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
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
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