Class 12 Continuity and Differentiability NCERT Solutions: Download Chapter 5 Exercise and Solutions PDF

• Continuity
• Differentiability 
• Exponential and Logarithmic Functions 
• Logarithmic Differentiation
• Derivatives of Functions in Parametric Forms 
• Second Order Derivative 

Check the NCERT Solutions Maths Class 12 Chapter 5 Continuity and Differentiability below.

Download Here: NCERT Solution for Class XII Maths Continuity and Differentiability PDF

Exercise – 5.1

Q1.Prove that the function f(x) = 5x 3f is continuous at x = 0, at x = – 3 and at x = 5.

A.1.Given, f(x) = 5x 3

At x = 0,  $\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}$ 5x 3 = 5 0 3 = 3.

So f is continuous at x = 1.

At x = 3,  $\pi +h$ 5x 3 = 5 ( 3) 3 = 15 3

= 18.

So f is continuous at x = 3.

At x = 5,  $\forall x\in ℝ$ .5x 3 = 5.5 3 = 25 3 = 22.

So, f is continuous at x = 5.

Q2.Examine the continuity of the function f(x) = 2x2 1  $$

A.2. Given, f(x) = 2x2 1

At x = 3

Lim f(x) =  $\frac{dy}{dx}=\frac{-\left(3x^2+2xy+y^2\right)}{\left(x^2+2xy+3y^2\right).}$ 2(3)2 1 = 18 1 = 17.

So, f is continuous at x = 3.

Q3.Examine the following functions for continuity.

$\begin{array}{l}\left(\right)f\left(x\right)=x-5\left(\right)f\left(x\right)=\frac{1}{x-5},x\ne 5\\ \\ \left(\right)f\left(x\right)=\frac{x^2-25}{x+5},x\ne -5\left(\right)f\left(x\right)=\left|x-5\right|\end{array}$

A.3. (a) Given, f (x) = x 5.

The given f x n is a polynernial f xn and as every pohyouraial f xn is continuous in its domain R we conclude that f (x) is continuous.

(b). Given, f(x) =  $\frac{1}{x-5},x\ne 5$

For any a  $=3{\left(5x\right)}^{3cos2x}\left[\frac{cos2x}{x}-2sin2xlog5x\right]$ {5},

$=\frac{x^2\times 1\times }{x-3}\frac{d}{dx}(x-3)+log(x-3)\cdot 2x$  $\frac{1}{(x-5)}=\frac{1}{a-5}.$

and f(a) =  $\frac{1}{a-5}$

i e,  $f(x)=-(x-1)+[-(x-2)]=-x+1-x+2=3-2x.$ f(x) = f(a).

Hence f is continuous in its domain.

(c) Given, f(x) =  $\frac{x^2-25}{x+5},x\ne -5$

For any a  $ℝ$ { 5}

$\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}$  $\frac{x^2-25}{x+5}=\frac{a^2-25}{a+5}=\frac{(a-5)(a+5)}{a+5}$ = a 5

And f(a) =  $\frac{a^2-25}{a+5}=\frac{(a-5)(a+5)}{a+5}.$

= a 5

$\therefore \underset{x\to a}{lim}$ f(x) = f(a).

So, f is continuous in its domain.

(d) Given f (a) =  $|x-5|=\{\begin{array}{cc}\hfill x-5\hfill & \hfill , if x-5>0\Rightarrow x\ge 5\hfill \\ \hfill -(x-5)\hfill & \hfill if x-5<0\Rightarrow x<5.\hfill \end{array}$

For x = c < 5.

f (c) = (c 5) = 5 c.

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ (x 5) = (c 5) = 5 c.

∴ f(c) =  $\underset{x\to c}{lim}$ f(x).

So f is continuous.

For x = c > 5.

f (c) = (x 5) = c 5

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ (x 5) = c 5.

∴ f(c) =  $\underset{x\to c}{lim}$ f(x)

So, f is continuous.

For x = c = 5,

f (5) = 5 5 = 0

$\underset{x\to 5^{}}{lim}$ f(a) =  $\underset{x\to 5^{}}{lim}$ (x 5) = (5 5) = 0

$\underset{x\to 5^{+}}{lim}$ f(x) =  $\underset{x\to 5^{+}}{lim}$ (x 5) = 5 5 = 0

∴  $\underset{x\to 5^{+}}{}$ + (x) =  $\underset{x\to c^{+}}{sin}$ + (x) = f (c)

Hence f is continuous.

Q4.Prove that the function  $f\left(x\right)=x^n$ continuous at  $x=n,$ where n is a positive integer.

A.4.Given, f(x) = x n > n = positive.

At x = 2,

(x) = n n.

$\underset{x\to n}{lim}$ f(x) =  $\underset{x\to n}{lim}$ x n = n n

∴  $\underset{x\to n}{lim}$ f(x) = f(x)

So f is continuous at x = n.

