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Limits Continuity and Differentiability question

2015 · Shift 1 · Q39
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  2. /JEE Advanced
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  4. /Limits Continuity and Differentiability
  5. /2015 · Shift 1 · Q39

Limits Continuity and Differentiability question

2015 · Shift 1 · Q39

JEE AdvancedMathematicsLimits Continuity and DifferentiabilityMultiple correct+4 / −2
Let g:R→Rg:R \to Rg:R→R be a differentiable function with g(0)=0g(0) = 0g(0)=0, g′(0)=0g'(0) = 0g′(0)=0 and g′(1)e0g'(1) e 0g′(1)e0. Let f(x)={x∣x∣g(x),xe00,x=0f(x) = \left\{ {\begin{matrix} {{x \over {|x|}}g(x),} & {x e 0} \\ {0,} & {x = 0} \\ \end{matrix} } \right.f(x)={∣x∣x​g(x),0,​xe0x=0​ and h(x)=e∣x∣h(x) = {e^{|x|}}h(x)=e∣x∣ for all x∈Rx \in Rx∈R. Let (f ∘ h)(x)(f\, \circ \,h)(x)(f∘h)(x) denote f(h(x))f(h(x))f(h(x)) and (h ∘ f)(x)(h\, \circ \,f)(x)(h∘f)(x) denote f(f(x))f(f(x))f(f(x)). Then which of the following is (are) true?
  1. A
    f is differentiable at x = 0.
  2. B
    h is differentiable at x = 0.
  3. C
    f ∘ hf\, \circ \,hf∘h is differentiable at x = 0.
  4. D
    h ∘ fh\, \circ \,fh∘f is differentiable at x = 0.
View written solutionFree

Correct answer: A, D

Step-by-step analysis of each option:

Given functions:

  1. g:R→Rg:R \to Rg:R→R is a differentiable function with g(0)=0g(0) = 0g(0)=0, g′(0)=0g'(0) = 0g′(0)=0, and g′(1)≠0g'(1) \ne 0g′(1)=0.
  2. f(x)={x∣x∣g(x),x≠00,x=0f(x) = \left\{ \begin{matrix} \frac{x}{|x|}g(x), & x \ne 0 \\ 0, & x = 0 \end{matrix} \right.f(x)={∣x∣x​g(x),0,​x=0x=0​. This can be rewritten as: f(x)={g(x),x>0−g(x),x<00,x=0f(x) = \left\{ \begin{matrix} g(x), & x > 0 \\ -g(x), & x < 0 \\ 0, & x=0 \end{matrix} \right.f(x)=⎩⎨⎧​g(x),−g(x),0,​x>0x<0x=0​.
  3. h(x)=e∣x∣h(x) = e^{|x|}h(x)=e∣x∣ for all x∈Rx \in Rx∈R. This can be rewritten as: h(x)={ex,x≥0e−x,x<0h(x) = \left\{ \begin{matrix} e^x, & x \ge 0 \\ e^{-x}, & x < 0 \end{matrix} \right.h(x)={ex,e−x,​x≥0x<0​.

A: f is differentiable at x = 0.

To check the differentiability of f(x) at x=0, we calculate the left-hand derivative (LHD) and right-hand derivative (RHD) at x=0.

RHD at x=0: f′(0+)=lim⁡k→0+f(0+k)−f(0)k=lim⁡k→0+f(k)−0kf'(0^+) = \lim_{k \to 0^+} \frac{f(0+k) - f(0)}{k} = \lim_{k \to 0^+} \frac{f(k) - 0}{k}f′(0+)=limk→0+​kf(0+k)−f(0)​=limk→0+​kf(k)−0​ For k>0k > 0k>0, f(k)=g(k)f(k) = g(k)f(k)=g(k). f′(0+)=lim⁡k→0+g(k)kf'(0^+) = \lim_{k \to 0^+} \frac{g(k)}{k}f′(0+)=limk→0+​kg(k)​ Since ggg is differentiable at 0, we know g′(0)=lim⁡k→0g(k)−g(0)kg'(0) = \lim_{k \to 0} \frac{g(k) - g(0)}{k}g′(0)=limk→0​kg(k)−g(0)​. Given g(0)=0g(0) = 0g(0)=0, this becomes g′(0)=lim⁡k→0g(k)kg'(0) = \lim_{k \to 0} \frac{g(k)}{k}g′(0)=limk→0​kg(k)​. Given g′(0)=0g'(0) = 0g′(0)=0, so f′(0+)=g′(0)=0f'(0^+) = g'(0) = 0f′(0+)=g′(0)=0.

