NCERT Solutions for Exercise 5.1 Class 12 Maths Chapter 5 - Continuity and Differentiability

In the previous class, you have already learned about limits, derivatives, limits, and derivatives of trigonometric functions. In this article, you will get NCERT solutions for Class 12 Maths chapter 5 exercise 5.1. NCERT book Class 12 Maths chapter 5 exercise 5.1 consists of questions related to finding whether a function is continuous or not.

Continuity of functions can't be learned without fundamental knowledge of limit which you have learned already. It is a fundamental concept of calculus that you must know to understand more concepts of calculus. Solving exercise 5.1 Class 12 Maths questions are very important to get conceptual clarity about continuity. There are different theorems to check the continuity of different types of functions mentioned in the NCERT syllabus of Class 12 Maths. You must go through these theorems also. Check here for NCERT Solutions.

Also, see

Continuity and Differentiability Exercise 5.2
Continuity and Differentiability Exercise 5.3
Continuity and Differentiability Exercise 5.4
Continuity and Differentiability Exercise 5.5
Continuity and Differentiability Exercise 5.6
Continuity and Differentiability Exercise 5.7
Continuity and Differentiability Exercise 5.8
Continuity and Differentiability Miscellaneous Exercise

Continuity and Differentiability Exercise: 5.1

Question:1. Prove that the function $f ( x) = 5 x -3$ is continuous at $x = 0, at\: \: x = - 3$ and at $x = 5$

Answer:

Given function is
$f ( x) = 5 x -3$
$f(0) = 5(0)-3 = -3$
$\lim_{x\rightarrow 0} f(x) = 5(0)-3 = -3$
$\lim_{x\rightarrow 0} f(x) =f(0)$
Hence, function is continous at x = 0

$f(-3)= 5(-3)-3=-15-3=-18\\\Rightarrow \lim_{x\rightarrow -3} f(x) = 5(-3)-3 = -15-3=-18\\\Rightarrow \lim_{x\rightarrow -3} f(x) = f(-3)$
Hence, function is continous at x = -3

$f(5)= 5(5)-3=25-3=22\\\Rightarrow \lim_{x\rightarrow 5} f(x) = 5(5)-3 = 25-3=-22\\ \Rightarrow \lim_{x\rightarrow 5} f(x) = f(5)$
Hence, function is continuous at x = 5

Question:2.Examine the continuity of the function $f (x) = 2x ^2 - 1 \: \: at\: \: x = 3.$

Answer:

Given function is
$f(x) = 2x^2-1$
at x = 3
$f(3) = 2(3)^2-1 = 2\times 9 - 1=18-1=17\\ \lim_{x\rightarrow 3}f(x) = 2(3)^2-1=2\times 9-1=18-1=17$
$\lim_{x\rightarrow 3}f(x) = f(3)$
Hence, function is continous at x = 3

Question:3 Examine the following functions for continuity.
$(a) f (x) = x - 5$

Answer:

Given function is
$f(x) = x-5$
Our function is defined for every real number say k
and value at x = k , $f(k) = k-5$
and also,
$\lim_{x\rightarrow k} f(x) = k -5\\ \lim_{x\rightarrow k} f(x) = f(k)$
Hence, the function $f(x) = x-5$ is continuous at every real number

Question:3 b) Examine the following functions for continuity.

$f (x) = \frac{1}{x-5} , x \neq 5$

Answer:

Given function is
$f(x ) = \frac{1}{x-5}$
For every real number k , $k \neq 5$
We get,
$f(k) = \frac{1}{k-5}\\ \lim_{x\rightarrow k}f(x ) = \frac{1}{k-5}\\ \lim_{x\rightarrow k}f(x ) = f(k)$
Hence, function $f(x ) = \frac{1}{x-5}$ continuous for every real value of x, $x \neq 5$

Question:3 c) Examine the following functions for continuity.