Q5.Is the function  $f$ defined by

$\begin{array}{l}f\left(x\right)=\{\begin{array}{l}x,x\le 1\\ 5,x\end{array}\\ xx\end{array}$

Find all points of discontinuity of  $f$ , where  $f$ is defined by

A.5.Given, f(a) =  $\{\begin{array}{ll}x, if x\le 1\hfill & 5, if x>1.\hfill \end{array}$

At x = 0,

(0) = 0

$\underset{x\to 0}{lim}$ f(x) =  $\underset{x\to 0}{lim}$ x = 0

∴  $\underset{x\to 0}{lim}$ f (x) = f(0)

So, f is continuous at x = 0.

At x = 1,

Left hand limit,

L.H.L =  $\underset{x\to 1}{lim}$ f(x) =  $\underset{x\to 1}{lim}$ x = 1.

Right hand limit,

R. H. L. =  $\underset{x\to 1^{+}}{lim}$ f(x) =  $\underset{x\to 1^{+}}{lim}$ 5 = 5.

L.H.L.  $\cancel{=}$ R.H.L.

So, f is not continuous at x = 1.

At x = 2,

f(2) = 5.

$\underset{x\to 1}{lim}$ f(x) =  $\underset{x\to 2}{lim}$ 5 = 5

$\underset{lim\to 2}{limf}$ (x) = f (2)

So f is continuous at x = 2.

Find all points of discontinuity of f, where f is defined by

Q6.  $f\left(x\right)=$  $\{\begin{array}{cc}\hfill 2x+3 if x\le 2\hfill & \hfill 2x-3 if x>2.\hfill \end{array}$

A.6. Given f(x) =  $\{\begin{array}{cc}\hfill 2x+3 if x\le 2\hfill & \hfill 2x-3 if x>2.\hfill \end{array}$

For x = c < 2,

F (c) = 2c + 3

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ 2x + 3 = 2c + 3

∴  $\underset{x\to c}{lim}$ f (x) = f(c)

So f is continuous at x  $\left|<\right|$ 2.

For x = c > 2.

F (c) = 2c 3

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ 2x 3 = 2c 3

∴  $\underset{x\to c}{lim}$ f(x) = f(c)

So f is continuous at x  $\left|>\right|$ 2.

For x = c = 2,

L.H.L. =  $\underset{x\to 2^{-}}{lim}$ f(x) =  $\underset{x\to 2^{-}}{lim}$ .2x + 3 = 2. 2 + 3 = 4 + 3 = 7.

R.H.L. =  $\underset{x\to 2^{+}}{lim}$ f(x) =  $\underset{x\to 2^{+}}{lim}$ 2x 3 = 2. 2 3 = 4 3 = 1.

∴ LHL  $\cancel{=}$ RHL

∴ f is not continuous at x = 2.i e, point of discontinuity

Q7.  $f\left(x\right)=$  $\{\begin{array}{lll}\left|x\right|+3 if x\le -3\hfill & -2x if -3<x<3\hfill & 6x+2 if x\ge 3\hfill \end{array}$

A.7. Given, f(x) =  $\{\begin{array}{lll}\left|x\right|+3 if x\le -3\hfill & -2x if -3<x<3\hfill & 6x+2 if x\ge 3\hfill \end{array}$

For x =  $\mid c<-3,$

f ( 3) = e + 3 (∴x< 3,  $\left|x\right|=-x$ )

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$  $\left|x\right|+3=-a+3.$

∴  $\underset{x\to c}{lim}$ f(x) = f(c)

So, f is continuous at x = c < 3.

For x = c > 3

f(3) = 6.3 + 2 = 18 + 2 = 20

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ 6x + 2 = 18 + 2 = 20

∴  $\underset{x\to c}{lim}$ f(x) = f(c).So f is continuous at x = c > 3.

For. C = 3,

f ( 3) = ( 3) + 3 = 6.

$\underset{x\to c^{-}}{lim}$ f(x) =  $\underset{x\to c^{-}}{lim}$ .x + 3 = ( 3) + 3 = 6.

$\underset{x\to c^{+}}{lim}$ f(x) =  $\underset{x\to c^{+}}{lim}$ ( 2x) = 2 ( 3) = 6.

∴  $\underset{x\to c^{-}}{lim}$ f(x) =  $\underset{x\to c^{-}}{lim}$ f(x) = f( 3)

So, f is continuous at x = c = 3.

For c = 3,

f(3) = 6.3 + 2 = 18 + = 20.

