Ex 5.1 Class 12 Maths Ncert Solutions

The given function is \( \mathrm{f}(x)=5 x-3 \)
At \( x=0, \mathrm{f}(0)=5 \times 0-3=-3 \)
\( \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(5 x-3)=5 \times 0-3=-3 \)
Thus, \( \lim _{x \rightarrow 0} f(x)=f(0) \)
Therefore, \( f \) is continuous at \( x=0 \)
At \( x=-3, f(-3)=5 \times(-3)-3=-18 \)
\( \lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow}(5 x-3)=5 \times(-3)-3=-18 \)
Thus, \( \lim _{x \rightarrow 3} f(x)=f(-3) \)
Therefore, \( f \) is continuous at \( x=-3 \)
At \( x=5, \mathrm{f}(5)=5 \times 5-3=22 \)
\( \lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5}(5 x-3)=5 \times 5-3=22 \)
Thus,
Therefore, \( f \) is continuous at \( x=5 \)

The given function is \( f(x)=2 x^{2}-1 \)
At \( x=3, f(x)=f(3)=2 \times 3^{2}-1=17 \)
Left hand limit (LHL):
Right hand \( \operatorname{limit}(\mathrm{RHL}) \) :
As, \( \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(3) \)
Therefore, f is continuous at \( x=3 \)

The given function is \( f(x)=x-5 \)
We know that \( f \) is defined at every real number \( k \) and its value at \( k \) is \( k- \) 5.
We can see that \( \lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(x-5)=\mathrm{k}-5=\mathrm{f}(\mathrm{k}) \)
Thus, \( \lim _{x \rightarrow k} f(x)=\mathrm{f}(\mathrm{k}) \)
Therefore, \( f \) is continuous at every real number and thus, it is continuous function.

The given function is \( \mathrm{f}(x)=\frac{1}{x-5}, x \neq 5 \)
For any real number \( \mathrm{k} \neq 5 \), we get,
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k} \frac{1}{x-5}=\lim _{(k-5)} \frac{1}{k-5}\)
Also, \( \mathrm{f}(\mathrm{k})=\frac{1}{k-5}( \) Ask \( \neq 5) \)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Therefore, \( f \) is continuous at point in the domain of \( f \) and thus, it is continuous function.

The given function is \( \mathrm{f}(x)=\frac{x^{2}-25}{x+5}, x \neq 5 \)
For any real number \( \mathrm{k} \neq 5 \), we get,
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k} \frac{x^{2}-25}{x+5}=\lim _{x \rightarrow k} \frac{(x-5)(x+5)}{x+5}=\lim _{x \rightarrow k}(x-5)=(\mathrm{k}-5)\)
Also, \( \mathrm{f}(\mathrm{k})=\lim _{x \rightarrow k} \frac{(k-5)(k+5)}{k+5}=\lim _{x \rightarrow k}(k-5)=(\mathrm{k}-5)( \text{ ask} \neq 5) \)
Thus, \( \lim _{x \rightarrow k} f(x)=\mathrm{f}(\mathrm{k}) \)
Therefore, \( f \) is continuous at point in the domain of \( f \) and thus, it is continuous function.

The given function is \( \mathrm{f}(x)=|x-5|=\left\{\begin{array}{l}5-x, \text { if } x < 5 \\ x-5, \text { if } x \geq 5\end{array}\right. \)
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.ee, \( k < 5 \), or \( k=5 \) or \( k > 5 \)
Now, Case I: \( k < 5 \)
Then, \( \mathrm{f}(\mathrm{k})=5-\mathrm{k} \)
\( \lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(5-x)=5-\mathrm{k}=\mathrm{f}(\mathrm{k}) \)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number less than 5 .
Case II: \( \mathrm{k}=5 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{f}(5)=5-5=0 \)
\(\begin{array}{l}
\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5}(5-x)=5-5=0 \\
\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(5-x)=5-5=0 \\
=\lim _{x \rightarrow k^{-}} f(x)=\lim _{x \rightarrow k^{+}} f(x)=f(k)
\end{array}\)
Hence, f is continuous at \( x=5 \).
Case III: \( \mathrm{k} > 5 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{k}-5 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(x-5)=\mathrm{k}-5=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number greater than 5 .
Therefore, \( f \) is a continuous function.

It is given that function \( f(x)=x^{n} \)
We can see that \( f \) is defined at all positive integers, \( n \) and the value of \( f \) at \(n\) is \( \mathrm{n}^{\mathrm{n}} \).
\(=\lim _{x \rightarrow n} f(n)=\lim _{x \rightarrow n}\left(x^{n}\right)=\mathrm{n}^{\mathrm{n}}\)
Thus, \( \lim _{x \rightarrow n} f(x)=f(n) \)
Therefore, f is continuous at \( x=\mathrm{n} \), where n is a positive integer.

It is given that \( \mathrm{f}(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5, \text { i } f x > 1\end{array}\right. \)
Case I: \( x=0 \)
We can see that f is defined at 0 and its value at 0 is 0 .
LHL
\(\lim _{x \rightarrow 0^{-}}=\lim _{h \rightarrow 0} f(0-h)\)
\(=\lim _{h \rightarrow 0}-h=0\)
RHL
\(\lim _{x \rightarrow 0^{+}}=\lim _{h \rightarrow 0} f(0+h)\)
\(=\lim _{h \rightarrow 0} h=0\)
LHL \( = \) RHL \( =f(0) \) Hence, \( f \) is continuous at \( x=0 \).
Case II: \( x=1 \)
We can see that f is defined at 1 and its value at 1 is 1 .
For \( x < 1 \)
\( \mathrm{f}(x)=x \) Hence, LHL:
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} x=1\)
For \( x > 1 \mathrm{f}(x)= \) 5therefore, RHL
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(5)=5\)
\(=\lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)\)
Hence, f is not continuous at \( x=1 \).
Case III: \( x=2 \)
As,
We can see that f is defined at 2 and its value at 2 is 5
LHL:
\(\lim _{x \rightarrow 2^{-}}=\lim _{h \rightarrow 0} f(2-h)\)
\(=\lim _{h \rightarrow 0} 5=5\)
here \( \mathrm{f}(2-\mathrm{h})=5 \), as \( \mathrm{h} \rightarrow 0 \Rightarrow 2-\mathrm{h} \rightarrow 2 \) RHL:
\(\lim _{x \rightarrow 2^{+}}=\lim _{h \rightarrow 0}(2+h)\)
\(=\lim _{h \rightarrow 0} 5=5\)
\( \mathrm{LHL}= \) RHL \( =\mathrm{f}(2) \)
here \( \mathrm{f}(2+\mathrm{h})=5 \), as \( \mathrm{h} \rightarrow 0 \Rightarrow 2+\mathrm{h} \rightarrow 2 \)
Hence, f is continuous at \( x=2 \).