LHD at x=0: f′(0−)=lim⁡k→0−f(0+k)−f(0)k=lim⁡k→0−f(k)−0kf'(0^-) = \lim_{k \to 0^-} \frac{f(0+k) - f(0)}{k} = \lim_{k \to 0^-} \frac{f(k) - 0}{k}f′(0−)=limk→0−​kf(0+k)−f(0)​=limk→0−​kf(k)−0​ For k<0k < 0k<0, f(k)=−g(k)f(k) = -g(k)f(k)=−g(k). f′(0−)=lim⁡k→0−−g(k)k=−lim⁡k→0−g(k)kf'(0^-) = \lim_{k \to 0^-} \frac{-g(k)}{k} = - \lim_{k \to 0^-} \frac{g(k)}{k}f′(0−)=limk→0−​k−g(k)​=−limk→0−​kg(k)​ This limit is also g′(0)g'(0)g′(0). So, f′(0−)=−g′(0)f'(0^-) = -g'(0)f′(0−)=−g′(0). Given g′(0)=0g'(0) = 0g′(0)=0, so f′(0−)=−0=0f'(0^-) = -0 = 0f′(0−)=−0=0.

Since f′(0+)=f′(0−)=0f'(0^+) = f'(0^-) = 0f′(0+)=f′(0−)=0, the function f(x) is differentiable at x = 0. Therefore, option A is true.

B: h is differentiable at x = 0.

To check the differentiability of h(x)=e∣x∣h(x) = e^{|x|}h(x)=e∣x∣ at x=0, we calculate its LHD and RHD.

RHD at x=0: For x>0x > 0x>0, h(x)=exh(x) = e^xh(x)=ex. The derivative is h′(x)=exh'(x) = e^xh′(x)=ex. h′(0+)=lim⁡x→0+ex=e0=1h'(0^+) = \lim_{x \to 0^+} e^x = e^0 = 1h′(0+)=limx→0+​ex=e0=1.

LHD at x=0: For x<0x < 0x<0, h(x)=e−xh(x) = e^{-x}h(x)=e−x. The derivative is h′(x)=−e−xh'(x) = -e^{-x}h′(x)=−e−x. h′(0−)=lim⁡x→0−−e−x=−e0=−1h'(0^-) = \lim_{x \to 0^-} -e^{-x} = -e^0 = -1h′(0−)=limx→0−​−e−x=−e0=−1.

Since h′(0+)≠h′(0−)h'(0^+) \ne h'(0^-)h′(0+)=h′(0−), the function h(x) is not differentiable at x = 0. Therefore, option B is false.

C: f o h is differentiable at x = 0.

Let F(x)=(f∘h)(x)=f(h(x))=f(e∣x∣)F(x) = (f \circ h)(x) = f(h(x)) = f(e^{|x|})F(x)=(f∘h)(x)=f(h(x))=f(e∣x∣). We check the differentiability of F(x)F(x)F(x) at x=0x=0x=0. Note that h(x)=e∣x∣≥1h(x) = e^{|x|} \ge 1h(x)=e∣x∣≥1 for all xxx. Since the argument of fff is always positive, we use the definition f(y)=g(y)f(y) = g(y)f(y)=g(y) for y>0y > 0y>0. So, F(x)=f(e∣x∣)=g(e∣x∣)F(x) = f(e^{|x|}) = g(e^{|x|})F(x)=f(e∣x∣)=g(e∣x∣).

Let's find the LHD and RHD of F(x)F(x)F(x) at x=0x=0x=0. F(0)=g(e∣0∣)=g(e0)=g(1)F(0) = g(e^{|0|}) = g(e^0) = g(1)F(0)=g(e∣0∣)=g(e0)=g(1).

