Question

If the function

& 2x,\text{ }\left| x \right|\le 1 \\\ & {{x}^{2}}+ax+b,\text{ }\left| x \right|>1 \\\ \end{aligned} \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 1

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

& 2x\text{, }\left| x \right|\le 1 \\\ & {{x}^{2}}+ax+b,\text{ }\left| x \right|>1 \\\ \end{aligned} \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{aligned} & x,\text{ }x\ge 0 \\\ & -x,\text{ }x<0 \\\ \end{aligned} \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{aligned} & {{x}^{2}}+ax+b,\text{ }x<-1 \\\ & 2x,\text{ }-1\le x\le 1 \\\ & {{x}^{2}}+ax+b,\text{ }x>1 \\\ \end{aligned} \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 $$\begin{aligned} & 2+b=1 \\\ & b=1-2 \\\ \end{aligned}$$ Therefore, we get $$b=-1$$ Therefore we get, $$f\left( x \right)=\left\\{ \begin{aligned} & {{x}^{2}}+2x-1,\text{ }x<-1 \\\ & 2x,\text{ }-1\le x\le 1 \\\ & {{x}^{2}}+2x-1,\text{ }x>1 \\\ \end{aligned} \right.$$ 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, $${{f}^{'}}\left( x \right)=\left\\{ \begin{aligned} & 2x+2,\text{ }x<-1 \\\ & 2,\text{ }-1 & 2x+2,\text{ }x>1 \\\ \end{aligned} \right.$$ 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