The given function is \( \mathrm{f}(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x \geq 2 \end{array}\right. \)
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e., \( \mathrm{k} < 2 \), or \( \mathrm{k}=2 \) or \( \mathrm{k} > 2 \)
Now, Case I: \( \mathrm{k} < 2 \)
Then, \( \mathrm{f}(\mathrm{k})=2 \mathrm{k}+3 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(2 x+3)=2 \mathrm{k}+3=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number less than 2 .
Case II: \( \mathrm{k}=2 \)
\(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}(2 x+3)=2 \times 2+3=7\)
\(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x-3)=2 \times 2-3=1\)
\(=\lim _{x \rightarrow k^{-}} f(k) \neq \lim _{x \rightarrow k^{+}} f(k)=f(k)\)
Hence, f is not continuous at \( x=2 \).
Case III: \( \mathrm{k} > 2 \)
Then, \( \mathrm{f}(\mathrm{k})=2 \mathrm{k}-3 \)
\( \lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(2 x-3)=2 \mathrm{k}-3=\mathrm{f}(\mathrm{k}) \)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number greater than 2 .
Therefore, \( x=2 \) is the only point of discontinuity of \( f \).

The given function is \( f(x)=\left\{\begin{array}{c}|x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3 < x < 3 \\ 6 x+2, \text { if } f \geq 3\end{array}\right. \)
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 5 cases i.e., \( \mathrm{k} < -3, \mathrm{k}=-3,-3 < \mathrm{k} < 3, \mathrm{k}=3 \) or \( \mathrm{k} > 3 \)
Now, Case I: \( \mathrm{k} < -3 \)
Then, \( \mathrm{f}(\mathrm{k})=-\mathrm{k}+3 \)
\( \lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(-x+3)=-\mathrm{k}+3=\mathrm{f}(\mathrm{k}) \)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number \( x < -3 \).
Case II: \( \mathrm{k}=-3 \)
\(\begin{array}{l}
f(-3)=-(-3)+3=6 \\
\lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3^{-}}(-x+3)=-(-3)+3=6 \\
\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3^{-}}(-2 x)=-2 \times(-3)=6 \\
=\lim _{x \rightarrow k^{-}} f(x)=\lim _{x \rightarrow k^{+}} f(x)=f(k)
\end{array}\)
Hence, f is continuous at \( x=-3 \).
Case III: \( -3 < \mathrm{k} < 3 \)
Then, \( \mathrm{f}(\mathrm{k})=-2 \mathrm{k} \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(-2 x)=-2 \mathrm{k}=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, \( f \) is continuous in \( (-3,3) \).
Case IV: \( \mathrm{k}=3 \)
\(\begin{array}{l}
\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(-2 x)=-2 \times(3)=-6 \\
\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(6 x+2)=6 \times 3+2=20 \\
=\lim _{x \rightarrow k^{-}} f(x) \neq \lim _{x \rightarrow k^{+}} f(x)
\end{array}\)
Hence, f is not continuous at \( x=3 \).
Case V: \( \mathrm{k} > 3 \)
Then, \( \mathrm{f}(\mathrm{k})=6 \mathrm{k}+2 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(6 x+2)=6 \mathrm{k}+2=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number \( x < 3 \).
Therefore, \( x=3 \) is the only point of discontinuity of \( f \).

The given function is \( \mathrm{f}(x)=\left\{\begin{array}{l}\frac{|x|}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right. \)
We know that if \( x > 0 \)
\(\Rightarrow|x|=-x \text { and } x > 0\)
\(\Rightarrow|x|=x\)
So, we can rewrite the given function as:
\(\mathrm{f}(x)=\left\{\begin{array}{l}
\frac{|x|}{x}=\frac{-x}{x}=-1, \text { if } x < 0 \\
\frac{|x|}{x}=\frac{x}{x}=1, \text { if } x > 0
\end{array}\right.\)
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e., \( \mathrm{k} < 0 \), or \( \mathrm{k}=0 \) or \( \mathrm{k} > 0 \).
Now, Case I: \( \mathrm{k} < 0 \)
Then, \( \mathrm{f}(\mathrm{k})=-1 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(-1)=-1=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number less than 0 .
Case II: \( \mathrm{k}=0 \)
\(\begin{array}{l}
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-1)=-1 \\
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(1)=1 \\
=\lim _{x \rightarrow k^{-}} f(x) \neq \lim _{x \rightarrow k^{+}} f(x)=f(k)
\end{array}\)
Hence, f is not continuous at \( x=0 \).
Case III: \( \mathrm{k} > 0 \)
Then, \( \mathrm{f}(\mathrm{k})=1 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(1)=1=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number greater than 1.
Therefore, \( x=0 \) is the only point of discontinuity of \( f \).

The given function is \( \mathrm{f}(x)=\left\{\begin{array}{l}\frac{x}{|x|}, \text { if } x < 0 \\ -1, \text { if } x \geq 0\end{array}\right. \)
We know that if \( x < 0 \)
\(\Rightarrow|x|=-x\)
So, we can rewrite the given function as:
\(\mathrm{f}(x)=\left\{\begin{array}{c}
\frac{x}{|x|}=\frac{x}{-x}=-1, \text { if } x < 0 \\
-1, \text { if } x \geq 0
\end{array}\right.\)
\( \Rightarrow \mathrm{f}(x)=-1 \) for all \( x \in \mathrm{R} \)
Let k be the point on a real line.
Then, \( \mathrm{f}(\mathrm{k})=-1 \)
\( \lim _{x \rightarrow k} f(k)=\lim _{x \rightarrow k}(-1)=-1=\mathrm{f}(\mathrm{k}) \)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Therefore, the given function is a continuous function.

The given function is \( \mathrm{f}(x)=\left\{\begin{array}{c}x+1, \text { if } x \geq 0 \\ x^{2}+1, \text { if } x < 0\end{array}\right. \)
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e., \( \mathrm{k} < 1 \), or \( \mathrm{k}=1 \) or \( \mathrm{k} > 1 \)
Now, Case I: \( \mathrm{k} < 1 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{k}^{2}+1 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}\left(x^{2}+1\right)=\mathrm{k}^{2}+1=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number less than 1.
Case II: \( \mathrm{k}=1 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{f}(1)=1+1=2 \)
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^{2}+1\right)=1^{2}+1=2\)
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x+1)=1+1=2\)
\(=\lim _{x \rightarrow k^{-}} f(x)=\lim _{x \rightarrow k^{+}} f(x)=f(k)\)
Hence, f is continuous at \( x=1 \).
Case III: \( \mathrm{k} > 1 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{k}+1 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(x+1)=\mathrm{k}+1=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number greater than 1.

The given function is \( \mathrm{f}(x)=\left\{\begin{array}{l}x^{3}-3, \text { if } x \leq 2 \\ x^{2}+1, \text { if } x > 2\end{array}\right. \)
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e., \( \mathrm{k} < 2 \), or \( \mathrm{k}=2 \) or \( \mathrm{k} > 2 \)
Now, Case I: \( \mathrm{k} < 2 \)
Then, \( f(k)=k^{3}-3 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}\left(x^{3}-3\right)=\mathrm{k}^{3}-3=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number less than 2 .
Case II: \( \mathrm{k}=2 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{f}(2)=2^{3}-3=5 \)
\(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(x^{3}-3\right)=2^{3}-3=5\)
\(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(x^{2}+1\right)=2^{2}+1=5\)
\(=\lim _{x \rightarrow k^{-}} f(x)=\lim _{x \rightarrow k^{+}} f(x)=f(k)\)
Hence, f is continuous at \( x=2 \).
Case III: \( \mathrm{k} > 2 \)
Then, \( \mathrm{f}(\mathrm{k})=2^{2}+1=5 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}\left(x^{2}+1\right)=2^{2}+1=5=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number greater than 2 .