$\underset{x\to 3^{-}}{lim}$ f(x) =  $\underset{x\to 3^{-}}{lim}$ 2x = 2 (3) = 6

$\underset{x\to 3^{+}}{lim}$ f(x) =  $\underset{x\to 3^{+}}{lim}$ (6x + 2) = 6.3 + 2 = 20

∴  $\underset{x\to 3^{-}}{lim}$ f(x)  $\cancel{=}$  $\underset{x\to 3^{-}}{lim}$ f(x).

f is not continuous at x = 3 point of discontinuity

Q8.  $f\left(x\right)=$  $\{\begin{array}{cc}\hfill \frac{\left|x\right|}{x}, if x\ne 0\hfill & \hfill 0, if x=0.\hfill \end{array}$

A.8.Given, f(x) =  $fxxx$

For x = c < 0,

f(c) = 1

$\underset{x\to 0}{lim}$ f(x) =  $\underset{x\to 0}{lim}$ 1 = 1

∴f(c) =  $\underset{x\to 0}{lim}$ f (x)

f is continuous at x  $\left|<\right|$ 0.

For x = c > 0,

F (c) = 1

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$  $\cancel{=}$ = 1.

∴f(c) =  $\underset{x\to c}{lim}$ f(x)

f is continuous at x > 0.

For x = c 0.

L.H.L. =  $\underset{x\to 0^{-}}{lim}$ f(x) =  $\underset{x\to 0^{-}}{lim}$ ( 1) = 1

R.H.L.  $\underset{x\to 0^{+}}{lim}$ f(x) =  $\underset{x\to 0^{+}}{lim}$ 1 = 1

∴ L.H.L.  $\cancel{=}$ R.H.L.

is now continuous at x = 0, point of discontinuity of f is at x = 0.

Q9.  $f\left(x\right)=$  $\{\begin{array}{cc}\hfill \frac{x}{\left|x\right|} ,if x<0\hfill & \hfill -1, if x\ge 0\hfill \end{array}$

A.9.Given, f(x) =  $\{\begin{array}{cc}\hfill \frac{x}{\left|x\right|} if x<0\hfill & \hfill -1 if x\ge 0\hfill \end{array}=\{\begin{array}{cc}\hfill \frac{x}{-x}=-1 if x<0\hfill & \hfill \frac{x}{2}-1 if x\ge 0.\hfill \end{array}$

So, f (x) = 1  $\forall x\in R.$

Hence,  $\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ 1 = 1  $\forall \in \in ℝ$

∴ f is continuous in its domain

Hence, f has no point of discontinuity.

Q10.  $f\left(x\right)=$  $\{\begin{array}{cc}\hfill x+1,if x\ge 1\hfill & \hfill x^2+1, if x<1\hfill \end{array}$

A.10.Given, f(x) =  $\{\begin{array}{cc}\hfill x+1,\pi x\ge 1\hfill & \hfill x^2+1, \pi x<1.\hfill \end{array}$

For x = c < 1,

$\underset{x\to c}{lim}$ f (x) =  $\underset{x\to c}{lim}$ x2 + 1 = c2 + 1

∴  $\underset{x\to c}{lim}$ f(x) = f(c)

So f is continuous at x = c < 1.

For x = c > 1,

F (c) = c + 1

$\underset{x\to c}{lim}$ f (x) =  $\underset{x\to c}{lim}$ x + 1 = c + 1

∴  $\underset{x\to c}{lim}$ f(x) = f(c)

So, f is continuous at x = c > 1.

For x = c = 1, + (1) = 1 + 1 = 2

L.H.L. =  $\underset{x\to 1^{-}}{lim}$ f(x) =  $\underset{x\to 1^{-}}{lim}$ x2 + 1 = 12 + 1 = 2.

R.H.L. =  $\underset{x\to 1^{+}}{lim}$ f(x) =  $\underset{x\to 1^{+}}{lim}$ x + 1 = .1 + 1 = 2

∴ L.H.L = R.H.L. = f(1)

So, f is continuous at x = 1.Hence f has no point of discontinuity.

Q11.  $f\left(x\right)=$  $\{\begin{array}{cc}\hfill x^3-3,if x\le 2\hfill & \hfill x^2+1 ,if x>2\hfill \end{array}$

A.11.Given f(x) =  $\{\begin{array}{cc}\hfill x^3-3if x\left|⩽\right|2\hfill & \hfill x^2+1 if x>2\hfill \end{array}$

For x = c < 2,

f (c) = c3 3

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ x3 3 = c3 3.

So f is continuous at x  $\left|<\right|$ 2.

For x = c > 2

f(c) = x2 + 1 = c2 + 1

$\underset{x\to 2}{lim}$ f(x) =  $\underset{x\to 2}{lim}$ x2 + 1 = c2 + 1 = f(c)

So, f is continuous at x  $\left|>\right|$ 2.

For x = c = 2, f(2) = 23 3 = 8 3 = 5.

L.H.L.  $\underset{x\to 2^{-}}{lim}$ f(x) =  $\underset{x\to 2^{-}}{lim}$ x3 3 = 23 3 = 5.

R.H.L.  $\underset{x\to 2^{+}}{lim}$ f(x) =  $\underset{x\to 2^{+}}{lim}$ x2 + 1 = 22 + 1 = 5

∴ R.H.L. = L.H.L. = f(2).

So, f is continuous at x = 2

Hence f has no point of discontinuity.