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

Answer:

Given function is
$f(x ) = \frac{x^2-25}{x+5}$
For every real number k , $k \neq -5$
We gwt,
$f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = f(k)$
Hence, function $f(x ) = \frac{x^2-25}{x+5}$ continuous for every real value of x , $x \neq -5$

Question:3 d) Examine the following functions for continuity. $f (x) = | x - 5|$

Answer:

Given function is
$f (x) = | x - 5|$
for x > 5 , f(x) = x - 5
for x < 5 , f(x) = 5 - x
SO, different cases are their
case(i) x > 5
for every real number k > 5 , f(x) = x - 5 is defined
$f(k) = k - 5\\ \lim_{x\rightarrow k }f(x) = k -5\\ \lim_{x\rightarrow k }f(x) = f(k)$
Hence, function f(x) = x - 5 is continous for x > 5

case (ii) x < 5
for every real number k < 5 , f(x) = 5 - x is defined
$f(k) = 5-k\\ \lim_{x\rightarrow k }f(x) = 5 -k\\ \lim_{x\rightarrow k }f(x) = f(k)$
Hence, function f(x) = 5 - x is continous for x < 5

case(iii) x = 5
for x = 5 , f(x) = x - 5 is defined
$f(5) = 5 - 5=0\\ \lim_{x\rightarrow 5 }f(x) = 5 -5=0\\ \lim_{x\rightarrow 5 }f(x) = f(5)$
Hence, function f(x) = x - 5 is continous for x = 5

Hence, the function $f (x) = | x - 5|$ is continuous for each and every real number

Question:4. Prove that the function $f (x) = x^n$ is continuous at x = n, where n is a positive integer

Answer:

GIven function is
$f (x) = x^n$
the function $f (x) = x^n$ is defined for all positive integer, n
$f(n) = n^n\\ \lim_{x\rightarrow n}f(x) = n^n\\ \lim_{x\rightarrow n}f(x) = f(n)$
Hence, the function $f (x) = x^n$ is continuous at x = n, where n is a positive integer

Question:5. Is the function f defined by
$f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.$
continuous at x = 0? At x = 1? At x = 2?

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.$
function is defined at x = 0 and its value is 0
$f(0) = 0\\ \lim_{x\rightarrow 0}f(x) = f(x) = 0\\ \lim_{x\rightarrow 0}f(x) = f(0)$
Hence , given function is continous at x = 0

given function is defined for x = 1
Now, for x = 1 Right-hand limit and left-hand limit are not equal
$f(1) = 1\\ \lim_{x\rightarrow 1^-}f(x) = f(x) = 1\\ \lim_{x\rightarrow 1^+}f(x) =f(5) = 5$
R.H.L $\neq$ L.H.L.
Therefore, given function is not continous at x =1
Given function is defined for x = 2 and its value at x = 2 is 5
$f(2) = 2\\ \lim_{x\rightarrow 2}f(x) = f(5) = 5\\\lim_{x\rightarrow 2}f(x) = f(2)$
Hence, given function is continous at x = 2

Question:6. Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 2
$f(k) = 2k-3\\ \lim_{x\rightarrow k}f(x) = 2k-3\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 2

case(ii) k < 2
$f(k) = 2k +3\\ \lim_{x\rightarrow k}f(x) = 2k+3\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k < 2

case(iii) x = 2

$\lim_{x\rightarrow 2^-}f(x) = 2x+3 = 2\times 2 + 3 = 4 + 3 = 7\\ \lim_{x\rightarrow 2^+}f(x) = 2x-3 = 2\times 2-3 = 4-3 = 1$
Right hand limit at x= 2 $\neq$ Left hand limit at x = 2
Therefore, x = 2 is the point of discontinuity