The given function is \( \mathrm{f}(x)=\left\{\begin{array}{c}x^{10}-1, \text { if } x \leq 1 \\ x^{2}, \text { if } x > 1\end{array}\right. \)
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e., \( \mathrm{k} < 1 \), or \( \mathrm{k}=1 \) or \( \mathrm{k} > 1 \)
Now,
Case I: \( \mathrm{k} < 1 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{k}^{10}-1 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}\left(x^{10}-1\right)=\mathrm{k}^{10}-1=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number less than 1.
Case II: \( \mathrm{k}=1 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{f}(1)=1^{10}-1=0 \)
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^{10}-1\right)=1^{10}-1=0\)
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(x^{2}\right)=1^{2}=1\)
\(=\lim _{x \rightarrow k^{-}} f(x) \neq \lim _{x \rightarrow k^{+}} f(x)\)
Hence, f is not continuous at \( x=1 \).
Case III: \( \mathrm{k} > 1 \)
Then, \( \mathrm{f}(\mathrm{k})=1^{2}=1 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}\left(x^{2}\right)=1^{2}=1=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number greater than 1.
Therefore, \( x=1 \) is the only point of discontinuity of \( f \).

The given function is \( \mathrm{f}(x)=\left\{\begin{array}{l}x+5, \text { if } x \leq 1 \\ x-5, \text { if } x > 1\end{array}\right. \)
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e., \( \mathrm{k} < 1 \), or \( \mathrm{k}=1 \) or \( \mathrm{k} > 1 \)
Now,
Case I: \( \mathrm{k} < 1 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{k}+5 \)
\( \lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(x+5)=\mathrm{k}+5=\mathrm{f}(\mathrm{k}) \)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number less than 1.
Case II: \( \mathrm{k}=1 \)
Then, \( \mathrm{f}(\mathrm{k})=\mathrm{f}(1)=1+5=6 \)
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x+5)=1+5=6\)
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-5)=1-5=-4\)
\(=\lim _{x \rightarrow k^{-}} f(x) \neq \lim _{x \rightarrow k^{+}} f(x)\)
Hence, f is not continuous at \( x=1 \).
Case III: \( \mathrm{k} > 1 \)
Then, \( f(k)=k-5 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(x-5)=\mathrm{k}-5\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all real number greater than 1.
Therefore, \( x=1 \) is the only point of discontinuity of \( f \).

The given function is \(\mathrm{f}(x)=\left\{\begin{array}{c}
3 \ \text{if} \ 0 \leq x \leq 1 \\
4 \ \text{if} \ 1 < x < 3 \\
5 \ \text{if} \ 3 \leq x \leq 10
\end{array}\right.\)
The function f is defined at all points of the interval \( [0,10] \).
Let k be the point in the interval \( [0,10] \).
Then, we have 5 cases i.e., \( 0 \leq \mathrm{k} < 1, \mathrm{k}=1,1 < \mathrm{k} < 3, \mathrm{k}=3 \) or \( 3 < \mathrm{k} \leq 10\) .
Now, Case I: \( 0 \leq \mathrm{k} < 1 \)
Then, \( \mathrm{f}(\mathrm{k})=3 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(3)=3=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous in the interval \( [0,10) \).
Case II: \( \mathrm{k}=1 \)
\(\mathrm{f}(1)=3\)
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(3)=3\)
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4)=4\)
\(=\lim _{x \rightarrow k^{-}} f(x) \neq \lim _{x \rightarrow k^{+}} f(x)\)
Hence, f is not continuous at \( x=1 \).
Case III: \( 1 < \mathrm{k} < 3 \)
Then, \( f(k)=4 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(4)=4=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous in \( (1,3) \).
Case IV: \( \mathrm{k}=3 \)
\(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(4)=4\)
\(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(5)=5\)
\(=\lim _{x \rightarrow k^{-}} f(x) \neq \lim _{x \rightarrow k^{+}} f(x)\)
Hence, f is not continuous at \( x=3 \).
Case V: \( 3 < \mathrm{k} \leq 10 \)
Then, \( f(k)=5 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(5)=5=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow} f(x)=f(k) \)
Hence, \( f \) is continuous at all points of the interval (3, 10].
Therefore, \( x=1 \) and 3 are the points of discontinuity of \( f \).

The given function is \( \mathrm{f}(x)=\left\{\begin{array}{c}2 x, \text { if } x \leq 0 \\ 0 \text { if } 0 \leq x \leq 1 \\ 4 x \text { if } x > 1\end{array}\right. \)
The function f is defined at all points of the real line.
Then, we have 5 cases i.e., \( \mathrm{k} < 0, \mathrm{k}=0,0 < \mathrm{k} < 1, \mathrm{k}=1 \) or \( \mathrm{k} < 1 \).
Now, Case I: \( \mathrm{k} < 0 \)
Then, \( \mathrm{f}(\mathrm{k})=2 \mathrm{k} \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(2 x)=2 \mathrm{k}=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all points \(x \), s.t. \( x < 0 \).
Case II: \( \mathrm{k}=0 \)
\(\mathrm{f}(0)=0\)
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x)=2 \times 0=0\)
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(0)=0\)
\(=\lim _{x \rightarrow k^{-}} f(x)=\lim _{x \rightarrow k^{+}} f(x)=f(k)\)
Hence, \( f \) is continuous at \( x=0 \).
Case III: \( 0 < \mathrm{k} < 1 \)
Then, \( \mathrm{f}(\mathrm{k})=0 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(0)=0=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, \( f \) is continuous in \( (0,1) \).
Case IV: \( \mathrm{k}=1 \)
Then \( \mathrm{f}(\mathrm{k})=\mathrm{f}(1)=0 \)
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(0)=0\)
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x)=4 \times 1=4\)
\(=\lim _{x \rightarrow k^{-}} f(x) \neq \lim _{x \rightarrow k^{+}} f(x)\)
Hence, f is not continuous at \( x=1 \).
Case V: \( \mathrm{k} < 1 \)
Then, \( \mathrm{f}(\mathrm{k})=4 \mathrm{k} \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(4 x)=4 \mathrm{k}=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all points \(x \), s.t. \( x > 1 \).
Therefore, \( x=1 \) is the only point of discontinuity of \( f \).