Q12.  $f\left(x\right)=$  $\{\begin{array}{ll}{x}^{10}-1, if x\le 1\hfill & x^2, if x>1\hfill \end{array}$

A.12.Given, f(x) =  $\{\begin{array}{ll}{x}^{10}-1, if x\le 1\hfill & x^2, if x>1.\hfill \end{array}$

For x = c < 1.

f(c) =  $\underset{x\to c}{lim}$ f(x) = c10 1.

So, f is continuous for x  $\left|>\right|$ 1.

For x = c > 1.

f(c) =  $\underset{x\to c}{lim}$ f(x) = c2

So, f is continuous for x  $\left|>\right|$ 1.

For x = c = 1,

L.H.L =  $\underset{x\to 1^{-}}{lim}$ f(x) =  $\underset{x\to 1^{-}}{lim}$ x10 1 110 1 = 0.

R.H.L. =  $\underset{x\to 1^{+}}{lim}$ f(x) =  $\underset{x\to 1^{+}}{lim}$ x2 = 12 = 1.

∴ L.H.L  $\cancel{=}$ R.H.L.

So, f is not continuous at x = 1.

Hence, f has point of discontinuity at x = 1.

Q13.Is the function defined by

$f\left(x\right)=$  $\{\begin{array}{ll}x+5, if x\le 1\hfill & x-5, if x>1\hfill \end{array}$ a continuous function?

Discuss the continuity of the function  $f$ , where  $f$ is defined by

A.13.Given, f(x) =  $\{\begin{array}{ll}x+5 if x\left|⩽\right|1\hfill & x-5 if x>1.\hfill \end{array}$

For x = c < 1.

F (c) = c + 5

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ f x + 5 = c + 5

∴  $\underset{x\to c}{lim}$ f(x) = f(c)

So, f is continuous at x  $\left|<\right|$ 1.

For x = c > 1

F (c) = c 5

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ x 5 = c 5.

$\underset{x\to c}{lim}$ f(x) = f(c)

So, f is continuous at x  $\left|>\right|$ 1.

For x = 1

L.H.L. =  $\underset{x\to 1^{-}}{lim}$ f(x) =  $\underset{x\to 1^{-}}{lim}$ x + 5 = 1 + 5 = 6.

R.H.L. =  $\underset{x\to 1^{+}}{lim}$ f(x) =  $\underset{x\to 1^{-}}{lim}$ x 5 = 1 5 = 4.

L.H.L.  $\cancel{=}$ R.H.L.

f is not continuous at x = 1

So, point of discontinuity of f is at x = 1.

Discuss the continuity of the function  $f$ , where  $f$ is defined by

Q14.  $f\left(x\right)=$  $\{\begin{array}{ccc}\hfill 3,if \pi 0\le x\le 1\hfill & \hfill 4,if \pi 1<x<3\hfill & \hfill 5,if \pi 3\le x\le 10\hfill \end{array}$

A.14.Given, f(x) =  $\{\begin{array}{ccc}\hfill 3 \pi 0\le x\le 1\hfill & \hfill 4 \pi 1<x<3\hfill & \hfill 5 \pi 3\le x\le 10.\hfill \end{array}$

For x = c such that  $0\le c<1$

f(c) = 3

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ 3 = 3 = f(c)

So, f is continuous in [0, 1].

For x = c = 1,

L.H.L. =  $\underset{x\to 1^{-}}{lim}$ f(x) =  $\underset{x\to 1^{-}}{lim}$ 3 = 3.

R.H.L.  $\underset{x\to 1^{+}}{lim}$ f(x) =  $\underset{x\to 1^{+}}{lim}$ 4 = 4

∴ L.H.L  $\cancel{=}$ R.H.L.

f is discontinuity at x = 1

for x = c such that  $1<c<3.$

f(c) = 4

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ 4 = 4 = f(c)

So, f is continuous in  $x\in (1,3)$

For x = c = 3

L.H.L.  $\underset{x\to 3^{-}}{lim}$ f(x) =  $\underset{x\to 3^{-}}{lim}$ 4 = 4

R.H.L.  $\underset{x\to 3^{+}}{lim}$ f(x) =  $\underset{x\to 3^{+}}{lim}$ 5 = 5.

So, f is discontinuous at x = 3.

For x = c such that  $3<c\le 10$

f (c) = 5.

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ 5 = 5 = f(c)

So, f is continuous in  $x\in (3,10]$

Q15.  $f\left(x\right)=$  $\{\begin{array}{ccc}\hfill 2x, if x<0\hfill & \hfill 0, if 0\le x\le 1\hfill & \hfill 4x, if x>1\hfill \end{array}$

A.15.Given f(x) =  $\{\begin{array}{ccc}\hfill 2x, if x<0\hfill & \hfill 0, if 0\le x\le 1\hfill & \hfill 4x, if x>1.\hfill \end{array}$

For (c) = c < 0,

f(c) = 2c.