Question:7. Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.$
GIven function is defined for every real number k
Different cases are their
case (i) k < -3
$f(k) = -k + 3\\ \lim_{x\rightarrow k}f(x) = -k + 3\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for every value of k < -3

case(ii) k = -3
$f(-3) = -(-3) + 3 = 6\\ \lim_{x\rightarrow -3^-}f(x) = -k + 3=-(-3)+3 = 6\\ \lim_{x\rightarrow -3^+}f(x) = -2x = -2(-3) = 6\\ R.H.L. = L.H.L. = f(-3)$
Hence, given function is continous for x = -3

case(iii) -3 < k < 3
$f(k) = -2k \\ \lim_{x\rightarrow k}f(x) = -2k\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in -3 < k < 3 given function is continous

case(iv) k = 3
$f(3) = 6x+2 = 6\times3+2 =18+2=20\\ \lim_{x\rightarrow 3^-}f(x) = -2x = -2(3) = -6\\ \lim_{x\rightarrow 3^+}f(x) = 6x+2 = 6\times3+2 = 20\\ R.H.L. = f(3) \neq L.H.L.$
Hence. x = 3 is the point of discontinuity

case(v) k > 3
$f(k) = 6k+2 \\ \lim_{x\rightarrow k}f(x) = 6k+2 \\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each and every value of k > 3

Question:8. Find all points of discontinuity of f, where f is defined by

$f (x )= \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.$

Answer:

Given function is
$f (x ) \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.$
if x > 0 , $f(x)=\frac{x}{x} = 1$
if x < 0 , $f(x)=\frac{-(x)}{x} = -1$
given function is defined for every real number k
Now,
case(i) k < 0
$f(k) = -1\\ \lim_{x\rightarrow k }f(x) = -1\\ \lim_{x\rightarrow k }f(x) = f(k)$
Hence, given function is continuous for every value of k < 0
case(ii) k > 0
$f(k) = 1\\ \lim_{x\rightarrow k }f(x) = 1\\ \lim_{x\rightarrow k }f(x) = f(k)$
Hence, given function is continuous for every value of k > 0
case(iii) x = 0
$f(0) = 0\\ \lim_{x\rightarrow 0^- }f(x) = -1\\ \lim_{x\rightarrow 0^+}f(x) = 1\\ f(0) \neq R.H.L. \neq L.H.L.$
Hence, 0 is the only point of discontinuity

Question:9. Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.$
if x < 0 , $f (x) =\frac{x }{|x|} = \frac{x}{-(x)} = -1$
Now, for any value of x, the value of our function is -1
Therefore, the given function is continuous for each and every value of x
Hence, no point of discontinuity

Question:10. Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right.$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right.$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
$f(k) = k+1\\ \lim_{x\rightarrow k}f(x) = k+1\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
$f(k) = k^2 ++1\\ \lim_{x\rightarrow k}f(x) = k^2+1\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

$\lim_{x\rightarrow 1^-}f(x) = x^2+1 = 1^2 + 1 = 1 + 1 = 2\\ \lim_{x\rightarrow 1^+}f(x) = x+1 = 1+1 = 2\\ f(1) = 1^2+1 = 2 \\ R.H.L. = L.H.L. = f(1)$

Hence, at x = 2 given function is continuous
Therefore, no point of discontinuity

Question:11. Find all points of discontinuity of f, where f is defined by

$f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.$

Answer:

Given function is
$f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 2
$f(k) = k^2+1\\ \lim_{x\rightarrow k}f(x) = k^2+1\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 2

case(ii) k < 2
$f(k) = k^3 -3\\ \lim_{x\rightarrow k}f(x) = k^3-3\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k < 2

case(iii) x = 2

$\lim_{x\rightarrow 2^-}f(x) = x^3-3 = 2^3- 3 = 8- 3 = 5\\ \lim_{x\rightarrow 2^+}f(x) = x^2+1= 2^2+1 = 4+1 = 5\\ f(2) = 2^3-3 = 8 - 3 = 5\\ f(2)=R.H.L.=L.H.L.$
Hence, given function is continuous at x = 2
There, no point of discontinuity