The given function is \( \mathrm{f}(x)=\left\{\begin{array}{c}-2, \text { if } x \leq-1 \\ 2 x, \text { if } 0 \leq x \leq 1 \\ 2, \text { if } x > 1\end{array}\right. \)
The function f is defined at all points of the real line.
Then, we have 5 cases i.e., \( \mathrm{k} < -1, \mathrm{k}=-1,-1 < \mathrm{k} < 1, \mathrm{k}=1 \) or \( \mathrm{k} > 1 \).
Now, Case I: \( \mathrm{k} < 0 \)
Then, \( \mathrm{f}(\mathrm{k})=-2 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k} f(x)=-2=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all points \(x \), s.t. \( x < -1 \).
Case II: \( \mathrm{k}=-1 \)
\(\mathrm{f}(\mathrm{k})=\mathrm{f}(=1)=-2\)
\(\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}}(-2)=-2\)
\(\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}}(2 x)=2 \times(-1)=-2\)
\(=\lim _{x \rightarrow k^{-}} f(x)=\lim _{x \rightarrow k^{+}} f(x)=f(k)\)
Hence, f is continuous at \( x=-1 \).
Case III: \( -1 < \mathrm{k} < 1 \)
Then, \( \mathrm{f}(\mathrm{k})=2 \mathrm{k} \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(-2)=2 \mathrm{k}=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, \( f \) is continuous in \( (-1,1) \).
\(\text { Case IV: } \mathrm{k}=1\)
\(\text { Then } \mathrm{f}(\mathrm{k})=\mathrm{f}(1)=2 \times 1=2\)
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(2 x)=2 \times 1=2\)
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(2)=2\)
\(=\lim _{x \rightarrow k^{-}} f(x)=\lim _{x \rightarrow k^{+}} f(x)=f(x)\)
Hence, f is continuous at \( x=1 \).
Case V: \( \mathrm{k} > 1 \)
Then, \( f(k)=2 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(2)=2=\mathrm{f}(\mathrm{k})\)
Thus, \( \lim _{x \rightarrow k} f(x)=f(k) \)
Hence, f is continuous at all points \(x \), s.t. \( x > 1 \).
Therefore, f is continuous at all points of the real line.

It is given function is \( \mathrm{f}(x)=\left\{\begin{array}{l}a x+1, \text { if } x \leq 3 \\ b x+3, \text { if } x > 3\end{array}\right. \)
It is given that f is continuous at \( x=3 \), then, we get,
\(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=\mathrm{f}(3) \ldots(1)\)
And
\(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(a x+1)=3 \mathrm{a}+1\)
\(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(b x+3)=3 \mathrm{~b}+1\)
\(\mathrm{f}(3)=3 \mathrm{a}+1\)
Thus, from (1), we get,
\(3 a+1=3 b+3=3 a+1\)
\(\Rightarrow 3 a+1=3 b+1\)
\(\Rightarrow 3 a=3 b\)
\(\Rightarrow a=b\)
Therefore, the required the relation is \( \mathrm{a}=\mathrm{b}\).

It is given that \( \mathrm{f}(x)=\left\{\begin{array}{c}\lambda\left(x^{2}-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \text { if } x > 0\end{array}\right. \)
It is given that f is continuous at \( x=0 \), then, we get,
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=\mathrm{f}(0)\)
And
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(\lambda\left(x^{2}-2 x\right)\right)=\lim _{x \rightarrow 0^{-}}\left(\lambda\left(0^{2}-2 \times 0\right)\right)=0\)
\(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(4 x+1)=4 \times 0+1=1\)
\(\therefore \lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)\)
Thus, there is no value of for which f is continuous at \( x=0 \)
\(\mathrm{f}(1)=4 x+1=4 \times 1+1=5\)
\(\lim _{x \rightarrow 1}(4 x+1)=4 \times 1+1=5\)
Then, \( \lim _{x \rightarrow 1} f(x)=f(1) \)
Hence, for any values of, \( f \) is continuous at \( x=1 \)

It is given that \( g(x)=x-[x] \)
We know that g is defined at all integral points.
Let k be ant integer.
Then,
\(\mathrm{g}(\mathrm{k})=\mathrm{k}-[-\mathrm{k}]=\mathrm{k}+\mathrm{k}=2 \mathrm{k}\)
\(\lim _{x \rightarrow k^{-}} g(k)=\lim _{x \rightarrow k^{-}}(x-[x])\)
\(\lim _{x \rightarrow k^{-}}(x)=\lim _{x \rightarrow k^{-}}[x]=\mathrm{k}-(\mathrm{k}-1)=1\)
And
\(\lim _{x \rightarrow k^{+}} g(x)=\lim _{x \rightarrow k^{+}}(x-[x])\)
\(\lim _{x \rightarrow k^{+}}(x)-\lim _{x \rightarrow k^{+}}[x]=\mathrm{k}-\mathrm{k}=0\)
\(\therefore \lim _{x \rightarrow k^{-}} f(x) \neq \lim _{x \rightarrow k^{+}} f(x)\)
Therefore, \(g\) is discontinuous at all integral points.

It is given that \( f(x)=x^{2}-\sin x+5 \)
We know that f is defined at \( x=\pi \)
So, at \( x=\pi \),
\(\mathrm{f}(x)=\mathrm{f}(\pi)=\pi^{2}-\sin \pi+5=\pi^{2}-0+5=\pi 2+5\)
Now, \( \lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} f\left(x^{2}-\sin x+5\right) \)
Let put \( x=\pi+\mathrm{h} \)
If \( x \rightarrow \pi \), then we know that \( \mathrm{k} \rightarrow 0 \)
\(\therefore \lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi}\left(x^{2}-\sin x+5\right)\)
\(=\lim _{k \rightarrow 0} f\left[(\pi+k)^{2}-\sin (\pi+k)+5\right]\)
\(=\lim _{k \rightarrow 0} f(\pi+k)^{2}-\lim _{k \rightarrow 0} \sin (\pi+k)+\lim _{k \rightarrow 0} 5\)
\(=(\pi+0)^{2}-\lim _{k \rightarrow 0}[\sin \pi \cos k+\cos \pi \sin k]+5\)
\(=\pi^{2}-\lim _{k \rightarrow 0} \sin \pi \cos k-\lim _{k \rightarrow 0} \cos \pi \sin k+5\)
\(=\pi^{2}-\sin \pi \cos 0-\cos \pi \sin 0+5\)
\(=\pi^{2}-0 \times 1(-1) \times 0+5\)
\(=\pi^{2}+5\)
Thus, \( \lim _{x \rightarrow \pi} f(x) f(\pi) \)
Therefore, the function f is continuous at \( x=\pi \).