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ 2x = 2c = f(c)

So, f is continuous at x  $\left|<\right|$ 0

For x = c > 1,

f(c) = 4c

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ 4x = 4c = f(c)

So, f is continuous at x> 1.

For x = 0

L.H.L. =  $\underset{x\to 0^{-}}{lim}$ f(x) =  $\underset{x\to 0^{-}}{lim}$ . 2x = 2 (0) = 0

R.H.L. =  $\underset{x\to 0^{+}}{lim}$ f(x) =  $\underset{x\to 0^{+}}{lim}$ . 0 = 0.

f(0) = 0.

∴ L.H.L. = R.H.L. = f(0).

So, f is continuous at x = 0.

For x = 1.

L.H.L. =  $\underset{x\to 1^{-}}{lim}$ f(x) =  $\underset{x\to 1^{-}}{lim}$ . 0 = 0

R.H.L. =  $\underset{x\to 1^{+}}{lim}$ f(x) =  $\underset{x\to 1^{+}}{lim}$ . 4x = 4 (1) = 4.

∴ L.H.L.  $\cancel{=}$ R.H.L.

So, f is discontinuous at x = 1.

Q16.  $f\left(x\right)=$  $\{\begin{array}{ccc}\hfill -2,\hfill & \hfill if x\le -1\hfill & \hfill 2x,\hfill \\ \hfill if -1<x\le 1\hfill & \hfill 2,\hfill & \hfill if x>1\hfill \end{array}$

A.16.Given, f(x) =  $\{\begin{array}{ccc}\hfill -2,\hfill & \hfill if x\le -1\hfill & \hfill 2x,\hfill \\ \hfill if -1<x\le 1\hfill & \hfill 2,\hfill & \hfill if x>1.\hfill \end{array}$

For x = c < 1,

f(c) = 2

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ ( 2) = 2 = f(c)

So, f is continuous at x< 1.

For x = c > 1,

f(c) = 2

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ . 2 = 2 = f(c)

So, f is continuous at x  $\left|>\right|$ 1.

For x = 1,

L.H.L. =  $\underset{x\to -1^{-}}{lim}$ f(x) =  $\underset{x\to -1^{-}}{lim}$ 2 = 2

R.H.L. =  $\underset{x\to -1^{+}}{lim}$ f(x) =  $\underset{x\to -1^{+}}{lim}$ . 2x = 2 ( 1) = 2

and f( 1) = 2

So, L.H.L. = R.H.L. = f( 1)

∴f is continuous at x = 1.

For x = 1,

L.H.L. =  $\underset{x\to 1^{-}}{lim}$ f(x) =  $\underset{x\to 1^{-}}{lim}$ . 2x = 2.1 = 2

R.H.L. =  $\underset{x\to 1^{+}}{lim}$ f(x) =  $\underset{x\to 1^{+}}{lim}$ . 2 = 2.

f(1) = 2

f(1) = L.H.L = R.H.L.

So, f is continuous at x = 1.

Q17.Find the relationship between a and b so that the function  $f$ defined by

$f\left(x\right)=$  $\{\begin{array}{ll}ax+1, if x\le 3\hfill & bx+3, if x>3\hfill \end{array}$ is continuous at  $x$ = 3.

A.17.Given, f(x) =  $\{\begin{array}{ll}ax+1, if x\le 3\hfill & bx+3, if x>3\hfill \end{array}$ is continuous at x = 3

So, f(3) = 3a + 1

L.H.L =  $\underset{x\to 3^{-}}{lim}$ f(x) =  $\underset{x\to 3^{-}}{lim}$ ax + 1 = 3a + 1

R.H.L =  $\underset{x\to 3^{+}}{lim}$ f(x) =  $\underset{x\to 3^{+}}{lim}$ b x + 3 = 3b + 3

for continuity at x = 3,

L.H.L = R.H.L. = f(3)

$\Rightarrow$ 3a + 1 = 3 + 3 = 3a + 1

So, 3a + 1 = 3b + 3

3a = 3b + 3 1

3a = 3b + 2.

a = b +  $\frac{2}{3}.$

Q18.Find the relationship between a and b so that the function  $f$ defined by

$f\left(x\right)=$  $\{\begin{array}{cc}\hfill \lambda \left(x^2-2x\right),\hfill & \hfill if x\le 0\hfill \\ \hfill 4x+1,\hfill & \hfill if x>0\hfill \end{array}$ continuous at  $x$ = 0? What about continuity at  $x$ = 1?

A.18.Given, f(x) =  $\{\begin{array}{cc}\hfill \lambda \left(x^2-2x\right)\hfill & \hfill if x\left|⩽\right|0\hfill \\ \hfill 4x+1\hfill & \hfill if x>0.\hfill \end{array}$

For continuity at x = 0,

$\underset{x\to 0^{-}}{lim}$ f(x) =  $\underset{x\to 0^{+}}{lim}$ f(x) = f(0).