RHD at x=0: F′(0+)=lim⁡k→0+F(k)−F(0)k=lim⁡k→0+g(e∣k∣)−g(1)kF'(0^+) = \lim_{k \to 0^+} \frac{F(k) - F(0)}{k} = \lim_{k \to 0^+} \frac{g(e^{|k|}) - g(1)}{k}F′(0+)=limk→0+​kF(k)−F(0)​=limk→0+​kg(e∣k∣)−g(1)​ For k>0k > 0k>0, ∣k∣=k|k|=k∣k∣=k, so F′(0+)=lim⁡k→0+g(ek)−g(1)kF'(0^+) = \lim_{k \to 0^+} \frac{g(e^k) - g(1)}{k}F′(0+)=limk→0+​kg(ek)−g(1)​. Using the chain rule, this is the derivative of g(ex)g(e^x)g(ex) at x=0x=0x=0. The derivative is g′(ex)⋅exg'(e^x) \cdot e^xg′(ex)⋅ex. At x=0x=0x=0, this is g′(e0)⋅e0=g′(1)⋅1=g′(1)g'(e^0) \cdot e^0 = g'(1) \cdot 1 = g'(1)g′(e0)⋅e0=g′(1)⋅1=g′(1).

LHD at x=0: F′(0−)=lim⁡k→0−F(k)−F(0)k=lim⁡k→0−g(e∣k∣)−g(1)kF'(0^-) = \lim_{k \to 0^-} \frac{F(k) - F(0)}{k} = \lim_{k \to 0^-} \frac{g(e^{|k|}) - g(1)}{k}F′(0−)=limk→0−​kF(k)−F(0)​=limk→0−​kg(e∣k∣)−g(1)​ For k<0k < 0k<0, ∣k∣=−k|k|=-k∣k∣=−k, so F′(0−)=lim⁡k→0−g(e−k)−g(1)kF'(0^-) = \lim_{k \to 0^-} \frac{g(e^{-k}) - g(1)}{k}F′(0−)=limk→0−​kg(e−k)−g(1)​. Using the chain rule, this is the derivative of g(e−x)g(e^{-x})g(e−x) at x=0x=0x=0. The derivative is g′(e−x)⋅(−e−x)g'(e^{-x}) \cdot (-e^{-x})g′(e−x)⋅(−e−x). At x=0x=0x=0, this is g′(e0)⋅(−e0)=g′(1)⋅(−1)=−g′(1)g'(e^0) \cdot (-e^0) = g'(1) \cdot (-1) = -g'(1)g′(e0)⋅(−e0)=g′(1)⋅(−1)=−g′(1).

For F(x)F(x)F(x) to be differentiable at x=0x=0x=0, we must have F′(0+)=F′(0−)F'(0^+) = F'(0^-)F′(0+)=F′(0−), which means g′(1)=−g′(1)g'(1) = -g'(1)g′(1)=−g′(1). This implies 2g′(1)=02g'(1) = 02g′(1)=0, so g′(1)=0g'(1)=0g′(1)=0. However, the problem states that g′(1)≠0g'(1) \ne 0g′(1)=0. Thus, F′(0+)≠F′(0−)F'(0^+) \ne F'(0^-)F′(0+)=F′(0−), and f o h is not differentiable at x = 0. Therefore, option C is false.

D: h o f is differentiable at x = 0.

Let G(x)=(h∘f)(x)=h(f(x))=e∣f(x)∣G(x) = (h \circ f)(x) = h(f(x)) = e^{|f(x)|}G(x)=(h∘f)(x)=h(f(x))=e∣f(x)∣. We check the differentiability of G(x)G(x)G(x) at x=0x=0x=0. G(0)=h(f(0))=h(0)=e∣0∣=1G(0) = h(f(0)) = h(0) = e^{|0|} = 1G(0)=h(f(0))=h(0)=e∣0∣=1.