We know that g and k are two continuous functions, then, \( \mathrm{g}+\mathrm{k}, \mathrm{g}-\mathrm{k} \) and \( \mathrm{g} . \mathrm{k} \) are also continuous.
First we have to prove that \( \mathrm{g}(x)=\sin x \) and \( \mathrm{k}(x)=\cos x \) are continuous functions.
Now, let \( \mathrm{g}(x)=\sin x \)
We know that \( \mathrm{g}(x)=\sin x \) is defined for every real number.
Let h be a real number. Now, put \( x=\mathrm{h}+\mathrm{k} \)
So, if \( x \rightarrow \mathrm{h} \) and \( \mathrm{k} \rightarrow 0 \)
\(\mathrm{g}(\mathrm{h})=\sin \mathrm{h}\)
\(\lim _{x \rightarrow h} g(x)=\lim _{x \rightarrow h} \sin x\)
\(=\lim _{x \rightarrow 0} \sin (h+k)\)
\(=\lim _{x \rightarrow 0}[\sin h \cos k+\cos h \sin k]\)
\(=\sin h \cos 0+\cos h \sin 0\)
\(=\sin h+0\)
\(=\sin h\)
Thus, \( \lim _{x \rightarrow h} g(x)=g(h) \)
Therefore, g is a continuous function \(\ldots(1)\)
Now, let \( \mathrm{k}(x)=\cos x \)
We know that \( \mathrm{k}(x)=\cos x \) is defined for every real number.
Let h be a real number. Now, put \( x=\mathrm{h}+\mathrm{k} \)
So, if \( x \rightarrow \mathrm{h} \) and \( \mathrm{k} \rightarrow 0 \)
Now \( k(\mathrm{~h})=\cos \mathrm{h} \)
\( \lim _{x \rightarrow h} k(x)=\lim _{x \rightarrow h} \cos x \)
\( =\lim _{k \rightarrow 0} \cos (h+k) \)
\( =\lim _{x \rightarrow 0}[\cos h \cos k-\sin h \sin k] \)
\( =\cos h \cos 0-\sin h \sin 0 \)
\( =\cos h -0 \)
\( =\cos h \)
Thus, \( \lim _{x \rightarrow h} k(x)=k(h) \)
Therefore, k is a continuous function \(\ldots(2)\)
So, from (1) and (2), we get,
(a) \( f(x)=g(x)+k(x)=\sin x+\cos x \) is a continuous function.
(b) \( f(x)=g(x)-k(x)=\sin x-\cos x \) is a continuous function.
(c) \( f(x)=g(x) \times k(x)=\sin x \times \cos x \) is a continuous function.

We know that if g and h are two continuous functions, then,
(i) \( \frac{h(x)}{g(x)}, \mathrm{g}(x) \neq 0 \) is continuous
(ii) \( \frac{1}{g(x)}, \mathrm{g}(x) \neq 0 \) is continuous
So, first we have to prove that \( \mathrm{g}(x)=\sin x \) and \( \mathrm{h}(x)=\cos x \) are continuous functions.
Let \( \mathrm{g}(x)=\sin x \)
We know that \( \mathrm{g}(x)=\sin x \) is defined for every real number.
Let h be a real number. Now, put \( x=\mathrm{k}+\mathrm{h} \)
So, if \( x \rightarrow \mathrm{k} \) and \( \mathrm{h} \rightarrow 0 \)
\(\mathrm{g}(\mathrm{k})=\operatorname{sink}\)
\(\lim _{x \rightarrow k} g(x)=\lim _{x \rightarrow k} \sin x\)
\(=\lim _{h \rightarrow 0} \sin (k+h)\)
\(=\lim _{h \rightarrow 0}[\sin k \cos h+\cos k \sin h]\)
\(=\sin k \cos 0+\cos k \sin 0\)
\(=\sin \mathrm{k}+0\)
\(=\sin k\)
Thus, \( \lim _{x \rightarrow k} g(x)=g(k) \)
Therefore, \( g \) is a continuous function ... (1)
Let \( \mathrm{h}(x)=\cos x \)
We know that \( \mathrm{h}(x)=\cos x \) is defined for every real number.
Let k be a real number. Now, put \( x=\mathrm{k}+\mathrm{h} \)
So, if \( x \rightarrow \mathrm{k} \) and \( \mathrm{h} \rightarrow 0 \)
\(\mathrm{h}(\mathrm{k})=\sin k\)
\(\lim _{x \rightarrow k} h(x)=\lim _{x \rightarrow k} \cos x\)
\(=\lim _{h \rightarrow 0} \cos (k+h)\)
\(=\lim _{h \rightarrow 0}[\cos k \cos h-\sin k \sin h]\)
\(=\cos k \cos 0-\sin k \sin 0\)
\(=\cos \mathrm{k}-0\)
\(=\cos \mathrm{k}\)
\(\text { Thus, } \lim _{x \rightarrow k} h(x)=h(k)\)
Therefore, g is a continuous function ... (2)
So, from (1) and (2), we get,
\( \operatorname{Cosec} x=\frac{1}{\sin x}, \sin x \neq 0 \) is continuous
\( =\operatorname{cosec} x, \neq \mathrm{n} \pi(\mathrm{n} \in \mathrm{Z}) 0 \) is continuous
Thus, cosecant is continuous except at \( x=n p,(n \in Z) \)
\( \operatorname{Sec} x=\frac{1}{\cos x}, \cos x \neq 0 \) is continuous
\( = \sec x, x \neq(2 n+1) \frac{\pi}{2}(n \in Z) \) is continuous
Thus, secant is continuous except at \( x=(2 \mathrm{n}+1) \frac{\pi}{2},(\mathrm{n} \in \mathrm{Z}) \)
\( \operatorname{Cot} x=\frac{\cos x}{\sin x}, \sin x \neq 0 \) is continuous
\( =\cot x, x \neq n \pi(n \in Z) 0 \) is continuous
Thus, cotangent is continuous except at \( x=n p,(n \in Z) \)

It is given that \( \mathrm{f}(x)=\left\{\begin{array}{l}\frac{\sin x}{x}, \text { if } x < 0 \\ x+1, \text { if } x \geq 0\end{array}\right. \)
We know that f is defined at all points of the real line.
Let k be a real number.
Case I: \( \mathrm{k} < 0 \),
Then \( \mathrm{f}(\mathrm{k})=\frac{\sin k}{k} \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}\left(\frac{\sin x}{x}\right)=\frac{\sin k}{k}\)
\(=\lim _{x \rightarrow k} f(x)=f(k)\)
Thus, f is continuous at all points \(x\) that is \( x < 0 \).
Case II: \( \mathrm{k} > 0 \),
Then \( \mathrm{f}(\mathrm{k})=\mathrm{c}+1 \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(x+1)=\mathrm{k}+1\)
\(=\lim _{x \rightarrow k}(x)=f(k)\)
Thus, f is continuous at all points \(x\) that is \( x > 0 \).
Case III: \( \mathrm{k}=0 \)
Then \( \mathrm{f}(\mathrm{k})=\mathrm{f}(0)=0+1=1 \)
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(\frac{\sin x}{x}\right)=1\)
\(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(x+1)=1\)
\(=\lim _{x \rightarrow k^{-}} f(x)=\lim _{x \rightarrow k^{+}} f(x)=f(x)\)
Hence, \( f \) is continuous at \( x=0 \).
Therefore, f is continuous at all points of the real line

It is given that \( \mathrm{f}(x)=\left\{\begin{array}{c}x^{2} \sin \frac{1}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right. \)
We know that f is defined at all points of the real line.
Let k be a real number.
Case I: \( \mathrm{k} \neq 0 \),
Then \( \mathrm{f}(\mathrm{k})=\mathrm{k}^{2} \sin \frac{1}{k} \)
\(\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}\left(x^{2} \sin \frac{1}{x}\right)=\mathrm{k}^{2} \sin \frac{1}{k}\)
\(\therefore \lim _{x \rightarrow k} f(x)=f(k)\)
Thus, f is continuous at all points \(x\) that is \( x \neq 0 \).
Case II: \( \mathrm{k}=0 \)
Then \( f(k)=f(0)=0 \)
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(x^{2} \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)\)
We know that \( -1 \leq x \leq 1, x \neq 0 \)
\(\Rightarrow x^{2} \leq x^{2} \sin \frac{1}{x} \leq 0\)
\(\Rightarrow \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0\)
\(\Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=0\)
Similarly, \( \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x^{2} \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0 \)
\(\lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x)\)
Therefore, f is continuous at \( x=0 \).
Therefore, f has no point of discontinuity.

