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

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
$f(3) = 2(3)^2-1 = 2\times 9 - 1=18-1=17\\ \lim\limits_{x\rightarrow 3}f(x) = 2(3)^2-1=2\times 9-1=18-1=17$
$\lim\limits_{x\rightarrow 3}f(x) = f(3)$
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,
$\lim\limits_{x\rightarrow k} f(x) = k -5\\ \lim\limits_{x\rightarrow k} f(x) = f(k)$
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,
$f(k) = \frac{1}{k-5}\\ \lim\limits_{x\rightarrow k}f(x ) = \frac{1}{k-5}\\ \lim\limits_{x\rightarrow k}f(x ) = f(k)$
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,
$f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim\limits_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim\limits_{x\rightarrow k}f(x ) = f(k)$
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
$f(k) = k - 5\\ \lim\limits_{x\rightarrow k }f(x) = k -5\\ \lim\limits_{x\rightarrow k }f(x) = f(k)$
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
$f(k) = 5-k\\ \lim\limits_{x\rightarrow k }f(x) = 5 -k\\ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, function f(x) = 5 - x is continous for x < 5

case(iii) x = 5
for x = 5 , f(x) = x - 5 is defined
$f(5) = 5 - 5=0\\ \lim\limits_{x\rightarrow 5 }f(x) = 5 -5=0\\ \lim\limits_{x\rightarrow 5 }f(x) = f(5)$
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
$f(n) = n^n\\ \lim\limits_{x\rightarrow n}f(x) = n^n\\ \lim\limits_{x\rightarrow n}f(x) = f(n)$
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
$f(0) = 0\\ \lim\limits_{x\rightarrow 0}f(x) = f(x) = 0\\ \lim\limits_{x\rightarrow 0}f(x) = f(0)$
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

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

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
$f(2) = 2\\ \lim\limits_{x\rightarrow 2}f(x) = f(5) = 5\\\lim\limits_{x\rightarrow 2}f(x) = f(2)$
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
$f(k) = 2k-3\\ \lim\limits_{x\rightarrow k}f(x) = 2k-3\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 2

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

case(iii) x = 2

$\lim\limits_{x\rightarrow 2^-}f(x) = 2x+3 = 2\times 2 + 3 = 4 + 3 = 7\\ \lim\limits_{x\rightarrow 2^+}f(x) = 2x-3 = 2\times 2-3 = 4-3 = 1$
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
$f(k) = -k + 3\\ \lim\limits_{x\rightarrow k}f(x) = -k + 3\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for every value of k < -3

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

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

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

case(v) k > 3
$f(k) = 6k+2 \\ \lim\limits_{x\rightarrow k}f(x) = 6k+2 \\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
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
$f(k) = -1\\ \lim\limits_{x\rightarrow k }f(x) = -1\\ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, given function is continuous for every value of k < 0
case(ii) k > 0
$f(k) = 1\\ \lim\limits_{x\rightarrow k }f(x) = 1\\ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, given function is continuous for every value of k > 0
case(iii) x = 0
$f(0) = 0\\ \lim\limits_{x\rightarrow 0^- }f(x) = -1\\ \lim\limits_{x\rightarrow 0^+}f(x) = 1\\ f(0) \neq R.H.L. \neq L.H.L.$
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
$f(k) = k+1\\ \lim\limits_{x\rightarrow k}f(x) = k+1\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 1

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

case(iii) x = 1

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

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
$f(k) = k^2+1\\ \lim\limits_{x\rightarrow k}f(x) = k^2+1\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 2

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

case(iii) x = 2

$\lim\limits_{x \rightarrow 2^-} f(x) = x^3 - 3 = 2^3 - 3 = 8 - 3 = 5$
$\lim\limits_{x \rightarrow 2^+} f(x) = x^2 + 1 = 2^2 + 1 = 4 + 1 = 5$
$f(2) = 2^3 - 3 = 8 - 3 = 5$
$f(2) = R.H.L. = L.H.L.$
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
$f(k) = k^2\\ \lim\limits_{x\rightarrow k}f(x) = k^2\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 1

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

case(iii) x = 1

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

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
$f(k) = k-5\\ \lim\limits_{x\rightarrow k}f(x) = k-5\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 1

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

case(iii) x = 1
$\lim\limits_{x \rightarrow 1^-} f(x) = x + 5 = 1 + 5 = 6$

$\lim\limits_{x \rightarrow 1^+} f(x) = x - 5 = 1 - 5 = -4$

$f(1) = x + 5 = 1 + 5 = 6$

$L.H.L. = f(1) \neq R.H.L.$

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
$f(k) = 3\\ \lim\limits_{x\rightarrow k}f(x) = 3\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continous for every value of k < 1

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

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

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

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

$f(k) = 4 \\ \lim\limits_{x\rightarrow k}f(x) = 4\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
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
$f(x) = \begin{cases}
2x & \text{if } x < 0 \\
0 & \text{if } 0 \leq x \leq 1 \\
4x & \text{if } x > 1
\end{cases}$
Given function is satisfies for the all real values of x
case (i) k < 0
$f(k) = 2k$
$\lim\limits_{x \to 0^-} f(x) = 2k = f(k)$
Hence, function is continuous for all values of x < 0

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

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

case (iv) k < 1

$f(k) = 0$

$\lim\limits_{x \to 1^-} f(x) = 0 = f(k)$

Hence , function is continuous for all values of x < 1

case (v) k > 1

$f(k) = 4k$

$\lim\limits_{x \to 1^+} f(x) = 4k = f(k)$
Hence , function is continuous for all values of x > 1

case (vi) x = 1

$f(1) = 0$

$\lim\limits_{x \to 1^-} f(1) = 0$

$\lim\limits_{x \to 1^+} f(1) = 4(1) = 4$

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

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

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

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

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

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

$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
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.
$\lim\limits_{x\rightarrow 3^-}f(x)= ax + 1 = 3a+1\\ \lim\limits_{x\rightarrow 3^+}f(x) = bx+3=3b+3$
For the function to be continuous
$\lim\limits_{x\rightarrow 3^-}f(x) = \lim\limits_{x\rightarrow 3^+}f(x) \\ 3a + 1= 3b+3\\ 3(a-b)=2\\ a-b = \frac{2}{3}\\ a = b+\frac{2}{3}$

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.
$\lim\limits_{x\rightarrow 0^-}f(x) = \lambda (x^2-2x) = 0\\ \lim\limits_{x\rightarrow 0^+}f(x) = 4x+1=1$
For the function to be continuous
$\lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0^+}f(x) \\ 0\neq 1$
Hence, for no value of function is continuous at x = 0

For x = 1
$f(1)=4x+1=4(1)+1=5\\ \lim\limits_{x\rightarrow 1}f(x) =4+1=5 \\\ \lim\limits_{x\rightarrow 1}f(x) = f(x)$
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

$\lim\limits_{x \rightarrow k^-} f(x) = k - (k - 1) = k - k + 1 = 1$

$\lim\limits_{x \rightarrow k^+} f(x) = k - k = 0$

$\lim\limits_{x \rightarrow k^-} f(x) \neq \lim\limits_{x \rightarrow k^+} f(x)$

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 =$\pi$
$f(\pi) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim\limits_{x\rightarrow \pi}f(x) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim\limits_{x\rightarrow \pi}f(x) = f(\pi)$
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 $x \rightarrow c , \ then \ h \rightarrow 0$

$g(c) = \sin c$
$\lim\limits_{x \rightarrow c} g(x) = \lim\limits_{x \rightarrow c} \sin x = \lim\limits_{h \rightarrow 0} \sin(c + h)$

We know that
$\sin(a + b) = \sin a \cos b + \cos a \sin b$

$\lim\limits_{h \rightarrow 0} \sin(c + h) = \lim\limits_{h \rightarrow 0} (\sin c \cos h + \cos c \sin h)$
$= \sin c \cdot \lim\limits_{h \rightarrow 0} \cos h + \cos c \cdot \lim\limits_{h \rightarrow 0} \sin h$
$= \sin c \cdot \cos 0 + \cos c \cdot \sin 0$
$= \sin c$

$\lim\limits_{x \rightarrow c} g(x) = g(c)$

Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$

$h(c) = \cos c$
$\lim\limits_{x \rightarrow c} h(x) = \lim\limits_{x \rightarrow c} \cos x = \lim\limits_{h \rightarrow 0} \cos(c + h)$

We know that
$\cos(a + b) = \cos a \cos b - \sin a \sin b$

$\lim\limits_{h \rightarrow 0} \cos(c + h) = \lim\limits_{h \rightarrow 0} (\cos c \cos h - \sin c \sin h)$
$= \cos c \cdot \lim\limits_{h \rightarrow 0} \cos h - \sin c \cdot \lim\limits_{h \rightarrow 0} \sin h$
$= \cos c \cdot \cos 0 - \sin c \cdot \sin 0$
$= \cos c$

$\lim\limits_{x \rightarrow c} h(x) = h(c)$

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 $x \rightarrow c , \ then \ h \rightarrow 0$

$g(c) = \sin c$
$\lim\limits_{x \rightarrow c} g(x) = \lim\limits_{x \rightarrow c} \sin x = \lim\limits_{h \rightarrow 0} \sin(c + h)$

We know that
$\sin(a + b) = \sin a \cos b + \cos a \sin b$

$\lim\limits_{h \rightarrow 0} \sin(c + h) = \lim\limits_{h \rightarrow 0} (\sin c \cos h + \cos c \sin h)$
$= \sin c \cdot \lim\limits_{h \rightarrow 0} \cos h + \cos c \cdot \lim\limits_{h \rightarrow 0} \sin h$
$= \sin c \cdot \cos 0 + \cos c \cdot \sin 0$
$= \sin c$

$\lim\limits_{x \rightarrow c} g(x) = g(c)$

Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$

$h(c) = \cos c$
$\lim\limits_{x \rightarrow c} h(x) = \lim\limits_{x \rightarrow c} \cos x = \lim\limits_{h \rightarrow 0} \cos(c + h)$

We know that
$\cos(a + b) = \cos a \cos b - \sin a \sin b$

$\lim\limits_{h \rightarrow 0} \cos(c + h) = \lim\limits_{h \rightarrow 0} (\cos c \cos h - \sin c \sin h)$
$= \cos c \cdot \lim\limits_{h \rightarrow 0} \cos h - \sin c \cdot \lim\limits_{h \rightarrow 0} \sin h$
$= \cos c \cdot \cos 0 - \sin c \cdot \sin 0$
$= \cos c$

$\lim\limits_{x \rightarrow c} h(x) = h(c)$

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 $x \rightarrow c , \ then \ h \rightarrow 0$

$g(c) = \sin c$

$\lim\limits_{x \rightarrow c} g(x) = \lim\limits_{x \rightarrow c} \sin x = \lim\limits_{h \rightarrow 0} \sin(c + h)$

We know that

$\sin(a + b) = \sin a \cos b + \cos a \sin b$

$\lim\limits_{h \rightarrow 0} \sin(c + h) = \lim\limits_{h \rightarrow 0} (\sin c \cos h + \cos c \sin h)$

$= \sin c \cdot \lim\limits_{h \rightarrow 0} \cos h + \cos c \cdot \lim\limits_{h \rightarrow 0} \sin h$

$= \sin c \cdot \cos 0 + \cos c \cdot \sin 0$

$= \sin c$

$\lim\limits_{x \rightarrow c} g(x) = g(c)$

Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$

$h(c) = \cos c$

$\lim\limits_{x \rightarrow c} h(x) = \lim\limits_{x \rightarrow c} \cos x = \lim\limits_{h \rightarrow 0} \cos(c + h)$

We know that

$\cos(a + b) = \cos a \cos b - \sin a \sin b$

$\lim\limits_{h \rightarrow 0} \cos(c + h) = \lim\limits_{h \rightarrow 0} (\cos c \cos h - \sin c \sin h)$

$= \cos c \cdot \lim\limits_{h \rightarrow 0} \cos h - \sin c \cdot \lim\limits_{h \rightarrow 0} \sin h$

$= \cos c \cdot \cos 0 - \sin c \cdot \sin 0$

$= \cos c$

$\lim\limits_{x \rightarrow c} h(x) = h(c)$

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
$\frac{g(x)}{h(x)}$, $h(x) \neq 0$ is continuous
$\frac{1}{h(x)}$, $h(x) \neq 0$ is continuous
$\frac{1}{g(x)}$, $g(x) \neq 0$ is continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$

$g(c) = \sin c$

$\lim\limits_{x \rightarrow c} g(x) = \lim\limits_{x \rightarrow c} \sin x = \lim\limits_{h \rightarrow 0} \sin(c + h)$

We know that

$\sin(a + b) = \sin a \cos b + \cos a \sin b$

$\lim\limits_{h \rightarrow 0} \sin(c + h) = \lim\limits_{h \rightarrow 0} (\sin c \cos h + \cos c \sin h)$

Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$

$h(c) = \cos c$

$\lim\limits_{x \rightarrow c} h(x) = \lim\limits_{x \rightarrow c} \cos x = \lim\limits_{h \rightarrow 0} \cos(c + h)$

We know that

$\cos(a + b) = \cos a \cos b - \sin a \sin b$

$\lim\limits_{h \rightarrow 0} \cos(c + h) = \lim\limits_{h \rightarrow 0} (\cos c \cos h - \sin c \sin h)$

$= \cos c \cdot \lim\limits_{h \rightarrow 0} \cos h - \sin c \cdot \lim\limits_{h \rightarrow 0} \sin h$

$= \cos c \cdot \cos 0 - \sin c \cdot \sin 0$

$= \cos c$

$\lim\limits_{x \rightarrow c} h(x) = h(c)$

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
$f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.$
$\lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0}\frac{\sin x}{x} = 1\\ \lim\limits_{x\rightarrow 0^+}f(x) = x + 1 = 1\\ \lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0^+}f(x)$
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
$f(0) = 0\\ \lim\limits_{x\rightarrow 0}f(x)=\lim\limits_{x\rightarrow 0}\left ( x^2\sin\frac{1}{x} \right )=\lim\limits_{x\rightarrow 0}\left ( \frac{x.\sin\frac{1}{x}}{\frac{1}{x}} \right ) = 0(1)=0 \ \ \ \ \ \ (\because\lim\limits_{x\rightarrow 0}\frac{\sin x}{x} = 1)$
$\lim\limits_{x\rightarrow 0}f(x) = f(0)$
Hence, function is continuous at x = 0
when $x \neq 0$
$f(k) = k^2\sin \frac{1}{k}\\ \lim\limits_{x\rightarrow k}f(x)=\lim\limits_{x\rightarrow k}\left ( x^2\sin\frac{1}{x} \right )=k^2\sin \frac{1}{k}\\ \lim\limits_{x\rightarrow k} = f(k)$
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:

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 $x \rightarrow c , \ then \ h \rightarrow 0$

$g(c) = \sin c$

$\lim\limits_{x \rightarrow c} g(x) = \lim\limits_{x \rightarrow c} \sin x = \lim\limits_{h \rightarrow 0} \sin(c + h)$

We know that

$\sin(a + b) = \sin a \cos b + \cos a \sin b$

$\lim\limits_{h \rightarrow 0} \sin(c + h) = \lim\limits_{h \rightarrow 0} (\sin c \cos h + \cos c \sin h)$

$= \sin c \cdot \lim\limits_{h \rightarrow 0} \cos h + \cos c \cdot \lim\limits_{h \rightarrow 0} \sin h$

$= \sin c \cdot \cos 0 + \cos c \cdot \sin 0$

$= \sin c$

$\lim\limits_{x \rightarrow c} g(x) = g(c)$

Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$

$h(c) = \cos c$

$\lim\limits_{x \rightarrow c} h(x) = \lim\limits_{x \rightarrow c} \cos x = \lim\limits_{h \rightarrow 0} \cos(c + h)$

We know that

$\cos(a + b) = \cos a \cos b - \sin a \sin b$

$\lim\limits_{h \rightarrow 0} \cos(c + h) = \lim\limits_{h \rightarrow 0} (\cos c \cos h - \sin c \sin h)$

$= \cos c \cdot \lim\limits_{h \rightarrow 0} \cos h - \sin c \cdot \lim\limits_{h \rightarrow 0} \sin h$

$= \cos c \cdot \cos 0 - \sin c \cdot \sin 0$

$= \cos c$

$\lim\limits_{x \rightarrow c} h(x) = h(c)$

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

When x = 0
$f (0) = -1\\ \lim\limits_{x\rightarrow 0^-}f(x) = \sin 0 - \cos 0 = -1\\ \lim\limits_{x\rightarrow 0^+}f(x) = \sin 0 - \cos 0 = -1 \\ R.H.L. = L.H.L. = f(0)$
Hence, function is also continuous at x = 0

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 $x = \frac{\pi}{2}$
$f(\frac{\pi}{2}) = 3\\let\ x=\pi +h\\ \lim\limits_{x\rightarrow \frac{\pi}{2}}f(x)= \lim\limits_{h\rightarrow 0}\frac{k\cos\left ( \frac{\pi}{2}+h \right )}{\pi-2\left ( \frac{\pi}{2}+h \right )} = k. \lim\limits_{h\rightarrow 0}\frac{-\sin h}{-2h} = \frac{k}{2}\\$
For the function to be continuous
$\lim\limits_{x\rightarrow \frac{\pi}{2}}f(x)= f(\frac{\pi}{2})\\ \frac{k}{2} = 3\\ k = 6$
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.
$f(2) = 4k\\ \lim\limits_{x\rightarrow 2^-}f(x)= 4k\\ \lim\limits_{x\rightarrow 2^+}f(x) = 3\\ f(2) = \lim\limits_{x\rightarrow 2^-}f(x) = \lim\limits_{x\rightarrow 2^+}f(x)\\ 4k = 3\\ k = \frac{3}{4}$
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.
$f(\pi) = k\pi + 1$
$\lim\limits_{x \rightarrow \pi^-} f(x) = k\pi + 1$
$\lim\limits_{x \rightarrow \pi^+} f(x) = \cos \pi = -1$
$f(\pi) = \lim\limits_{x \rightarrow \pi^-} f(x) = \lim\limits_{x \rightarrow \pi^+} f(x)$
$k\pi + 1 = -1$
$k = \frac{-2}{\pi}$
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.

$f(5) = 5k + 1$

$\lim\limits_{x \rightarrow 5^-} f(x) = 5k + 1$

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

$f(5) = \lim\limits_{x \rightarrow 5^-} f(x) = \lim\limits_{x \rightarrow 5^+} f(x)$

$5k + 1 = 10$

$k = \frac{9}{5}$

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

$f(x) = \left\{
\begin{matrix}
5 & \text{if } x \leq 2 \\
ax + b & \text{if } 2 < x < 10 \\
21 & \text{if } x > 10
\end{matrix}
\right.$

The function is continuous, so
$\lim\limits_{x \rightarrow 2^-} f(x) = \lim\limits_{x \rightarrow 2^+} f(x)$
and
$\lim\limits_{x \rightarrow 10^-} f(x) = \lim\limits_{x \rightarrow 10^+} f(x)$

$\lim\limits_{x \rightarrow 2^-} f(x) = 5$
$\lim\limits_{x \rightarrow 2^+} f(x) = ax + b = 2a + b$
$2a + b = 5 \quad \quad \quad \quad \quad \quad \quad \quad \text{(i)}$

and

$\lim\limits_{x \rightarrow 10^-} f(x) = ax + b = 10a + b$
$\lim\limits_{x \rightarrow 10^+} f(x) = 21$
$10a + b = 21 \quad \quad \quad \quad \quad \text{(ii)}$
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 $x\rightarrow k , \ then \ h \rightarrow 0$
$f(k) = \cos k^2\\ \lim\limits_{x \rightarrow k}f(x) = \lim\limits_{x \rightarrow k}\cos x^2 = \lim\limits_{h \rightarrow 0}\cos (k+h)^2 = \cos k^2\\ \lim\limits_{x \rightarrow k}f(x) = f(k)$
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
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

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

case (iii) k = 0
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim\limits_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
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 $x \rightarrow c , \ then \ h \rightarrow 0$

$h(c) = \cos c$

$\lim\limits_{x \rightarrow c} h(x) = \lim\limits_{x \rightarrow c} \cos x = \lim\limits_{h \rightarrow 0} \cos(c + h)$

We know that

$\cos(a + b) = \cos a \cos b - \sin a \sin b$

$\lim\limits_{h \rightarrow 0} \cos(c + h) = \lim\limits_{h \rightarrow 0} (\cos c \cos h - \sin c \sin h) = \lim\limits_{h \rightarrow 0} \cos c \cos h - \lim\limits_{h \rightarrow 0} \sin c \sin h$

$= \cos c \cdot \cos 0 - \sin c \cdot \sin 0 = \cos c$

$\lim\limits_{x \rightarrow c} h(x) = h(c)$

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
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

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

case (iii) k = 0
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim\limits_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
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 $x \rightarrow c , \ then \ h \rightarrow 0$

$h(c) = \sin c$

$\lim\limits_{x \rightarrow c} h(x) = \lim\limits_{x \rightarrow c} \sin x = \lim\limits_{h \rightarrow 0} \sin(c + h)$

We know that

$\sin(a + b) = \sin a \cos b + \cos a \sin b$

$\lim\limits_{h \rightarrow 0} \sin(c + h) = \lim\limits_{h \rightarrow 0} (\sin c \cos h + \cos c \sin h) = \lim\limits_{h \rightarrow 0} \sin c \cos h + \lim\limits_{h \rightarrow 0} \cos c \sin h$

$= \sin c \cdot \cos 0 + \cos c \cdot \sin 0 = \sin c$

$\lim\limits_{x \rightarrow c} h(x) = h(c)$

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
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

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

case (iii) k = 0
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim\limits_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
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
$h(k) = -(k+1)\\ \lim\limits_{x\rightarrow k}h(x) = -(k+1)\\ \lim\limits_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k < -1

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

case (iii) k = -1
$h(-1) = 0\\ \lim\limits_{x\rightarrow -1^-}h(x) = -(x-1) = 0\\ \lim\limits_{x\rightarrow -1^+}h(x ) = x+1 = 0\\ \lim\limits_{x\rightarrow -1^-}h(x) = h(0) = \lim\limits_{x\rightarrow -1^+}h(x )$
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