$\underset{x\to 0^{-}}{lim}$  $\lambda \left(x^2-2x\right)$ =  $\underset{x\to 0^{+}}{lim}$ 4x + 1 =  $\lambda \left(0^2-2.0\right)$

0 = 1 = 0 which is not true

Hence, f is not continuous for any value of  $\lambda .$

For x = 1,

$\underset{x\to 1}{lim}$ f(x) = f(1).

$\underset{x\to 1}{lim}$ 4x + 1 = 4 (1) + 1

$\Rightarrow$ 4 + 1 = 4 + 1

$\Rightarrow$ 5 = 5.

So, f is continuous at x = 1 value of  $\lambda$

Q19.Show that the function defined by  $g\left(x\right)=x-\left[x\right]$ is discontinuous at all integral

points. Here  $\left[x\right]$ denotes the greatest integer less than or equal to  $x$ .

A.19.Given, g(x) = x [x].

For  $n\in z,$

g(n) = n [n] = nn = 0

$\underset{x\to n^{-}}{lim}$ f(x) =  $\underset{x\to n^{-}}{lim}$ (x [x]) = n [n 1] = n n + 1 = 1

$\underset{x\to n^{+}}{lim}$ g (x) =  $\underset{x\to n^{+}}{lim}$ x [x] = n [n] = 0

So,  $\underset{x\to n^{-}}{lim}$ g(x)  $\cancel{=}$  $\underset{x\to n^{+}}{lim}$ g(x).

g(x) is d is continuous at all x  $\in z.$

Q20.Is the function defined by  $f\left(x\right)=x^2-sinx+5$ continuous at  $x=\pi ?$

A.20. Given f(x) = x2 sin x + 5.

At x = .

$f(\pi )={\pi }^2-sin\pi +5={\pi }^2-0+5={\pi }^2+5$

$\underset{x\to \pi }{lim}$ f(x) =  $\underset{x\to \pi }{lim}$ [x2 sin x + 5]

If x =  $\pi +h$ then as x , h 0, so,

$\underset{x\to \pi }{lim}$ f(x) =  $\underset{x\to 0}{lim}$ [( + h)2 sin ( + h) + 5]

= ( + 0)2  $\underset{h\to 0}{lim}$  $\begin{array}{rr}\hfill [sin\pi cosh+cos\pi \cdot sina]& \hfill +5.\end{array}$

= 2  $\underset{h\to 0}{lim}$ sin cos h  $\underset{h\to 0}{lim}$ cos sin h + 5

= x2 0 × (1) ( 1) 0 + 5.

= 2 + 5 = f(x)

So, f is continuous at x = .

Q21.Discuss the continuity of the following functions:

$\begin{array}{l}\left(\right)f\left(x\right)=sinx+cosx\left(\right)f\left(x\right)=sinx-cosx\\ \left(\right)f\left(x\right)=sinx.cosx\end{array}$

A.21. (a) Given f(x) = sin x + cos x

(b). Given, f(x) = sin x cos x

(c). Given, f(x) = sin x .cos x.

Let g(x) = sin x and h(x) = cos x.

If g or h are continuous f x then

g + h

g h

g h are also continuous.

As g(x) = sin x is defined for all real number x.

Let  $c\in ℝ$ , and putting x = c + h. we see that as  $x\to c,h\to 0.$

Then g(c) = sin c

$\underset{x\to c}{lim}$ g(x) =  $\underset{x\to c}{lim}$ sin x =  $\underset{h\to 0}{lim}$ sin (c + h).

=  $\underset{h\to 0}{lim}$ (sin c cos h + cos c sin h )

= sin c. cos 0 + cos c. sin 0

= sin c 1 + 0

= sin c

= g (c)

So, g is continuous x R.

And h (c) = cos c

=  $\underset{h\to 0}{lim}$ g(x) =  $\underset{x\to c}{lim}$ sin x =  $\underset{x\to c}{lim}$ cos (c + h)

= cos c .cos 0 sin c. sin 0

= cos c .1 0.

= cos c = h(c).

As g and h are continuous x R.

The f x f(x)

Q22.Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

A.22. For two continuous fxn f(x) and g(x),  $\frac{f(x)}{g(x)},\frac{g(x)}{f(x)},$

$\frac{1}{f(x)}\frac{1}{g(x)}$ are also continuous

Let f(x) = sin x is defined x R.

Let C E R such that x = c + h. so, as x c, h 0

now, f(c) = sin c.

$\underset{x\to c}{lim}$ f(i) =  $\underset{x\to c}{lim}$ sin x =  $\underset{h\to 0}{lim}$ sin (c + h).

=  $\underset{h\to 0}{lim}$ (sin c cos h + cos c sin h)

= sin c cos 0 + cos c sin 0

= sin c 1 + 0

= sin c

= f(c)

So, f is continuous.