Given function is
\(f(x)=\left\{\begin{array}{ll}\sin x-\cos x, & \text { if } x \neq 0 \\-1, & \text { if } x=0\end{array}\right.\)
Lef's find the left hand and right hand limits at \( x=0 \).
\( \begin{array}{l}\text { At } x=0 \text {, L.H.L. }=\lim _{x \rightarrow 0{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} f(-h) \\ \Rightarrow \lim _{t \rightarrow 0} \sin (-h)-\cos (-h)=\lim _{h \rightarrow 0}(-\sin h-\cos h)=-0-1=-1\end{array} \)
\( \begin{array}{l}\text { R.H.L. }=\lim _{x \rightarrow 0^{\prime}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} f(h) \\ \Rightarrow \lim _{h \rightarrow 0} \sin (h)-\cos (h)=\lim _{h \rightarrow 0}(\sin h-\cos h)=0-1=-1 \\ \text { And. given } f(0)=-1\end{array} \)
Thus, \( \lim _{x \rightarrow} f(x)=\lim _{f \rightarrow} f(x)=f(0) \)
Therefore, \( f(x) \) is continuous at \( x=0 \).

It is given that \( \mathrm{f}(x)=\left\{\begin{array}{c}\frac{k \cos x}{\pi-2 x}, \text { if } f \neq \frac{\pi}{2} \\ \text { at } x=\frac{\pi}{2} \\ 3, \text { if } x=\frac{\pi}{2}\end{array}\right. \)
Also, it is given that function f is continuous at \( x=\frac{\pi}{2} \),
So, if \( f \) is defined at \( x=\frac{\pi}{2} \) and if the value of the \( f \) at \( x= \) equals the limit of \( \frac{\pi}{2} \mathrm{f} \) at \( x=\frac{\pi}{2} \).
We can see that f is defined at \( x=\frac{\pi}{2} \) and \( \mathrm{f}\left(\frac{\pi}{2}\right)=3 \)
\(\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}\)
Now, let put \( x=\frac{\pi}{2}+h \)
Then, \( x \rightarrow \frac{\pi}{2}=\mathrm{h} \rightarrow 0 \therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)} \)
\(=k \lim_{h \rightarrow 0} \frac{-\sin h}{-2 h}=\frac{k}{2} \lim _{h \rightarrow 0} \frac{-\sin h}{-2 h}=\frac{k}{2} \cdot 1=\frac{k}{2}\)
\(\therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)\)
\(\Rightarrow \frac{k}{2}=3\)
\(\Rightarrow \mathrm{k}=6\)
Therefore, the value of k is 6 .

It is given that \( \mathrm{f}(x)=\left\{\begin{array}{r}k x^{2}, \text { if } x \leq 2 \\ \text { at } x=2 \\ 3, \text { if } x > 2\end{array}\right. \)
Also, it is given that function f is continuous at \( x=2 \),
So, if \( f \) is defined at \( x=2 \) and if the value of the \( f \) at \( x=2 \) equals the limit of f at \( x=2 \).
We can see that f is defined at \( x=2 \) and
\(\mathrm{f}(2)=\mathrm{k}(2)^{2}=4 \mathrm{k}\)
\(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(2)\)
\(\Rightarrow \lim _{x \rightarrow 2^{-}}\left(k x^{2}\right)=\lim _{x \rightarrow 0^{+}}(3)\)
\(\Rightarrow \mathrm{k} \times 2^{2}=3=4 \mathrm{k}\)
\(\Rightarrow 4 \mathrm{k}=3=4 \mathrm{k}\)
\(\Rightarrow 4 \mathrm{k}=3\)
\(\Rightarrow \mathrm{k}=\frac{3}{4}\)
Therefore, the required value of k is \( \frac{3}{4} \).

It is given that \( \mathrm{f}(x)=\left\{\begin{array}{c}k x+1, \text { if } x \leq \pi \\ \cos x, \text { if } x > \pi\end{array}\right. \) at \( x=\pi \)
Also, it is given that function f is continuous at \( x=\mathrm{k} \),
So, if \( f \) is defined at \( x=p \) and if the value of the \( f \) at \( x=k \) equals the limit of \( f \) at \( x=k \).
We can see that f is defined at \( x=\mathrm{p} \) and
\(\mathrm{f}(\pi)=\mathrm{k} \pi+1\)
\(\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{+}} f(x)=f(\pi)\)
\(\Rightarrow \lim _{x \rightarrow \pi^{-}}(k x+1)=\lim _{x \rightarrow \pi^{+}}(\cos x)=\mathrm{k} \pi+1\)
\(\Rightarrow \mathrm{k} \pi+1=\cos \pi=\mathrm{k} \pi+1\)
\(\Rightarrow \mathrm{k} \pi+1=-1=\mathrm{k} \pi+1\)
\(\Rightarrow \mathrm{k}=-\frac{2}{\pi}\)
Therefore, the required value of k is \( -\frac{2}{\pi} \).

It is given that \( \mathrm{f}(x)=\left\{\begin{array}{l}k x+1, \text { if } x \leq 5 \\ 3 x-5, \text { if } x > 5\end{array}\right. \) at \( x=5 \)
Also, it is given that function f is continuous at \( x=5 \),
So, if \( f \) is defined at \( x=5 \) and if the value of the \( f \) at \( x=5 \) equals the limit of f at \( x=5 \).
We can see that f is defined at \( x=5 \) and
\(\mathrm{f}(5)=\mathrm{k}x+1=5 \mathrm{k}+1\)
\(\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{+}} f(x)=f(\pi)\)
\(=\lim _{x \rightarrow 5^{-}}(k x+1)=\lim _{x \rightarrow 5^{+}}(3 x-5)=5 \mathrm{k}+1\)
\(\Rightarrow 5 \mathrm{k}+1=15-5=5 \mathrm{k}+1\)
\(\Rightarrow 5 \mathrm{k}+1=10\)
\(\Rightarrow 5 \mathrm{k}=9\)
\(\Rightarrow \mathrm{k}=\frac{9}{5}\)
Therefore, the required value of k is \( \frac{9}{5} \).