Question:12. Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
$f(k) = k^2\\ \lim_{x\rightarrow k}f(x) = k^2\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
$f(k) = k^{10} -1\\ \lim_{x\rightarrow k}f(x) = k^{10}-1\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

$\lim_{x\rightarrow 1^-}f(x) = x^{10}-1 = 1^{10} - 1 = 1 - 1 = 0\\ \lim_{x\rightarrow 1^+}f(x) = x^2 = 1^2 = 1\\ f(1) = x^{10}-1 = 0\ f(1) = L.H.L. \neq R.H.L.$

Hence, x = 1 is the point of discontinuity

Question:13. Is the function defined by

$f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.$

a continuous function?

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
$f(k) = k-5\\ \lim_{x\rightarrow k}f(x) = k-5\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
$f(k) = k+5\\ \lim_{x\rightarrow k}f(x) = k+5\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

$\lim_{x\rightarrow 1^-}f(x) = x+5 = 1 + 5 = 1 + 5 = 6\\ \lim_{x\rightarrow 1^+}f(x) = x-5 = 1-5 = -4\\ f(1) = x+5 =1+5= 6 \\ L.H.L. = f(1) \neq R.H.S.$

Hence, x = 1 is the point of discontinuity

Question:14. Discuss the continuity of the function f, where f is defined by

$f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.$

Answer:

Given function is
$f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.$
GIven function is defined for every real number k
Different cases are their
case (i) k < 1
$f(k) = 3\\ \lim_{x\rightarrow k}f(x) = 3\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continous for every value of k < 1

case(ii) k = 1
$f(1) = 3 \\ \lim_{x\rightarrow 1^-}f(x) = 3\\ \lim_{x\rightarrow 1^+}f(x) = 4\\ R.H.L. \neq L.H.L. = f(1)$
Hence, given function is discontinous at x = 1
Therefore, x = 1 is he point od discontinuity

case(iii) 1 < k < 3
$f(k) = 4 \\ \lim_{x\rightarrow k}f(x) = 4\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in 1 < k < 3 given function is continous

case(iv) k = 3
$f(3) =5\\ \lim_{x\rightarrow 3^-}f(x) = 4\\ \lim_{x\rightarrow 3^+}f(x) =5\\ R.H.L. = f(3) \neq L.H.L.$
Hence. x = 3 is the point of discontinuity

case(v) k > 3
$f(k) = 5 \\ \lim_{x\rightarrow k}f(x) = 5 \\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continous for each and every value of k > 3
case(vi) when k < 3

$f(k) = 4 \\ \lim_{x\rightarrow k}f(x) = 4\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in k < 3 given function is continous

Question:15 Discuss the continuity of the function f, where f is defined by $f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$

Answer:

Given function is
$f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$
Given function is satisfies for the all real values of x
case (i) k < 0
$f(k) = 2k$
$\lim_{k \rightarrow 0^-}f(x)= 2k = f(k)$
Hence, function is continuous for all values of x < 0

case (ii) x = 0
$f(0 )= 0$
L.H.L at x= 0
$\lim_{x\rightarrow 0^-}f(x)= 2(0)= 0$
R.H.L. at x = 0
$\lim_{x\rightarrow 0^+}f(x)= 0$
L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0

case (iii) k > 0
$f(k)=0$
$\lim_{k\rightarrow 0^+}f(x)= 0= f(k)$
Hence , function is continuous for all values of x > 0

case (iv) k < 1
$f(k) = 0$
$\lim_{x\rightarrow 1^-}f(x)= 0 = f(k)$
Hence , function is continuous for all values of x < 1

case (v) k > 1
$f(k) = 4k$
$\lim_{x\rightarrow 1^+}f(x)= 4k = f(k)$
Hence , function is continuous for all values of x > 1

case (vi) x = 1
$f(1)= 0$
$\lim_{x\rightarrow 1^-}f(1)= 0$
$\lim_{x\rightarrow 1^+}f(1)= 4(1) = 4$
Hence, function is not continuous at x = 1