Then,  $\frac{1}{f(x)}$ is also continuous

$\Rightarrow \frac{1}{sin(x)}$ is also continuous

$\Rightarrow$ cosec x is also continuous

Let g(x) = cos x is defined x R.

Then, g(c) = cos c

$\underset{x\to c}{lim}$ g(x) =  $\underset{x\to c}{lim}$ . cos x

=  $\underset{h\to 0}{lim}$ cos (c + h).

=  $\underset{h\to 0}{lim}$ (cos c cos h sin c sin h.)

= cos c cos h sin c sin h

= cos c.

= g(c)

So, g is continuous

Then,  $\frac{1}{g(x)}$ is also continuous

$\Rightarrow \frac{1}{cosx}$ is also continuous

$\Rightarrow$ cos x is also continuous.

Hence,  $\frac{g(x)}{f(x)}$ is also continuous

$$  $$  $\Rightarrow \frac{cosx}{sinx}$ is also continuous

$\Rightarrow$ cot x is also continuous .

Q23.Find all points of discontinuity of  $f$ , where

$f\left(x\right)=$  $\{\begin{array}{ll}\frac{sinx}{x},\hfill & if x<0\hfill \\ x+1,\hfill & if x\ge 0\hfill \end{array}$

A.23. Given f(x) =  $\{\begin{array}{ll}\frac{sinx}{x},\hfill & if x<0.\hfill \\ x+1,\hfill & if x\ge 0.\hfill \end{array}$

For x = c < 0,

f(c) =  $\frac{sinc}{c}$

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$  $\frac{sinx}{x}=\frac{sinc}{c}=f(c)$

So, f is continuous for x < 0

For x = c > 0

f(c) = c + 1

$\underset{x\to c}{lim}$ f(x) =  $\underset{x\to c}{lim}$ x + 1 = c + 1 = f(c)

So, f is continuous for x > 0.

For x = 0.

L.H.L. =  $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{sinx}{x}=1.$

R.H.S. =  $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}x+1=0+1=1$

And f(0) = 0 + 1 = 1

L.H.L = R.H.L. = f(0)

So, f is continuous at x = 1.

Hence, discontinuous point of x does not exit.

Q24.Determine if  $f$ defined by

$f\left(x\right)=$  $\{\begin{array}{ll}{x}^{2}sin\frac{1}{x}, if x\ne 0\hfill & 0, if x=0\hfill \end{array}$ is a continuous function?

A.24.Given f(x) =  $\{\begin{array}{ll}{x}^{2}sin\frac{1}{x}, if x\ne 0.\hfill & 0 if x=0.\hfill \end{array}$

For x = c  $\cancel{=}$ 0,

f(c) =  $c^{2}sin\frac{1}{c}$

$\underset{x\to c}{lim}f(x)=\underset{x\to c}{lim}{x}^{2}sin\frac{1}{x}=c^{2}sin\frac{1}{c}.$

So, f is continuous for  $x\ne 0.$

For x = 0,

f(0) = 0

$\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}\left(x^{2}sin\frac{1}{x}\right)$

As we have sin  $\frac{1}{x}\in [-1,1]$

$\underset{x\to 0}{lim}$ f(x) = 02 a where  $a\in [-1,1]$

= 0 = f(0).

∴ f is also continuous at x = 0.

Q25.Examine the continuity of  $f$ , where  $f$ is defined by

$f\left(x\right)=$  $\{\begin{array}{cc}\hfill sinx-cosx,\hfill & \hfill if x\ne 0\hfill \\ \hfill -1,\hfill & \hfill if x=0\hfill \end{array}$

A.25. Given, f(x) =  $\{\begin{array}{cc}\hfill sinx-cosx,\hfill & \hfill if x\ne 0\hfill \\ \hfill -1,\hfill & \hfill if x=0.\hfill \end{array}$

For x = c  $\cancel{=}$ 0,

f(c) = sin c cos c.

$\underset{x\to c}{lim}$ f (x) =  $\underset{x\to c}{lim}$ (sin x cos x) = sin c cos c = f(c)

So, f is continuous at  $x\ne 0$

For x = 0,

f(0) = 1

$\underset{x\to 0}{lim}$ f (x) =  $\underset{x\to 0}{lim}$ (sin x cos x) = sin 0 cos 0 = 0 1 = 1

$\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}(sinx-corx)=sin0-cos0=-1.$

∴  $\underset{x\to 0-}{lim}$ f(x) =  $\underset{x\to 0+}{lim}$ f (x) = f (0)

So, f is continuous at x = 0.

Find the values of  $k$ so that the function  $f$ is continuous at the indicated point in Exercises 26 to 29.