It is given function is \(\mathrm{f}(x)=\left\{\begin{array}{l}
5, \text { if } x \leq 2 \\
a x+b, \text { if } 2 \leq x \leq 10 \\
21, \text { if } x \geq 10
\end{array}\right.\)
We know that the given function f is defined at all points of the real line.
Thus, f is continuous at \( x=2 \), we get,
\(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)\)
\(\Rightarrow \lim _{x \rightarrow 2^{-}}(5)=\lim _{x \rightarrow 2^{+}}(a x+b)=5\)
\(\Rightarrow 5=2 \mathrm{a}+\mathrm{b}=5\)
\(\Rightarrow 2 \mathrm{a}+\mathrm{b}=5 \ldots(1)\)
Thus, f is continuous at \( x=10 \), we get,
\(\lim _{x \rightarrow 10^{-}} f(x)=\lim _{x \rightarrow 10^{+}} f(x)=f(10)\)
\(=\lim _{x \rightarrow 10^{-}}(a x+b)=\lim _{x \rightarrow 10^{+}}(21)=21\)
\(\Rightarrow 10 \mathrm{a}+\mathrm{b}=21=21\)
\(\Rightarrow 10 \mathrm{a}+\mathrm{b}=21 \ldots(2)\)
On subtracting eq(1.) from eq(2.), we get,
\(8 a=16\)
\(\Rightarrow a=2\)
Thus, putting \( \mathrm{a}=2 \) in eq(1.), we get,
\(2 \times 2+b=5\)
\(\Rightarrow 4+b=5\)
\(\Rightarrow b=1\)
Therefore, the values of \( a \) and \( b \) for which \( f \) is a continuous function are 2 and 1 resp.

It is given function is \( \mathrm{f}(x)=\cos \left(x^{2}\right) \)
This function \( f \) is defined for every real number and \( f \) can be written as the composition of two function as,
\( \mathrm{f}= \) goh, where, \( \mathrm{g}(x)=\cos x \) and \( \mathrm{h}(x)=x^{2} \)
First we have to prove that \( \mathrm{g}(x)=\cos x \) and \( \mathrm{h}(x)=x^{2} \) are continuous functions.
We know that g is defined for every real number.
Let k be a real number.
Then, \( g(k)=\cos k \)
Now, put \( x=\mathrm{k}+\mathrm{h} \)
If \( x \rightarrow k \), then \( h \rightarrow 0 \)
\(\lim _{x \rightarrow k} g(k)=\lim _{x \rightarrow k} \cos x\)
\(=\lim _{h \rightarrow 0} \cos (k+h)\)
\(=\lim _{h \rightarrow 0} \cos [\cos k \cos h-\sin k \sin h]\)
\(=\lim _{h \rightarrow 0} \cos k \cos h-\lim _{h \rightarrow 0} \sin k \sin h\)
\(=\operatorname{cosk} \cos 0-\operatorname{sink} \sin 0\)
\(=\cos \mathrm{k} \times 1-\sin \times 0\)
\(=\cos \mathrm{k}\)
\(\therefore \lim _{x \rightarrow k} g(x)=g(k)\)
Thus, \( \mathrm{g}(x)=\cos x \) is continuous function.
Now, \( h(x)=x^{2} \)
So, \( h \) is defined for every real number.
Let c be a real number, then \( \mathrm{h}(\mathrm{c})=\mathrm{c}^{2} \)
\(\lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c} x^{2}\)
\(\lim _{x \rightarrow c} h(x)=h(c)\)
Therefore, h is a continuous function.
We know that for real valued functions g and h,
Such that (fog) is continuous at c.
Therefore, \( \mathrm{f}(x)=(\mathrm{goh})(x)=\cos \left(x^{2}\right) \) is a continuous function.

It is given that \( f(x)=|\cos x| \)
The given function f is defined for real number and f can be written as the composition of two functions, as
\( \mathrm{f}= \) goh, where \( \mathrm{g}(x)=|x| \) and \( \mathrm{h}(x)=\cos x \)
First we have to prove that \( \mathrm{g}(x)=|x| \) and \( \mathrm{h}(x)=\cos x \) are continuous functions.
\( \mathrm{g}(x)=|x| \) can be written as
\(\mathrm{g}(x)=\left\{\begin{array}{c}
-x, \text { if } x < 0 \\
x, \text { if } x \geq 0
\end{array}\right.\)
Now, g is defined for all real number.
Let k be a real number.
Case I: If \( \mathrm{k} < 0 \),
Then \( g(k)=-k \)
And \( \lim _{x \rightarrow k} g(x)=\lim _{x \rightarrow k}(-x)=-\mathrm{k} \)
Thus, \( \lim _{x \rightarrow k} g(x)=g(k) \)
Therefore, g is continuous at all points x , i.e., \( x > 0 \)
Case II: If \( \mathrm{k} > 0 \),
Then \( \mathrm{g}(\mathrm{k})=\mathrm{k} \) and
\(\lim _{x \rightarrow k} g(x)=\lim _{x \rightarrow k} x=\mathrm{k}\)
Thus, \( \lim _{x \rightarrow k} g(x)=g(k) \)
Therefore, g is continuous at all points \(x \), i.e., \( x < 0 \).
Case III: If \( \mathrm{k}=0 \),
Then, \( \mathrm{g}(\mathrm{k})=\mathrm{g}(0)=0 \)
\(\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0\)
\(\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0\)
\(\therefore \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}} g(x)=g(0)\)
Therefore, \(g\) is continuous at \( x=0 \)
From the above 3 cases, we get that g is continuous at all points.
\(h(x)=\cos x\)
We know that h is defined for every real number.
Let k be a real number.
\(\text { Now, put } x=\mathrm{k}+\mathrm{h}\)
\(\text { If } x \rightarrow \mathrm{k} \text {, then } \mathrm{h} \rightarrow 0\)
\(\lim _{x \rightarrow k} h(x)=\lim _{x \rightarrow k} \cos x\)
\(=\lim _{h \rightarrow 0} \cos (k+h)\)
\(=\lim _{h \rightarrow 0} \cos [\cos k \cos h-\sin k \sin h]\)
\(=\lim _{h \rightarrow 0} \cos k \cos h-\lim _{h \rightarrow 0} \sin k \sin h\)
\(=\cos k \cos 0-\operatorname{sink} \sin 0\)
\(=\cos \mathrm{k} \times 1-\sin \times 0\)
\(=\cos \mathrm{k}\)
\(\therefore \lim _{x \rightarrow k} h(k)=h(k)\)
Thus, \( \mathrm{h}(x)=\cos x \) is continuous function.
We know that for real valued functions \( g \) and \( h \), such that (goh) is defined at \( k \), if \( g \) is continuous at \( k \) and if \( f \) is continuous at \( g(k) \),
Then (fog) is continuous at k .
Therefore, \( \mathrm{f}(x)=(\mathrm{gof})(x)=\mathrm{g}(\mathrm{h}(x))=\mathrm{g}(\cos x)=|\cos x| \) is a continuous function.