Question:16. Discuss the continuity of the function f, where f is defined by

$f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.$

Answer:

Given function is
$f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.$
GIven function is defined for every real number k
Different cases are their
case (i) k < -1
$f(k) = -2\\ \lim_{x\rightarrow k}f(x) = -2\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for every value of k < -1

case(ii) k = -1
$f(-1) = -2 \\ \lim_{x\rightarrow -1^-}f(x) = -2\\ \lim_{x\rightarrow -1^+}f(x) = 2x = 2(-1) = -2\\ R.H.L. =L.H.L. = f(-1)$
Hence, given function is continous at x = -1

case(iii) k > -1
$f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continous for all values of x > -1

case(vi) -1 < k < 1
$f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in -1 < k < 1 given function is continous

case(v) k = 1
$f(1) =2x = 2(1)=2\\ \lim_{x\rightarrow 1^-}f(x) = 2x=2(1)=2\\ \lim_{x\rightarrow 1^+}f(x) =2\\ R.H.L. = f(1) = L.H.L.$
Hence.at x =1 function is continous

case(vi) k > 1
$f(k) = 2 \\ \lim_{x\rightarrow k}f(x) = 2 \\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continous for each and every value of k > 1
case(vii) when k < 1

$f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in k < 1 given function is continuous

Therefore, continuous at all points

Question:17. Find the relationship between a and b so that the function f defined by
$f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.$
is continuous at x = 3.

Answer:

Given function is
$f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.$
For the function to be continuous at x = 3 , R.H.L. must be equal to L.H.L.
$\lim_{x\rightarrow 3^-}f(x)= ax + 1 = 3a+1\\ \lim_{x\rightarrow 3^+}f(x) = bx+3=3b+3$
For the function to be continuous
$\lim_{x\rightarrow 3^-}f(x) = \lim_{x\rightarrow 3^+}f(x) \\ 3a + 1= 3b+3\\ 3(a-b)=2\\ a-b = \frac{2}{3}\\ a = b+\frac{2}{3}$

Question:18. For what value of l is the function defined by
$f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.$
continuous at x = 0? What about continuity at x = 1?

Answer:

Given function is
$f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.$
For the function to be continuous at x = 0 , R.H.L. must be equal to L.H.L.
$\lim_{x\rightarrow 0^-}f(x) = \lambda (x^2-2x) = 0\\ \lim_{x\rightarrow 0^+}f(x) = 4x+1=1$
For the function to be continuous
$\lim_{x\rightarrow 0^-}f(x) = \lim_{x\rightarrow 0^+}f(x) \\ 0\neq 1$
Hence, for no value of function is continuous at x = 0

For x = 1
$f(1)=4x+1=4(1)+1=5\\ \lim_{x\rightarrow 1}f(x) =4+1=5 \\\ \lim_{x\rightarrow 1}f(x) = f(x)$
Hence, given function is continuous at x =1

Question:19. Show that the function defined by $g (x) = x- [x]$ is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

Answer:

Given function is
$g (x) = x- [x]$
Given is defined for all real numbers k
$\lim_{x\rightarrow k^-}f(x) = k - (k-1) = k-k+1 =1\\ \lim_{x\rightarrow k^+}f(x) = k - k = 0\\ \lim_{x\rightarrow k^-}f(x) \neq \lim_{x\rightarrow k^+}f(x)$
Hence, by this, we can say that the function defined by $g (x) = x- [x]$ is discontinuous at all integral points

Question:20. Is the function defined by $f (x) = x^2 - sin x + 5$ continuous at x = $\pi$ ?

