log_{7} x \\ = \frac{logx}{\log 7} [by change of base formula] \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} (\frac{logx}{\log 7}) \\ = \frac{1}{\log 7}, \frac{1}{x} \\ = \frac{1}{xlog 7} \end{array}$

Q.47 Prove that sine function is continuous.

Ans

$\begin{array}{l} Since, \lim_{x \to 0} sinx = 0 \\ Let f (x) = \sin x is defined for every real number . \\ Let c be a real number . Put c = c + h . \\ If x \to c, we know that h \to 0 . Therefore, \\ \lim_{x \to c} f (x) = \lim_{x \to c} sinx \\ = \lim_{h \to 0} \sin (c + h) \\ = \lim_{h \to 0} (sinc \cosh - cosc \sinh) \\ = \lim_{h \to 0} (sinc \cosh) + \lim_{h \to 0} (cosc \sinh) \\ = sinc + 0 \\ = sinc = f (c) \\ Therefore, \\ \lim_{x \to c^{-}} f (x) = f (c) \\ and hence f is a continuous function . \end{array}$

Q.48 Prove that cosine function is continuous.

Ans

$\begin{array}{l} Let f (x) = cosx is defined for every real number . \\ Let c be a real number . Put c = c + h . \\ If x \to c, we know that h \to 0 . Therefore, \\ \lim_{x \to c^{+}} f (x) = \lim_{x \to c^{+}} cosx \\ = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosc \cosh - sinc \sinh) \\ = \lim_{h \to 0} (cosc \cosh) + \lim_{h \to 0} (sinc \sinh) \\ = cosc + 0 [\begin{array}{l} \lim_{h \to 0} \cosh = 1 and \\ \underset{h \to 0}{\lim \sinh} = 0 \end{array}] \\ = cosc \\ Similarly, \\ \lim_{x \to c^{-}} f (x) = f (c) \\ ∴ \lim_{x \to c^{+}} f (x) = \lim_{x \to c^{-}} f (x) = f (c) \\ Hence f (x) is a continuous at x = c . \\ Since, c is arbitrary real number, so f (x) is \\ everywhere continuous . \end{array}$

Q.49 Prove that the function is defined by g(x) = x – [x] is discontinuous at all integral points.

Ans

$\begin{array}{l} Let a be an integer, then [a - h] = a - 1, [a + h] = a \\ and [a] = a \\ At x = a, \\ LHL = \lim_{x \to a^{-}} g (x) \\ = \lim_{x \to a^{-}} (x - [x]) \\ Putting x = a - h, as x \to a^{+}, h \to 0 \\ ∴ \lim_{x \to a^{-}} (x - [x]) = \lim_{h \to 0} (a - h - (a - 1)) [[a - h] = (a - 1)] \\ = \lim_{h \to 0} (1 - h) \\ = 1 \\ RHL = \lim_{x \to a} (x - [x]) \\ [Putting x = a + h as x \to a^{+} when h \to 0] \\ = \lim_{h \to 0} (a + h - (a + h)) \\ = \lim_{h \to 0} (a + h - a) [[a + h] = a] \\ = \lim_{h \to 0} (h) \\ = 0 \\ ∴ RHL \neq LHL \\ Thus, g (x) is discontinuous at all integral points . \end{array}$

Q.50

$\begin{array}{l} f (x) = \frac{\sqrt{a^{2} - ax + x^{2}} - \sqrt{a^{2} + ax + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}} becomes \\ continuous for all x . \end{array}$

Ans

$\begin{array}{l} f (x) = \frac{\sqrt{a^{2} - ax + x^{2}} - \sqrt{a^{2} + ax + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}} \\ = \frac{\sqrt{a^{2} - ax + x^{2}} - \sqrt{a^{2} + ax + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}} \times \frac{\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} + ax + x^{2}}}{\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} + ax + x^{2}}} \\ \times \frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}} \\ = \frac{(a^{2} - ax + x^{2}) - (a^{2} + ax + x^{2})}{\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} - ax + x^{2}}} \times \frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} \\ = \frac{- 2 ax (\sqrt{a + x} + \sqrt{a - x})}{2 x (\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} - ax + x^{2}})} \\ \lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{- a (\sqrt{a + x} + \sqrt{a - x})}{(\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} - ax + x^{2}})} \\ = \frac{- a (\sqrt{a + 0} + \sqrt{a - 0})}{(\sqrt{a^{2} - a) (0) + 0^{2}} + \sqrt{a^{2} - a {+ 0}^{2}})} \\ = \frac{- a (2 \sqrt{a})}{(2 a)} = - \sqrt{a} \end{array}$

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FAQs (Frequently Asked Questions)

1. Which chapter notes can I get for Class 12 Mathematics on Extramarks?

Extramarks, an online learning platform, provides detailed and well-structured notes that include all the topics and sub-topics of all the chapters for Mathematics Class 12. Students are advised to study with the help of these academic notes. Chapter-wise notes such as the Class 12 Mathematics chapter 5 notes are really helpful and beneficial for students to get an in-depth understanding of every chapter.

