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Hint: Use the definition of continuity and differentiability of the functions and check each of the given options. Also, take any function which satisfies the given conditions and then check if the given statement is true or false.

Complete step-by-step answer:

We have to check if each of the given options is true or false based on the definition of continuity.

We will check the first statement which says that if \[f\left( x \right)\] is not continuous at \[x=a\], then it is not differentiable at \[x=a\].

We know that a function is differentiable at \[x=a\] if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}\] exists and \[f'\left( a \right)=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}\].

We can rewrite \[f\left( x \right)\] as \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,f'\left( a \right)\left( x-a \right)+f\left( a \right)\]. As \[f\left( x \right)\] is not continuous, \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\] does not exist.

Thus, \[\underset{x\to a}{\mathop{\lim }}\,f'\left( a \right)\left( x-a \right)+f\left( a \right)\] does not exist as well. As \[\underset{x\to a}{\mathop{\lim }}\,\left( x-a \right)\] and \[\underset{x\to a}{\mathop{\lim }}\,f\left( a \right)\] exists, it means \[\underset{x\to a}{\mathop{\lim }}\,f'\left( a \right)\] does not exist.

We can also prove this by considering functions \[f\left( x \right)=2\] for \[x\ge 0\] and \[f\left( x \right)=x+1\] for \[x<0\].

As we can see that \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2\], \[f\left( x \right)\] is not continuous at \[x=0\].

Also, we observe \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 0 \right)}{x}=0\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 0 \right)}{x}=1\]. Thus, \[f\left( x \right)\] is not differentiable at \[x=0\].

Hence, the first statement is true.

Now, we will consider the second statement which states that if \[f\left( x \right)\] is continuous at \[x=a\], then it is differentiable at \[x=a\].

This statement is not always true. Consider the function \[f\left( x \right)=\left| x \right|\]. We observe that \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=0=f\left( 0 \right)\]. Thus, \[f\left( x \right)\] is continuous at \[x=0\]. However, we have \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 0 \right)}{x}=-1\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 0 \right)}{x}=1\]. Thus, \[f\left( x \right)\] is not differentiable at \[x=0\].

Hence, the second statement is not always true.

We will now consider third statement which states that if \[f\left( x \right)\] and \[g\left( x \right)\] are differentiable at \[x=a\], then \[f\left( x \right)+g\left( x \right)\] is also differentiable at \[x=a\].

This statement is always true as we have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=f'\left( a \right),\underset{x\to a}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( a \right)}{x-a}=g'\left( a \right)\], thus, we have \[f'\left( a \right)+g'\left( a \right)=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)+g\left( x \right)-f\left( a \right)-g\left( a \right)}{x-a}\]. So, \[f\left( x \right)+g\left( x \right)\] is differentiable as well.

Consider the function \[f\left( x \right)=x,g\left( x \right)={{x}^{2}}\]. We have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=f'\left( a \right)=1,\underset{x\to a}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( a \right)}{x-a}=g'\left( a \right)=2a\]. Also, we observe that \[f'\left( a \right)+g'\left( a \right)=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)+g\left( x \right)-f\left( a \right)-g\left( a \right)}{x-a}=1+2a\]. Thus, \[f\left( x \right)+g\left( x \right)\] is differentiable as well.

Hence, the third statement is always true.

We will now consider fourth statement which says that if a function \[f\left( x \right)\] is continuous at \[x=a\], then \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\] exists.

This is the definition of continuity of a function which states that \[f\left( x \right)\] is continuous at \[x=a\] if \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)\].

We can also prove this by considering the function \[f\left( x \right)=x\]. As \[f\left( x \right)=x\] is continuous at \[x=a\], we have \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)=a\]. Thus, \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\] exists as well.

Hence, the fourth statement is always true.

Thus, the only statement which is not always true is the second statement, which is option (b).

Note: To solve this question, one must keep the definition of continuity and differentiability of any function in mind. A function \[f\left( x \right)\] is continuous at \[x=a\] if \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)\] holds. A function \[f\left( x \right)\] is differentiable at \[x=a\] if \[f'\left( a \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}\] holds.