Answer:

Given function is
$f (x) = x^2 - sin x + 5$
Clearly, Given function is defined at x = $\pi$
$f(\pi) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim_{x\rightarrow \pi}f(x) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim_{x\rightarrow \pi}f(x) = f(\pi)$
Hence, the function defined by $f (x) = x^2 - sin x + 5$ continuous at x = $\pi$

Question:21. Discuss the continuity of the following functions:
a) $f (x) = \sin x + \cos x$

Answer:

Given function is
$f (x) = \sin x + \cos x$
Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) + h(x) = sin x + cos x is also a continuous function

Question:21. b) Discuss the continuity of the following functions:
$f (x) = \sin x - \cos x$

Answer:

Given function is
$f (x) = \sin x - \cos x$
Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) - h(x) = sin x - cos x is also a continuous function

Question:21 c) Discuss the continuity of the following functions:
$f (x) = \sin x \cdot \cos x$

Answer:

Given function is
$f (x) = \sin x . \cos x$
Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x).h(x) = sin x .cos x is also a continuous function

Question:22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Answer:

We, know that if two function g(x) and h(x) are continuous then
$\frac{g(x)}{h(x)} , h(x) \neq0\ is \ continuous\\ \frac{1}{h(x)} , h(x) \neq 0\ is \ continuous\\ \frac{1}{g(x)} , g(x) \neq0\ is \ continuous\\$
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, the function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x is a continous function
So, we can say that
cosec x = $\frac{1}{\sin x} = \frac{1}{g(x)}$ is also continuous except at $x=n\pi$
sec x = $\frac{1}{\cos x} = \frac{1}{h(x)}$ is also continuous except at $x=\frac{(2n+1) \pi}{2}$
cot x = $\frac{\cos x}{\sin x} = \frac{h(x)}{g(x)}$ is also continuous except at $x=n\pi$

Question:23. Find all points of discontinuity of f, where

$f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.$

Answer:

Given function is
$f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.$
$\lim_{x\rightarrow 0^-}f(x) = \lim_{x\rightarrow 0}\frac{\sin x}{x} = 1\\ \lim_{x\rightarrow 0^+}f(x) = x + 1 = 1\\ \lim_{x\rightarrow 0^-}f(x) = \lim_{x\rightarrow 0^+}f(x)$
Hence, the function is continuous
Therefore, no point of discontinuity

Question:24. Determine if f defined by
$f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.$
is a continuous function?

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.$
Given function is defined for all real numbers k
when x = 0
$f(0) = 0\\ \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\left ( x^2\sin\frac{1}{x} \right )=\lim_{x\rightarrow 0}\left ( \frac{x.\sin\frac{1}{x}}{\frac{1}{x}} \right ) = 0(1)=0 \ \ \ \ \ \ (\because\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1)$
$\lim_{x\rightarrow 0}f(x) = f(0)$
Hence, function is continuous at x = 0
when $x \neq 0$
$f(k) = k^2\sin \frac{1}{k}\\ \lim_{x\rightarrow k}f(x)=\lim_{x\rightarrow k}\left ( x^2\sin\frac{1}{x} \right )=k^2\sin \frac{1}{k}\\ \lim_{x\rightarrow k} = f(k)$
Hence, the given function is continuous for all points

Question:25. Examine the continuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} \sin x - \cos x & if x \neq 0 \\ -1 & if x = 0 \end{matrix}\right.$

Answer:

Question:26. Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} \frac{k \cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right. \: \: \: at \: \: x = \pi /2$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} \frac{k \cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right.$
When $x = \frac{\pi}{2}$
$f(\frac{\pi}{2}) = 3\\let\ x=\pi +h\\ \lim_{x\rightarrow \frac{\pi}{2}}f(x)= \lim_{h\rightarrow 0}\frac{k\cos\left ( \frac{\pi}{2}+h \right )}{\pi-2\left ( \frac{\pi}{2}+h \right )} = k. \lim_{h\rightarrow 0}\frac{-\sin h}{-2h} = \frac{k}{2}\\$
For the function to be continuous
$\lim_{x\rightarrow \frac{\pi}{2}}f(x)= f(\frac{\pi}{2})\\ \frac{k}{2} = 3\\ k = 6$
Therefore, the values of k so that the function f is continuous is 6