It is given that \( f(x)=\sin |x| \)
The given function f is defined for real number and f can be written as the composition of two functions, as
\( \mathrm{f}= \) goh, where \( \mathrm{g}(x)=|x| \) and \( \mathrm{h}(x)=\sin x \)
First we have to prove that \( \mathrm{g}(x)=|x| \) and \( \mathrm{h}(x)=\sin x \) are continuous functions.
\( \mathrm{g}(x)=|x| \) can be written as
\(\mathrm{g}(x)=\left\{\begin{array}{c}
-x, \text { if } x < 0 \\
x, \text { if } x \geq 0
\end{array}\right.\)
Now, g is defined for all real number.
Let k be a real number.
Case I: If \( \mathrm{k} < 0 \),
Then \( g(k)=-k \)
And \( \lim _{x \rightarrow k} g(x)=\lim _{x \rightarrow k}(-x)=-\mathrm{k} \)
Thus, \( \lim _{x \rightarrow k} g(x)=g(k) \)
Therefore, \(g\) is continuous at all points \(x \), i.e., \( x > 0 \)
Case II: If \( \mathrm{k} > 0 \),
Then \( \mathrm{g}(\mathrm{k})=\mathrm{k} \) and
\(\lim _{x \rightarrow k} g(x)=\lim _{x \rightarrow k} x=\mathrm{k}\)
Thus, \( \lim _{x \rightarrow k} g(x)=g(k) \)
Therefore, \(g\) is continuous at all points \(x \), i.e., \( x < 0 \).
Case III: If \( \mathrm{k}=0 \),
Then, \( \mathrm{g}(\mathrm{k})=\mathrm{g}(0)=0 \)
\(\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0\)
\(\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0\)
\(\therefore \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}} g(x)=g(0)\)
Therefore, g is continuous at \( x=0 \)
From the above 3 cases, we get that g is continuous at all points.
\(h(x)=\sin x\)
We know that h is defined for every real number.
Let k be a real number.
Now, put \( x=\mathrm{k}+\mathrm{h} \)
If \( x \rightarrow k \), then \( h \rightarrow 0 \)
\(\lim _{x \rightarrow k} h(x)=\lim _{x \rightarrow k} \sin x\)
\(=\lim _{h \rightarrow 0} \sin (k+h)\)
\(=\lim _{h \rightarrow 0}[\sin k \cos h+\cos k \sin h]\)
\(=\lim _{h \rightarrow 0} \sin k \cos h+\lim _{h \rightarrow 0} \cos k \sin h\)
\(=\sin k \cos 0+\cos k \sin 0\)
\(=\sin k\)
\(\therefore \lim _{x \rightarrow k} h(x)=g(k)\)
Thus, \( \mathrm{h}(x)=\cos x \) is continuous function.
We know that for real valued functions \( g \) and \( h \), such that (goh) is defined at \( k \), if \( g \) is continuous at \( k \) and if \( f \) is continuous at \( g(k) \),
Then (fog) is continuous at k .
Therefore, \( \mathrm{f}(x)=(\mathrm{gof})(x)=\mathrm{g}(\mathrm{h}(x))=\mathrm{g}(\sin x)=|\sin x| \) is a continuous function.

It is given that \( f(x)=|x|-|x+1| \)
The given function f is defined for real number and f can be written as the composition of two functions, as
\( \mathrm{f}= \) goh, where \( \mathrm{g}(x)=|x| \) and \( \mathrm{h}(x)=|x+1| \)
Then, \( \mathrm{f}=\mathrm{g}-\mathrm{h} \)
First we have to prove that \( \mathrm{g}(x)=|x| \) and \( \mathrm{h}(x)=|x+1| \) are continuous functions.
\( \mathrm{g}(x)=|x| \) can be written as
\(\mathrm{g}(x)=\left\{\begin{array}{c}
-x, \text { if } x < 0 \\
x, \text { if } x \geq 0
\end{array}\right.\)
Now, \( g \) is defined for all real number.
Let k be a real number.
Case I: If \( \mathrm{k} < 0 \),
Then \( g(k)=-k \)
And \( \lim _{x \rightarrow k} g(x)=\lim _{x \rightarrow k}(-x)=-\mathrm{k} \)
Thus, \( \lim _{x \rightarrow k} g(x)=g(k) \)
Therefore, \(g\) is continuous at all points \(x \), i.e., \( x > 0 \)
Case II: If \( \mathrm{k} > 0 \),
Then \( \mathrm{g}(\mathrm{k})=\mathrm{k} \) and
\(\lim _{x \rightarrow k} g(x)=\lim _{x \rightarrow k} x=\mathrm{k}\)
Thus, \( \lim _{x \rightarrow k} g(x)=g(k) \)
Therefore, \(g\) is continuous at all points \(x \), i.e., \( x < 0 \).
Case III: If \( \mathrm{k}=0 \),
Then, \( \mathrm{g}(\mathrm{k})=\mathrm{g}(0)=0 \)
\(\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0\)
\(\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0\)
\(\therefore \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0\)
Therefore, \(g\) is continuous at \( x=0 \)
From the above 3 cases, we get that \(g\) is continuous at all points.
\( \mathrm{g}(x)=|x+1| \) can be written as
\(g(x)=\left\{\begin{array}{c}
-(x+1), \text { if } x < -1 \\
x+1, \text { if } x \geq-1
\end{array}\right.\)
Now, \(h\) is defined for all real number.
Let k be a real number.
Case I: If \( \mathrm{k} < -1 \),
Then \( h(k)=-(k+1) \)
And \( \lim _{x \rightarrow k} h(x)=\lim _{x \rightarrow k}[-(x+1)]=-(\mathrm{k}+1) \)
Thus, \( \lim _{x \rightarrow k} h(x)=h(k) \)
Therefore, \(h\) is continuous at all points \(x \), i.e., \( x < -1 \)
Case II: If \( \mathrm{k} > -1 \),
Then \( \mathrm{h}(\mathrm{k})=\mathrm{k}+1 \) and
\(\lim _{x \rightarrow k} h(x)=\lim _{x \rightarrow k}(x+1)=\mathrm{k}+1\)
Thus, \( \lim _{x \rightarrow k} h(x)=h(k) \)
Therefore, \(h\) is continuous at all points \(x \), i.e., \( x > -1 \).
Case III: If \( \mathrm{k}=-1 \),
Then, \( \mathrm{h}(\mathrm{k})=\mathrm{h}(-1)=-1+1=0 \)
\(\lim _{x \rightarrow 1^{-}} h(x)=\lim _{x \rightarrow 1^{-}}[-(x+1)]=-(1+1)=0\)
\(\lim _{x \rightarrow 1^{+}} h(x)=\lim _{x \rightarrow 1^{+}}(x+1)=(-1+1)=0\)
\(\therefore \lim _{x \rightarrow 1^{-}} h(x)=\lim _{x \rightarrow 1^{+}} h(x)=h(-1)\)
Therefore, \(g\) is continuous at \( x=-1 \)
From the above 3 cases, we get that h is continuous at all points.
Hence, \(g\) and \(h\) are continuous function.
Therefore, \( \mathrm{f}=\mathrm{g}-\mathrm{h} \) is also a continuous function.