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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) \[a=-2,b=-1\text{ and }f\text{ is not differentiable at }x=-1,1\]
Answer
![Verified]()
Verified
Verified
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, \[a+b=1....\left( v \right)\]
\[a-b=3....\left( vi \right)\]
Adding these 2 equations,
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
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, \[a+b=1....\left( v \right)\]
\[a-b=3....\left( vi \right)\]
Adding these 2 equations,
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, \[{{f}^{'}}\left( x \right)=\left\{ \begin{align}
& 2x+2,\text{ }x<-1 \\
& 2,\text{ }-1
\end{align} \right.\]
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
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