Question:27. Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right. \: \: at \: \: x = 2$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right.$
When x = 2
For the function to be continuous
f(2) = R.H.L. = LH.L.
$f(2) = 4k\\ \lim_{x\rightarrow 2^-}f(x)= 4k\\ \lim_{x\rightarrow 2^+}f(x) = 3\\ f(2) = \lim_{x\rightarrow 2^-}f(x) = \lim_{x\rightarrow 2^+}f(x)\\ 4k = 3\\ k = \frac{3}{4}$
Hence, the values of k so that the function f is continuous at x= 2 is $\frac{3}{4}$

Question:28. Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right. \: \: at \: \: x = \pi$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right.$
When x = $\pi$
For the function to be continuous
f( $\pi$ ) = R.H.L. = LH.L.
$f(\pi) = k\pi+1\\ \lim_{x\rightarrow \pi^-}f(x)= k\pi+1\\ \lim_{x\rightarrow \pi^+}f(x) = \cos \pi = -1\\ f(\pi) = \lim_{x\rightarrow \pi^-}f(x) = \lim_{x\rightarrow \pi^+}f(x)\\ k\pi+1 = -1\\ k = \frac{-2}{\pi}$
Hence, the values of k so that the function f is continuous at x= $\pi$ is $\frac{-2}{\pi}$

Question:29 Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right. \: \: at x = 5$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right.$
When x = 5
For the function to be continuous
f(5) = R.H.L. = LH.L.
$f(5) = 5k+1\\ \lim_{x\rightarrow 5^-}f(x)= 5k+1\\ \lim_{x\rightarrow 5^+}f(x) = 3(5)-5 = 15-5=10\\ f(5) = \lim_{x\rightarrow 5^-}f(x) = \lim_{x\rightarrow 5^+}f(x)\\ 5k+1 = 10\\ k = \frac{9}{5}$
Hence, the values of k so that the function f is continuous at x= 5 is $\frac{9}{5}$

Question:30 Find the values of a and b such that the function defined by
$f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.$
is a continuous function.

Answer:

Given continuous function is
$f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.$
The function is continuous so
$\lim_{x\rightarrow 2^-}f(x) = \lim_{x\rightarrow 2^+}f(x)\\ and\\ \lim_{x\rightarrow 10^-}f(x)=\lim_{x\rightarrow 10^+}f(x)$
$\lim_{x\rightarrow 2^-}f(x) = 5\\ \lim_{x\rightarrow 2^+}f(x)=ax+b=2a+b\\ 2a+b = 5 \ \ \ \ \ \ \ \ \ \ \ -(i)\\ and\\ \lim_{x\rightarrow 10^-}f(x)=ax+b=10a+b\\ \lim_{x\rightarrow 10^+}f(x)=21\\ 10a+b=21 \ \ \ \ \ \ \ \ -(ii)$
By solving equation (i) and (ii)
a = 2 and b = 1
Hence, values of a and b such that the function defined by $f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.$ is a continuous function is 2 and 1 respectively

Question:31. Show that the function defined by $f (x) = \cos (x^2 )$ is a continuous function.

Answer:

Given function is
$f (x) = \cos (x^2 )$
given function is defined for all real values of x
Let x = k + h
if $x\rightarrow k , \ then \ h \rightarrow 0$
$f(k) = \cos k^2\\ \lim_{x \rightarrow k}f(x) = \lim_{x \rightarrow k}\cos x^2 = \lim_{h \rightarrow 0}\cos (k+h)^2 = \cos k^2\\ \lim_{x \rightarrow k}f(x) = f(k)$
Hence, the function $f (x) = \cos (x^2 )$ is a continuous function

Question:32. Show that the function defined by $f (x) = |\cos x |$ is a continuous function.

