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Limits Continuity and Differentiability question

2019 · 9 Apr · Shift 1 · Q25
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  5. /2019 · 9 Apr · Shift 1 · Q25

Limits Continuity and Differentiability question

2019 · 9 Apr · Shift 1 · Q25

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let ƒ(x) = 15 – |x – 10|; x ∈\in∈ R. Then the set of all values of x, at which the function, g(x) = ƒ(ƒ(x)) is not differentiable, is :
  1. A
    {10,15}
  2. B
    {5,10,15,20}
  3. C
    {10}
  4. D
    {5,10,15}
View written solutionFree

Correct answer: D

  1. Given function

We have f(x)=15−∣x−10∣.f(x)=15-|x-10|.f(x)=15−∣x−10∣.

First write f(x)f(x)f(x) in piecewise form:

\begin{cases} 15-(10-x)=x+5, & x<10,\\[4pt] 15-(x-10)=25-x, & x\ge 10. \end{cases}$$ So $f(x)$ is not differentiable at $$x=10$$ because of the absolute value corner. --- 2. **Define $g(x)$** $$g(x)=f(f(x)).$$ A composition $f(f(x))$ can fail to be differentiable when: - the inner function $f(x)$ is not differentiable, i.e. at $x=10$, or - the outer function $f$ is applied at its non-differentiable input, i.e. when $$f(x)=10.$$ So we must find all $x$ such that $$f(x)=10.$$ --- 3. **Solve $f(x)=10$** Given $$15-|x-10|=10,$$ we get $$|x-10|=5.$$ Hence $$x=5 \quad \text{or} \quad x=15.$$ So possible non-differentiability points are $$x\in\{5,10,15\}.$$ --- 4. **Verify by explicit piecewise computation** Let us compute $g(x)$ on intervals. Since $$f(x)=15-|x-10|,$$ its range is $(-\infty,15]$. Now check whether $f(x)<10$ or $f(x)\ge 10$. From above: - if $x<5$ or $x>15$, then $f(x)<10$, - if $5\le x\le 15$, then $f(x)\ge 10$. Now use $$f(y)=\begin{cases} y+5, & y<10,\\ 25-y, & y\ge 10. \end{cases}$$ So: - For $x<5$ or $x>15$, $f(x)<10$, hence $$g(x)=f(x)+5=(15-|x-10|)+5=20-|x-10|.$$ - For $5\le x\le 15$, $f(x)\ge 10$, hence $$g(x)=25-f(x)=25-(15-|x-10|)=10+|x-10|.$$ Now split further: $$g(x)=\begin{cases} 20-(10-x)=x+10, & x<5,\\ 10+(10-x)=20-x, & 5\le x<10,\\ 10+(x-10)=x, & 10\le x\le 15,\\ 20-(x-10)=30-x, & x>15. \end{cases}$$ Thus slopes are: - for $x<5$: slope $=1$, - for $5<x<10$: slope $=-1$, - for $10<x<15$: slope $=1$, - for $x>15$: slope $=-1$. Hence derivative changes at $$x=5,\;10,\;15.$$ So $g(x)$ is not differentiable exactly at $$\{5,10,15\}.$$ --- 5. **Check options** - **A:** $\{10,15\}$ — missing $5$, so false. - **B:** $\{5,10,15,20\}$ — extra $20$, so false. - **C:** $\{10\}$ — incomplete, so false. - **D:** $\{5,10,15\}$ — correct. --- 6. **Final answer** The set of points where $g(x)=f(f(x))$ is not differentiable is $$\boxed{\{5,10,15\}}.$$
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