Chapter 1: Relations and Functions

Chapter 2: Inverse Trigonometric Functions

Chapter 3: Matrices

Chapter 4: Determinants

Chapter 5: Continuity and Differentiability

Chapter 6: Applications of Derivatives

Chapter 7: Integrals

Chapter 8: Application of Integrals

Chapter 9: Differential Equations

Chapter 10: Vector Algebra

Chapter 11: Three Dimensional Geometry

Chapter 12: Linear programming

Chapter 13: Probability

2. How can I use the Class 12 Mathematics Chapter 5 notes to help me in my exam preparation?

The notes provided by Extramarks help students in solving and practising problems so that they can tackle questions of any difficulty level easily in no time. The Class 12 Mathematics Chapter 5 notes enable an optimum level of concentration and focus. The notes include information about repeated questions, mostly liked questions and all important details about the chapter. All information in the Class 12 Mathematics Chapter 5 notes follows the latest guidelines issued by the CBSE board.

3. Will the Class 12 Mathematics Chapter 5 notes include important questions?

The CBSE Class 12th board exams include many questions based on differentiation and continuity. The Class 12 Mathematics Chapter 5 notes help students with this. The notes include several questions on continuous and differentiable functions, logarithmic differentiation, and problems based on the Mean value Theorem and Rolle’s Theorem. The Class 12 Mathematics Chapter 5 notes help students to know the most important sections of the chapter.

4. Are there any tips to study and practice Mathematics during the exam?

Some of the best tips students may consider are below.

Be familiar with and understand the syllabus properly: The CBSE syllabus is very vast. It is important for students to know the important topics so that they can spend time and concentrate on those concepts.
Use high-quality study material: Studying with the help of the right study material is crucial to scoring high marks. Students are advised to use study materials such as the Class 12 Mathematics chapter 5 notes to know the type of questions asked in the examination. The NCERT books help students to get an in-depth insight into the chapter.
Create a study plan: Following a timetable will help the students to be focused and enhance the CBSE exam preparation. A well-structured study plan will help them to organise their studies and make time for other activities as well.
Revise regularly: Students are advised to regularly revise particular topics and concepts of the chapter. Revising with the help of class 12 chapter 5 Mathematics notes is very important in preparing for CBSE exams. Students can emphasise weaker areas and revise formulas and derivations.
Practice mock tests: In the CBSE exams, students have to attempt many different types of questions in approx. 120 mins. This makes time management an essential skill to possess. Practising mock tests will help enhance the examination preparation.

CBSE Class 12 Maths Revision Notes Chapter 5

Class 12 Mathematics Chapter 5 Notes

List Of The Topics Covered In Class 12 Mathematics Chapter 5 Notes

Key Topics Covered In Class 12 Mathematics Chapter 5 Notes

Definition of Continuity at a point:

Definition of Continuity in an Interval (INTERMEDIATE VALUE THEOREM):

Algebra of Continuous Functions

Definition and formula of differentiability:

Types of Discontinuity:

REMOVABLE DISCONTINUITY:

NON-REMOVABLE DISCONTINUITY:

Relation between continuity and differentiability:

Rules of Differentiation:

Second-Order Derivative:

Derivative of infinite series:

Rolle’s Theorem:

Lagrange’s Mean Value Theorem:

Useful substitutions for finding derivatives:

Derivatives of Trigonometric functions:

Derivatives of Inverse Trigonometric functions:

Derivatives of standard functions:

List of Continuous Functions:

Class 12 Mathematics chapter 5 notes Exercises & Answer Solutions.

NCERT Exemplar Class 12 Mathematics

Key Features of Class 12 Mathematics chapter 5 notes

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FAQs (Frequently Asked Questions)

1. Which chapter notes can I get for Class 12 Mathematics on Extramarks?

2. How can I use the Class 12 Mathematics Chapter 5 notes to help me in my exam preparation?

3. Will the Class 12 Mathematics Chapter 5 notes include important questions?

4. Are there any tips to study and practice Mathematics during the exam?

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