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JEE Advanced 2014 · Continuity and Differentiability

Let f:R→Rf: R \to R and g:R→Rg: R \to R be respectively given by f(x)=∣x∣+1f(x) = |x| + 1 and g(x)=x2+1.g(x) = x^{2} + 1. Define h:R→Rh: R \to R by h(x)=max⁡{f(x),g(x)}h(x) = \max\{f(x), g(x)\} if x≤0,x \le 0, and h(x)=min⁡{f(x),g(x)}h(x) = \min\{f(x), g(x)\} if x>0.x > 0. The number of points at which h(x) is not differentiable is ____.

Answer: 33

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Let f:R → R and g:R → R be respectively given by f(x)=|x|+1 and g(x)=x²+1. Define h:R → R by h(x)= max{f(x),g(x)} if x ≤ 0, min{f(x),g(x)} if x > 0. The number of points at which h(x) is not differentiable is ______. f(x)=∣x∣+1,g(x)=x2+1f(x) = |x| + 1, \quad g(x) = x^2 + 1 h(x)={max⁡(f,g),x≤0min⁡(f,g),x>0h(x) = \begin{cases} \max(f,g), & x \le 0 \\ \min(f,g), & x > 0 \end{cases} Find: number of points where h is not differentiable\text{Find: number of points where } h \text{ is not differentiable} f(x)=g(x)⇒∣x∣=x2⇒x=−1,0,1f(x) = g(x) \Rightarrow |x| = x^2 \Rightarrow x = -1, 0, 1 g−f=x2−∣x∣={x(x+1),x≤0x(x−1),x≥0g - f = x^2 - |x| = \begin{cases} x(x+1), & x \le 0 \\ x(x-1), & x \ge 0 \end{cases} h(x)={x2+1,x≤−11−x,−1≤x≤0x2+1,0<x≤1x+1,x≥1h(x) = \begin{cases} x^2+1, & x \le -1 \\ 1-x, & -1 \le x \le 0 \\ x^2+1, & 0 < x \le 1 \\ x+1, & x \ge 1 \end{cases} h is smooth inside each piece; check the junctions x=−1,0,1h \text{ is smooth inside each piece; check the junctions } x = -1, 0, 1 x=−1: h−′(−1)=2x∣−1=−2,h+′(−1)=−1x = -1: \ h'_-(-1) = 2x\big|_{-1} = -2, \quad h'_+(-1) = -1 −2≠−1⇒not differentiable-2 \neq -1 \Rightarrow \text{not differentiable} x=0: h−′(0)=−1,h+′(0)=2x∣0=0x = 0: \ h'_-(0) = -1, \quad h'_+(0) = 2x\big|_0 = 0 −1≠0⇒not differentiable-1 \neq 0 \Rightarrow \text{not differentiable} x=1: h−′(1)=2x∣1=2,h+′(1)=1x = 1: \ h'_-(1) = 2x\big|_1 = 2, \quad h'_+(1) = 1 2≠1⇒not differentiable2 \neq 1 \Rightarrow \text{not differentiable} h not differentiable at x=−1,0,1⇒number=3h \text{ not differentiable at } x = -1, 0, 1 \Rightarrow \text{number} = 3

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