Q26.  $f\left(x\right)=$  $\{\begin{array}{cc}\hfill \frac{kcosx}{\pi -2x},\hfill & \hfill if x\ne \frac{\pi }{2}\hfill \\ \hfill 3,\hfill & \hfill if x=\frac{\pi }{2}\hfill \end{array}\frac{\pi }{2}$

A.26. Given, f(x) =  $\{\begin{array}{cc}\hfill \frac{\pi cosx}{\pi -2x}\hfill & \hfill if x\ne \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 3\hfill & \hfill if x=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \end{array}$

For continuity at  $x=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\underset{x\to \raisebox{1ex}{${\pi }^{-}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}f(x)=\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}f(x)=f\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right).$

$\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{xcosx}{\pi -2x}=\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2^{+}$}\right.}{lim}\frac{xcosx}{\pi -2x}=3.$

Take  $\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{xcosx}{x-2x}=3$ .

Putting x =  $\frac{\pi }{2}+h$ such that as  $x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.,h\to 0.$

So  $\underset{h\to 0}{lim}\frac{xcos(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+h)}{x-2(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+n)}=\underset{h\to 0}{lim}\frac{x(-sinx)}{-2h}$

$=\underset{h\to 0}{lim}\frac{xsinh}{2h}$

$=\frac{k}{2}\underset{h\to 0}{lim}\frac{sinh}{h}=\frac{k}{2}.$

i e,  $\frac{k}{2}=3$

$\Rightarrow$ k = 6

Similarly from  $\underset{x\to \raisebox{1ex}{${\pi }^{+}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{xcosx}{\pi -2x}=3$

$\Rightarrow \underset{h\to 0}{lim}\frac{hcos\left(\frac{\pi }{2}+h\right)}{\pi -2\left(\frac{\pi }{2}+h\right)}=\underset{h\to 0}{lim}\frac{-hsinh}{-2h}$

$=\frac{}{2}\underset{h\to 0}{lim}\frac{sinh}{h}$

$=\frac{x}{2}$

So,  $\frac{x}{2}=3$

$\Rightarrow$ k = 6

Q27.  $f\left(x\right)=$  $\{\begin{array}{ll}k{x}^2,\hfill & if x\le 2\hfill \\ 3,\hfill & if x>2\hfill \end{array}x$

A.27. Given f(x) =  $\{\begin{array}{ll}k{x}^2\hfill & if x\le 2.\hfill \\ 3\hfill & if x>2.\hfill \end{array}$

For continuous at x = 2,

f(2) = k (2)2 = 4x.

L.H.L. =  $\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{-}}{lim}{x}^2=4x$

R.H.L. =  $\underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2^{+}}{lim}3=3$

Then, L.H.L = R.H.L. = f(2)

i e, 4x = 3

$\Rightarrow x=\frac{3}{4}.$

Q28.  $f\left(x\right)=$ Parsing Error

A.28. Given, f(x) = Parsing Error

For continuity at x =  $\pi$ . f(x) = k  $\pi$ + 1

L.H.S. =  $\underset{x\to {\pi }^{-}}{lim}f(x)=\underset{x\to {\pi }^{-}}{lim}kx+1=\pi +1$

R.H.L. =  $\underset{x\to x^{+}}{lim}f(x)=\underset{x\to {\pi }^{+}}{lim}cosx=cosx=-1.$

Then L.H.L = R.H.L. = f()

$\Rightarrow$ k + 1 = 1

$\Rightarrow$ k = 2

$\Rightarrow x=-\frac{2}{\pi }.$

Q29.  $f\left(x\right)=$  $\{\begin{array}{ll}kx+1,\hfill & if x\le 5\hfill \\ 3x-5,\hfill & if x>5\hfill \end{array}x=5$

A.29.Given, f (x)  $\{\begin{array}{ll}kx+1,\hfill & if x\le 5\hfill \\ 3x-5,\hfill & if x>5.\hfill \end{array}$

For continuity at x = 5,

$\underset{x\to 5^{-}}{lim}f(x)=\underset{x\to 5^{-}}{lim},kx+1=5x+1.$

$\underset{x\to 5^{+}}{lim}f(x)=\underset{x\to 5^{+}}{lim}3x-5=15-5=10$

f(5) = 5k + 1

So,  $\underset{x\to 5^{-}}{lim}f(x)=\underset{x\to 5^{+}}{lim}f(x)=f(5).$

i e, 5k + 1 = 10

$\Rightarrow$ 5k = 10 1

$\Rightarrow$ k =  $\frac{9}{5}.$

Q30.Find the values of  $a$ and  $b$ such that the function defined by

$f\left(x\right)=$  $\{\begin{array}{ccc}\hfill 5,\hfill & \hfill if x\le 2\hfill & \hfill ax+b,\hfill \\ \hfill if 2<x<10\hfill & \hfill 21,\hfill & \hfill if x\ge 10\hfill \end{array}$ is a continuous function.

A.30. Given, f(x) =  $\{\begin{array}{ccc}\hfill 5\hfill & \hfill if x\le 2\hfill & \hfill ax+b\hfill \\ \hfill if 2<x<10\hfill & \hfill 21\hfill & \hfill if x\ge 10\hfill \end{array}$

For continuity at x = 2,

$\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{+}}{lim}f(x)=f(2)$