The derivative at x=0x=0x=0 is given by: G′(0)=lim⁡k→0G(k)−G(0)k=lim⁡k→0e∣f(k)∣−1kG'(0) = \lim_{k \to 0} \frac{G(k) - G(0)}{k} = \lim_{k \to 0} \frac{e^{|f(k)|} - 1}{k}G′(0)=limk→0​kG(k)−G(0)​=limk→0​ke∣f(k)∣−1​.

We can rewrite the limit as: G′(0)=lim⁡k→0(e∣f(k)∣−1∣f(k)∣⋅∣f(k)∣k)G'(0) = \lim_{k \to 0} \left( \frac{e^{|f(k)|} - 1}{|f(k)|} \cdot \frac{|f(k)|}{k} \right)G′(0)=limk→0​(∣f(k)∣e∣f(k)∣−1​⋅k∣f(k)∣​) Since fff is differentiable at x=0x=0x=0, it is continuous at x=0x=0x=0. As k→0k \to 0k→0, f(k)→f(0)=0f(k) \to f(0) = 0f(k)→f(0)=0. Let u=∣f(k)∣u = |f(k)|u=∣f(k)∣. As k→0k \to 0k→0, u→0u \to 0u→0. The first part of the limit becomes lim⁡u→0eu−1u=1\lim_{u \to 0} \frac{e^u - 1}{u} = 1limu→0​ueu−1​=1. So, we need to evaluate the limit of the second part: L=lim⁡k→0∣f(k)∣kL = \lim_{k \to 0} \frac{|f(k)|}{k}L=limk→0​k∣f(k)∣​.

Let's check the RHD and LHD for this limit.

RHD for L: lim⁡k→0+∣f(k)∣k=lim⁡k→0+∣g(k)∣k\lim_{k \to 0^+} \frac{|f(k)|}{k} = \lim_{k \to 0^+} \frac{|g(k)|}{k}limk→0+​k∣f(k)∣​=limk→0+​k∣g(k)∣​. Since k>0k>0k>0, this is lim⁡k→0+∣g(k)∣∣k∣=lim⁡k→0+∣g(k)k∣=∣g′(0)∣=∣0∣=0\lim_{k \to 0^+} \frac{|g(k)|}{|k|} = \lim_{k \to 0^+} \left| \frac{g(k)}{k} \right| = |g'(0)| = |0| = 0limk→0+​∣k∣∣g(k)∣​=limk→0+​​kg(k)​​=∣g′(0)∣=∣0∣=0.

LHD for L: lim⁡k→0−∣f(k)∣k=lim⁡k→0−∣−g(k)∣k=lim⁡k→0−∣g(k)∣k\lim_{k \to 0^-} \frac{|f(k)|}{k} = \lim_{k \to 0^-} \frac{|-g(k)|}{k} = \lim_{k \to 0^-} \frac{|g(k)|}{k}limk→0−​k∣f(k)∣​=limk→0−​k∣−g(k)∣​=limk→0−​k∣g(k)∣​. Since k<0k<0k<0, k=−∣k∣k=-|k|k=−∣k∣, this is lim⁡k→0−∣g(k)∣−∣k∣=−lim⁡k→0−∣g(k)k∣=−∣g′(0)∣=−∣0∣=0\lim_{k \to 0^-} \frac{|g(k)|}{-|k|} = -\lim_{k \to 0^-} \left| \frac{g(k)}{k} \right| = -|g'(0)| = -|0| = 0limk→0−​−∣k∣∣g(k)∣​=−limk→0−​​kg(k)​​=−∣g′(0)∣=−∣0∣=0.

Since both the left and right hand limits are 0, L=lim⁡k→0∣f(k)∣k=0L = \lim_{k \to 0} \frac{|f(k)|}{k} = 0L=limk→0​k∣f(k)∣​=0. Therefore, G′(0)=1⋅L=1⋅0=0G'(0) = 1 \cdot L = 1 \cdot 0 = 0G′(0)=1⋅L=1⋅0=0. Since the derivative exists, h o f is differentiable at x = 0. Therefore, option D is true.

Conclusion:

Options A and D are true. Options B and C are false.

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