Answer:

Given function is
$f (x) = |\cos x |$
given function is defined for all values of x
f = g o h , g(x) = |x| and h(x) = cos x
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < 0
$g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

case (ii) k > 0
$g(k) = k\\ \lim_{x\rightarrow k}g(x) = k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) k = 0
$g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, f(x) = g o h is also continuous

Question:33. Examine that sin | x| is a continuous function.

Answer:

Given function is
f(x) = sin |x|
f(x) = h o g , h(x) = sin x and g(x) = |x|
Now,

$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < 0
$g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

case (ii) k > 0
$g(k) = k\\ \lim_{x\rightarrow k}g(x) = k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) k = 0
$g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) = sin x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \sin c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \sin x$ is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, f(x) = h o g is also continuous

Question:34. Find all the points of discontinuity of f defined by $f (x) = | x| - | x + 1|.$

Answer:

Given function is
$f (x) = | x| - | x + 1|$
Let g(x) = |x| and h(x) = |x+1|
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < 0
$g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

case (ii) k > 0
$g(k) = k\\ \lim_{x\rightarrow k}g(x) = k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) k = 0
$g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x

Now,
$h(x)\begin{cases} -(x+1) & \text{ if } x<-1 \\ 0 & \text{ if } x= -1\\ (x+1)& \text{ if } x>-1 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < -1
$h(k) = -(k+1)\\ \lim_{x\rightarrow k}h(x) = -(k+1)\\ \lim_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k < -1

case (ii) k > -1
$h(k) = k+1\\ \lim_{x\rightarrow k}h(x) = k+1\\ \lim_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k > -1

case (iii) k = -1
$h(-1) = 0\\ \lim_{x\rightarrow -1^-}h(x) = -(x-1) = 0\\ \lim_{x\rightarrow -1^+}h(x ) = x+1 = 0\\ \lim_{x\rightarrow -1^-}h(x) = h(0) = \lim_{x\rightarrow -1^+}h(x )$
Hence, h(x) is continuous when k = -1
Therefore, h(x) = |x+1| is continuous for all real values of x
g(x) is continuous and h(x) is continuous
Therefore, f(x) = g(x) - h(x) = |x| - |x+1| is also continuous

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1:-

Class 12 Maths chapter 5 exercise 5.1 solutions are described in a detailed manner, so you will get the concept of continuity easily.
In Class 12 Maths chapter 5 exercise 5.1 solutions, you will get different ways to approach the problem.
There are some properties of continuous functions which make it easy to check the continuity of the given functions.
In Class 12 Maths chapter 5 exercise 5.1 solutions, you will learn about the algebra of continuous functions.

Also see-

NCERT Solutions for Class 12 Maths Chapter 5
NCERT Exemplar Solutions Class 12 Maths Chapter 5

NCERT Solutions of Class 12 Subject Wise

NCERT Solutions for Class 12 Maths
NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Chemistry
NCERT Solutions for Class 12 Biology

Subject Wise NCERT Exampler Solutions

NCERT Exemplar Solutions for Class 12th Maths
NCERT Exemplar Solutions for Class 12th Physics
NCERT Exemplar Solutions for Class 12th Chemistry
NCERT Exemplar Solutions for Class 12th Biology

Happy learning!!!

Use These Tricks To Solve JEE Main Maths MCQs In Seconds 3 min read Mar 05, 2022 Read More Can You Predict Whether A Cricket Ball Will Hit The Stumps With Class 11 Maths? Learn How 3 min read Mar 05, 2022 Read More

NCERT Solutions for Exercise 5.1 Class 12 Maths Chapter 5 - Continuity and Differentiability

Continuity and Differentiability Exercise: 5.1

More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1:-

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1:-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions