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

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

Answer:

Given function is
$f(x)=2x^2-1$
at x = 3
$$\begin{gathered} f(3)=2(3{)}^2-1=2\times 9-1=18-1=17 \\ \underset{x\to 3}{lim}f(x)=2(3{)}^2-1=2\times 9-1=18-1=17 \end{gathered}$$
$\underset{x\to 3}{lim}f(x)=f(3)$
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,
$$\begin{gathered} \underset{x\to k}{lim}f(x)=k-5 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, the function $f(x)=x-5$ is continuous at every real number

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

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

Answer:

Given function is
$f(x)=\frac{1}{x-5}$
For every real number k , $k\ne 5$
We get,
$$\begin{gathered} f(k)=\frac{1}{k-5} \\ \underset{x\to k}{lim}f(x)=\frac{1}{k-5} \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, function $f(x)=\frac{1}{x-5}$ continuous for every real value of x, $x\ne 5$

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

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

Answer:

Given function is
$f(x)=\frac{x^2-25}{x+5}$
For every real number k , $k\ne -5$
We gwt,
$$\begin{gathered} f(k)=\frac{k^2-5^2}{k+5}=\frac{(k+5)(k-5)}{k+5}=k-5 \\ \underset{x\to k}{lim}f(x)=\frac{k^2-5^2}{k+5}=\frac{(k+5)(k-5)}{k+5}=k-5 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, function $f(x)=\frac{x^2-25}{x+5}$ continuous for every real value of x , $x\ne -5$

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
$$\begin{gathered} f(k)=k-5 \\ \underset{x\to k}{lim}f(x)=k-5 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
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
$$\begin{gathered} f(k)=5-k \\ \underset{x\to k}{lim}f(x)=5-k \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, function f(x) = 5 - x is continous for x < 5

case(iii) x = 5
for x = 5 , f(x) = x - 5 is defined
$$\begin{gathered} f(5)=5-5=0 \\ \underset{x\to 5}{lim}f(x)=5-5=0 \\ \underset{x\to 5}{lim}f(x)=f(5) \end{gathered}$$
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
$$\begin{gathered} f(n)=n^n \\ \underset{x\to n}{lim}f(x)=n^n \\ \underset{x\to n}{lim}f(x)=f(n) \end{gathered}$$
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
$f(x)=\{\begin{array}{cc}x,& ifx\le 1\\ 5& ifx\ge 1\end{array}$
continuous at x = 0? At x = 1? At x = 2?

Answer:

Given function is
$f(x)=\{\begin{array}{cc}x,& ifx\le 1\\ 5& ifx\ge 1\end{array}$
function is defined at x = 0 and its value is 0
$$\begin{gathered} f(0)=0 \\ \underset{x\to 0}{lim}f(x)=f(x)=0 \\ \underset{x\to 0}{lim}f(x)=f(0) \end{gathered}$$
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

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

R.H.L $\ne$ 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
$$\begin{gathered} f(2)=2 \\ \underset{x\to 2}{lim}f(x)=f(5)=5 \\ \underset{x\to 2}{lim}f(x)=f(2) \end{gathered}$$
Hence, given function is continous at x = 2

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

$f(x)=\{\begin{array}{cc}2x+3& ifx\le 2\\ 2x-3& ifx\ge 2\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}2x+3& ifx\le 2\\ 2x-3& ifx\ge 2\end{array}$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 2
$$\begin{gathered} f(k)=2k-3 \\ \underset{x\to k}{lim}f(x)=2k-3 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for each value of k > 2

case(ii) k < 2
$$\begin{gathered} f(k)=2k+3 \\ \underset{x\to k}{lim}f(x)=2k+3 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for each value of k < 2

case(iii) x = 2

$$\begin{gathered} \underset{x\to 2^{-}}{lim}f(x)=2x+3=2\times 2+3=4+3=7 \\ \underset{x\to 2^{+}}{lim}f(x)=2x-3=2\times 2-3=4-3=1 \end{gathered}$$
Right hand limit at x= 2 $\ne$ 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

$f(x)=\{\begin{array}{ccc}|x|+3& ifx\le -3& \\ -2x& if-3<x<3& \\ 6x+2& ifx\ge 3& \end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{ccc}|x|+3& ifx\le -3& \\ -2x& if-3<x<3& \\ 6x+2& ifx\ge 3& \end{array}$
GIven function is defined for every real number k
Different cases are their
case (i) k < -3
$$\begin{gathered} f(k)=-k+3 \\ \underset{x\to k}{lim}f(x)=-k+3 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for every value of k < -3

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

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

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

case(v) k > 3
$$\begin{gathered} f(k)=6k+2 \\ \underset{x\to k}{lim}f(x)=6k+2 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
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

$f(x)=\{\begin{array}{cc}\frac{|x|}{x}& ifx\ne 0\\ 0& ifx=0\end{array}$

Answer:

Given function is
$f(x)\{\begin{array}{cc}\frac{|x|}{x}& ifx\ne 0\\ 0& ifx=0\end{array}$
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
$$\begin{gathered} f(k)=-1 \\ \underset{x\to k}{lim}f(x)=-1 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for every value of k < 0
case(ii) k > 0
$$\begin{gathered} f(k)=1 \\ \underset{x\to k}{lim}f(x)=1 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for every value of k > 0
case(iii) x = 0
$$\begin{gathered} f(0)=0 \\ \underset{x\to 0^{-}}{lim}f(x)=-1 \\ \underset{x\to 0^{+}}{lim}f(x)=1 \\ f(0)\ne R.H.L.\ne L.H.L. \end{gathered}$$
Hence, 0 is the only point of discontinuity

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

$f(x)=\{\begin{array}{cc}\frac{x}{|x|}& ifx<0\\ -1& ifx\ge 0\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}\frac{x}{|x|}& ifx<0\\ -1& ifx\ge 0\end{array}$
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

$f(x)=\{\begin{array}{cc}x+1& ifx\ge 1\\ x^2+1& ifx<1\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}x+1& ifx\ge 1\\ x^2+1& ifx<1\end{array}$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
$$\begin{gathered} f(k)=k+1 \\ \underset{x\to k}{lim}f(x)=k+1 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for each value of k > 1

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

case(iii) x = 1

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

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

$f(x)=\{\begin{array}{cc}{x}^3-3& ifx\le 2\\ x^2+1& ifx>2\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}{x}^3-3& ifx\le 2\\ x^2+1& ifx>2\end{array}$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 2
$$\begin{gathered} f(k)=k^2+1 \\ \underset{x\to k}{lim}f(x)=k^2+1 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for each value of k > 2

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

case(iii) x = 2

$\underset{x\to 2^{-}}{lim}f(x)=x^3-3=2^3-3=8-3=5$
$\underset{x\to 2^{+}}{lim}f(x)=x^2+1=2^2+1=4+1=5$
$f(2)=2^3-3=8-3=5$
$f(2)=R.H.L.=L.H.L.$
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

$f(x)=\{\begin{array}{cc}{x}^{10}-1& ifx\le 1\\ x^2& x>1\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}{x}^{10}-1& ifx\le 1\\ x^2& x>1\end{array}$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
$$\begin{gathered} f(k)=k^2 \\ \underset{x\to k}{lim}f(x)=k^2 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for each value of k > 1

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

case(iii) x = 1

$\underset{x\to 1^{-}}{lim}f(x)=x^{10}-1=1^{10}-1=1-1=0$
$\underset{x\to 1^{+}}{lim}f(x)=x^2=1^2=1$
$f(1)=x^{10}-1=0$
$f(1)=L.H.L.\ne R.H.L.$

Hence, x = 1 is the point of discontinuity

Question:13. Is the function defined by

$f(x)=\{\begin{array}{cc}x+5& ifx\le 1\\ x-5& ifx>1\end{array}$

a continuous function?

Answer:

Given function is
$f(x)=\{\begin{array}{cc}x+5& ifx\le 1\\ x-5& ifx>1\end{array}$
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
$$\begin{gathered} f(k)=k-5 \\ \underset{x\to k}{lim}f(x)=k-5 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
$$\begin{gathered} f(k)=k+5 \\ \underset{x\to k}{lim}f(x)=k+5 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for each value of k < 1

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

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

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

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

Hence, x = 1 is the point of discontinuity

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

$f(x)\{\begin{array}{cc}3& if0\le x\le 1\\ 4& if1<x<3\\ 5& if3\le x\le 10\end{array}$

Answer:

Given function is
$f(x)\{\begin{array}{cc}3& if0\le x\le 1\\ 4& if1<x<3\\ 5& if3\le x\le 10\end{array}$
GIven function is defined for every real number k
Different cases are their
case (i) k < 1
$$\begin{gathered} f(k)=3 \\ \underset{x\to k}{lim}f(x)=3 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continous for every value of k < 1

case(ii) k = 1
$$\begin{gathered} f(1)=3 \\ \underset{x\to 1^{-}}{lim}f(x)=3 \\ \underset{x\to 1^{+}}{lim}f(x)=4 \\ R.H.L.\ne L.H.L.=f(1) \end{gathered}$$
Hence, given function is discontinous at x = 1
Therefore, x = 1 is he point od discontinuity

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

case(iv) k = 3
$$\begin{gathered} f(3)=5 \\ \underset{x\to 3^{-}}{lim}f(x)=4 \\ \underset{x\to 3^{+}}{lim}f(x)=5 \\ R.H.L.=f(3)\ne L.H.L. \end{gathered}$$
Hence. x = 3 is the point of discontinuity

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

$$\begin{gathered} f(k)=4 \\ \underset{x\to k}{lim}f(x)=4 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
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 $f(x)\{\begin{array}{ccc}2x& if& x<0\\ 0& if& 0\le x\le 1\\ 4x& if& x>1\end{array}$

Answer:

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

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

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

case (iv) k < 1

$f(k)=0$

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

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

case (v) k > 1

$f(k)=4k$

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

case (vi) x = 1

$f(1)=0$

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

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

Hence, function is not continuous at x = 1

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

$f(x)=\{\begin{array}{cc}-2& ifx\le -1\\ 2x& if-1<x\le 1\\ 2& ifx>1\end{array}$

Answer:

Given function is
$f(x)=\{\begin{array}{cc}-2& ifx\le -1\\ 2x& if-1<x\le 1\\ 2& ifx>1\end{array}$
GIven function is defined for every real number k
Different cases are their
case (i) k < -1
$$\begin{gathered} f(k)=-2 \\ \underset{x\to k}{lim}f(x)=-2 \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continuous for every value of k < -1

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

case(iii) k > -1
$$\begin{gathered} f(k)=2k \\ \underset{x\to k}{lim}f(x)=2k \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
Hence, given function is continous for all values of x > -1

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

case(v) k = 1
$$\begin{gathered} f(1)=2x=2(1)=2 \\ \underset{x\to 1^{-}}{lim}f(x)=2x=2(1)=2 \\ \underset{x\to 1^{+}}{lim}f(x)=2 \\ R.H.L.=f(1)=L.H.L. \end{gathered}$$
Hence.at x =1 function is continous

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

$$\begin{gathered} f(k)=2k \\ \underset{x\to k}{lim}f(x)=2k \\ \underset{x\to k}{lim}f(x)=f(k) \end{gathered}$$
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
$f(x)=\{\begin{array}{cc}ax+1,& ifx<3\\ bx+3& ifx>3\end{array}$
is continuous at x = 3.

Answer:

Given function is
$f(x)=\{\begin{array}{cc}ax+1,& ifx<3\\ bx+3& ifx>3\end{array}$
For the function to be continuous at x = 3 , R.H.L. must be equal to L.H.L.
$$\begin{gathered} \underset{x\to 3^{-}}{lim}f(x)=ax+1=3a+1 \\ \underset{x\to 3^{+}}{lim}f(x)=bx+3=3b+3 \end{gathered}$$
For the function to be continuous
$$\begin{gathered} \underset{x\to 3^{-}}{lim}f(x)=\underset{x\to 3^{+}}{lim}f(x) \\ 3a+1=3b+3 \\ 3(a-b)=2 \\ a-b=\frac{2}{3} \\ a=b+\frac{2}{3} \end{gathered}$$

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

Answer:

Given function is
$f(x)=\{\begin{array}{cc}\lambda (x^2-2x)& ifx\le 0\\ 4x+1& ifx>0\end{array}$
For the function to be continuous at x = 0 , R.H.L. must be equal to L.H.L.
$$\begin{gathered} \underset{x\to 0^{-}}{lim}f(x)=\lambda (x^2-2x)=0 \\ \underset{x\to 0^{+}}{lim}f(x)=4x+1=1 \end{gathered}$$
For the function to be continuous
$$\begin{gathered} \underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{+}}{lim}f(x) \\ 0\ne 1 \end{gathered}$$
Hence, for no value of function is continuous at x = 0

For x = 1
$$\begin{gathered} f(1)=4x+1=4(1)+1=5 \\ \underset{x\to 1}{lim}f(x)=4+1=5 \\ \underset{x\to 1}{lim}f(x)=f(x) \end{gathered}$$
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

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

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

$\underset{x\to k^{-}}{lim}f(x)\ne \underset{x\to k^{+}}{lim}f(x)$

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 $f(x)=x^2-sinx+5$ continuous at x = $\pi$?

Answer:

Given function is
$f(x)=x^2-sinx+5$
Clearly, Given function is defined at x =$\pi$
$$\begin{gathered} f(\pi )={\pi }^2-\sin \pi +5={\pi }^2-0+5={\pi }^2+5 \\ \underset{x\to \pi }{lim}f(x)={\pi }^2-\sin \pi +5={\pi }^2-0+5={\pi }^2+5 \\ \underset{x\to \pi }{lim}f(x)=f(\pi ) \end{gathered}$$
Hence, the function defined by $f(x)=x^2-sinx+5$ 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 $x\to c,thenh\to 0$

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

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

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

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

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

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

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

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

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

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 $x\to c,thenh\to 0$

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

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

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

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