Complete step-by-step answer:

We have to check if each of the given options is true or false based on the definition of continuity.

We will check the first statement which says that if \[f\left( x \right)\] is not continuous at \[x=a\], then it is not differentiable at \[x=a\].

We know that a function is differentiable at \[x=a\] if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}\] exists and \[f'\left( a \right)=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}\].

We can rewrite \[f\left( x \right)\] as \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,f'\left( a \right)\left( x-a \right)+f\left( a \right)\]. As \[f\left( x \right)\] is not continuous, \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\] does not exist.

Thus, \[\underset{x\to a}{\mathop{\lim }}\,f'\left( a \right)\left( x-a \right)+f\left( a \right)\] does not exist as well. As \[\underset{x\to a}{\mathop{\lim }}\,\left( x-a \right)\] and \[\underset{x\to a}{\mathop{\lim }}\,f\left( a \right)\] exists, it means \[\underset{x\to a}{\mathop{\lim }}\,f'\left( a \right)\] does not exist.

We can also prove this by considering functions \[f\left( x \right)=2\] for \[x\ge 0\] and \[f\left( x \right)=x+1\] for \[x<0\].

As we can see that \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2\], \[f\left( x \right)\] is not continuous at \[x=0\].

Also, we observe \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 0 \right)}{x}=0\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 0 \right)}{x}=1\]. Thus, \[f\left( x \right)\] is not differentiable at \[x=0\].

Hence, the first statement is true.

Now, we will consider the second statement which states that if \[f\left( x \right)\] is continuous at \[x=a\], then it is differentiable at \[x=a\].

This statement is not always true. Consider the function \[f\left( x \right)=\left| x \right|\]. We observe that \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=0=f\left( 0 \right)\]. Thus, \[f\left( x \right)\] is continuous at \[x=0\]. However, we have \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 0 \right)}{x}=-1\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( 0 \right)}{x}=1\]. Thus, \[f\left( x \right)\] is not differentiable at \[x=0\].

Hence, the second statement is not always true.

We will now consider third statement which states that if \[f\left( x \right)\] and \[g\left( x \right)\] are differentiable at \[x=a\], then \[f\left( x \right)+g\left( x \right)\] is also differentiable at \[x=a\].

This statement is always true as we have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=f'\left( a \right),\underset{x\to a}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( a \right)}{x-a}=g'\left( a \right)\], thus, we have \[f'\left( a \right)+g'\left( a \right)=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)+g\left( x \right)-f\left( a \right)-g\left( a \right)}{x-a}\]. So, \[f\left( x \right)+g\left( x \right)\] is differentiable as well.

Consider the function \[f\left( x \right)=x,g\left( x \right)={{x}^{2}}\]. We have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=f'\left( a \right)=1,\underset{x\to a}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( a \right)}{x-a}=g'\left( a \right)=2a\]. Also, we observe that \[f'\left( a \right)+g'\left( a \right)=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)+g\left( x \right)-f\left( a \right)-g\left( a \right)}{x-a}=1+2a\]. Thus, \[f\left( x \right)+g\left( x \right)\] is differentiable as well.

Hence, the third statement is always true.

We will now consider fourth statement which says that if a function \[f\left( x \right)\] is continuous at \[x=a\], then \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\] exists.

This is the definition of continuity of a function which states that \[f\left( x \right)\] is continuous at \[x=a\] if \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)\].

We can also prove this by considering the function \[f\left( x \right)=x\]. As \[f\left( x \right)=x\] is continuous at \[x=a\], we have \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)=a\]. Thus, \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\] exists as well.

Hence, the fourth statement is always true.

Thus, the only statement which is not always true is the second statement, which is option (b).

Note: To solve this question, one must keep the definition of continuity and differentiability of any function in mind. A function \[f\left( x \right)\] is continuous at \[x=a\] if \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)\] holds. A function \[f\left( x \right)\] is differentiable at \[x=a\] if \[f'\left( a \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( a \right)}{x-a}\] holds.

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