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# If the function f\left( x \right)=\left\{ \begin{align} & 2x,\text{ }\left| x \right|\le 1 \\ & {{x}^{2}}+ax+b,\text{ }\left| x \right|>1 \\ \end{align} \right.Is continuous for all real x, then(a) $a=2,b=-1\text{ and }f\text{ is differentiable for all }x$(b) $a=-2,b=1\text{ and }f\text{ is not differentiable at }x=-1,1$(c) $a=2,b=-1\text{ and }f\text{ is not differentiable at }x=-1,1$(d) $a=-2,b=-1\text{ and }f\text{ is not differentiable at }x=-1,1$ Verified
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Hint: First find a and b by putting $f\left( {{1}^{-}} \right)=f\left( {{1}^{+}} \right)=f\left( 1 \right)$and $f\left( -{{1}^{-}} \right)=f\left( -{{1}^{+}} \right)=f\left( -1 \right)$and then check if ${{f}^{'}}\left( {{1}^{-}} \right)={{f}^{'}}\left( {{1}^{+}} \right)$and ${{f}^{'}}\left( -{{1}^{-}} \right)={{f}^{'}}\left( -{{1}^{+}} \right)$

We are given that
f\left( x \right)=\left\{ \begin{align} & 2x\text{, }\left| x \right|\le 1 \\ & {{x}^{2}}+ax+b,\text{ }\left| x \right|>1 \\ \end{align} \right.
Is continuous for all real x.
We have to check the differentiability of $f\left( x \right)$and also find the values of a and b.
As we know that, \left| x \right|=\left\{ \begin{align} & x,\text{ }x\ge 0 \\ & -x,\text{ }x<0 \\ \end{align} \right.
Therefore, $\left| x \right|\le 1\text{ means }-1\le x\le 1$
And $\left| x \right|>1\text{ means }x>1\text{ and }x<-1$
Therefore, we get
f\left( x \right)=\left\{ \begin{align} & {{x}^{2}}+ax+b,\text{ }x<-1 \\ & 2x,\text{ }-1\le x\le 1 \\ & {{x}^{2}}+ax+b,\text{ }x>1 \\ \end{align} \right.
As we are given that f (x) is continuous for all $x\in R,\text{ therefore }f\left( x \right)$would be continuous for $x=1\text{ and }x=-1$as well.
For $f\left( x \right)$to be continuous at $x=1$
$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 1 \right)....\left( i \right)$
We are given that for $x>1,\text{ }f\left( x \right)={{x}^{2}}+ax+b$
Therefore, $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{\left( 1 \right)}^{2}}+a\left( 1 \right)+b=1+a+b$
Also, we are given that for $-1\le x\le 1\text{ }f\left( x \right)=2x$
Therefore, $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2\left( 1 \right)=2$
Also, $f\left( 1 \right)=2\left( 1 \right)=2$
By putting these values in equation (i)
We get, $1+a+b=2$
Or, $a+b=2-1$
Hence, we get $a+b=1.....\left( ii \right)$
Now, for $f\left( x \right)$to be continuous at $x=-1$
$\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( -1 \right)....\left( iii \right)$
We are given that for $x\ge -1,\text{ }f\left( x \right)=2x$
Therefore, $\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=2\left( -1 \right)=-2$
Also, $f\left( -1 \right)=2\left( -1 \right)=-2$
Also, we are given that for $x<-1,\text{ }f\left( x \right)={{x}^{2}}+ax+b$
Therefore,$\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,={{\left( -1 \right)}^{2}}+a\left( -1 \right)+b$$=1-a+b$
By putting these values in equation (iii)
We get, $-2=1-a+b=-2$
Or, $1-a+b=-2$
$a-b=3....\left( iv \right)$
Taking equation (ii) and (iv) together
That is, $a+b=1....\left( v \right)$
$a-b=3....\left( vi \right)$
We get $\left( a+b \right)+\left( a-b \right)=4$
$\Rightarrow 2a=4$
Therefore, we get $a=2$
By putting the values of a in equation (v), we get
\begin{align} & 2+b=1 \\ & b=1-2 \\ \end{align}
Therefore, we get $b=-1$
Therefore we get, f\left( x \right)=\left\{ \begin{align} & {{x}^{2}}+2x-1,\text{ }x<-1 \\ & 2x,\text{ }-1\le x\le 1 \\ & {{x}^{2}}+2x-1,\text{ }x>1 \\ \end{align} \right.
Now to check the differentiability of f (x), we will differentiate f (x) with respect to x.
Since, we know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
We get, {{f}^{'}}\left( x \right)=\left\{ \begin{align} & 2x+2,\text{ }x<-1 \\ & 2,\text{ }-1 & 2x+2,\text{ }x>1 \\ \end{align} \right.
For f (x) to be differentiable at x = -1
$\underset{x\to {{\left( -1 \right)}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{\left( -1 \right)}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\text{finite quantity}$
For $x<-1,\text{ }{{f}^{'}}\left( x \right)=2x+2$
Therefore, $\underset{x\to {{\left( -1 \right)}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=2\left( -1 \right)+2=0$
For, $x>-1,\text{ }{{f}^{'}}\left( x \right)=2$
Therefore, $\underset{x\to {{\left( -1 \right)}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=2$
Since, $\underset{x\to {{\left( -1 \right)}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\ne \underset{x\to {{\left( -1 \right)}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)$, therefore f (x) is not differentiable at x = -1.
Also, for f (x) to be differentiable at x = 1
$\underset{x\to {{\left( 1 \right)}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{\left( 1 \right)}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\text{finite quantity}$
For $x<1,\text{ }{{f}^{'}}\left( x \right)=2$
Therefore, $\underset{x\to {{\left( 1 \right)}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=2$
$x>1,\text{ }{{f}^{'}}\left( x \right)=2x+2$
Therefore, $\underset{x\to {{\left( 1 \right)}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=2+2=4$
Since, $\underset{x\to {{\left( 1 \right)}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\ne \underset{x\to {{\left( 1 \right)}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)$, therefore f (x) is not differentiable at x = 1.
Therefore, a = 2, b = -1 and f is not differentiable at x = -1, 1
Hence, option (c) is correct.

Note: Students should always remember to expand the modulus function first because they often mistake taking |x|< 1 as x < 1 and |x|> 1 as x > 1 but actually |x|< 1 means -1 < x < 1 and |x| > 1 means x > 1 and x < -1