Let f : R \to R be a function defined by f(x) = max x, x2. Let S denote the set of all points in R, where f is not differentiable. Then :

Limits Continuity and Differentiability PYQ — JEE Main 2020 · 6 Sep · Shift 2 · Q24 with Solution

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Correct answer: A

We are given $f(x)=\max\{x,x^2\}$ and we need the set of points where $f$ is not differentiable.

1. Compare $x$ and $x^2$

To determine $f(x)$ , we check when $x \ge x^2$ or equivalently, $x-x^2 \ge 0$ $x(1-x)\ge 0$ This holds for $x\in [0,1].$

Thus:

for $x\in [0,1]$ , we have $x\ge x^2$ , so $f(x)=x$
for $x<0$ or $x>1$ , we have $x^2>x$ , so $f(x)=x^2$

Hence,

\begin{cases} x^2, & x<0 \\ x, & 0\le x\le 1 \\ x^2, & x>1 \end{cases}$$ ## 2. Check differentiability on each interval On the open intervals: - $(-\infty,0)$, $f(x)=x^2$, which is differentiable. - $(0,1)$, $f(x)=x$, which is differentiable. - $(1,\infty)$, $f(x)=x^2$, which is differentiable. So possible non-differentiable points are only where the formula changes: $x=0$ and $x=1$. ## 3. Check differentiability at $x=0$ ### Left derivative: For $x<0$, $f(x)=x^2$, so $$f'_-(0)=\left.\frac{d}{dx}(x^2)\right|_{x=0}=0.$$ ### Right derivative: For $x>0$, $f(x)=x$, so $$f'_+(0)=\left.\frac{d}{dx}(x)\right|_{x=0}=1.$$ Since $$f'_-(0)\ne f'_+(0),$$ $f$ is **not differentiable at $x=0$**. ## 4. Check differentiability at $x=1$ ### Left derivative: For $x<1$ (near 1), $f(x)=x$, so $$f'_-(1)=1.$$ ### Right derivative: For $x>1$, $f(x)=x^2$, so $$f'_+(1)=\left.2x\right|_{x=1}=2.$$ Since $$f'_-(1)\ne f'_+(1),$$ $f$ is **not differentiable at $x=1$**. ## 5. Final set Therefore, the set of all points where $f$ is not differentiable is $$S=\{0,1\}.$$ So the correct option is **A**.

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1. Compare

x

and

x^2

To determine

f(x)

, we check when

x \ge x^2

or equivalently,

x-x^2 \ge 0

x(1-x)\ge 0

This holds for

x\in [0,1].

Thus:

for

x\in [0,1]

, we have

x\ge x^2

, so

f(x)=x

for

x<0

x>1

, we have

x^2>x

, so

f(x)=x^2

Hence,

\begin{cases} x^2, & x<0 \\ x, & 0\le x\le 1 \\ x^2, & x>1 \end{cases}$$ ## 2. Check differentiability on each interval On the open intervals: - $(-\infty,0)$, $f(x)=x^2$, which is differentiable. - $(0,1)$, $f(x)=x$, which is differentiable. - $(1,\infty)$, $f(x)=x^2$, which is differentiable. So possible non-differentiable points are only where the formula changes: $x=0$ and $x=1$. ## 3. Check differentiability at $x=0$ ### Left derivative: For $x<0$, $f(x)=x^2$, so $$f'_-(0)=\left.\frac{d}{dx}(x^2)\right|_{x=0}=0.$$ ### Right derivative: For $x>0$, $f(x)=x$, so $$f'_+(0)=\left.\frac{d}{dx}(x)\right|_{x=0}=1.$$ Since $$f'_-(0)\ne f'_+(0),$$ $f$ is **not differentiable at $x=0$**. ## 4. Check differentiability at $x=1$ ### Left derivative: For $x<1$ (near 1), $f(x)=x$, so $$f'_-(1)=1.$$ ### Right derivative: For $x>1$, $f(x)=x^2$, so $$f'_+(1)=\left.2x\right|_{x=1}=2.$$ Since $$f'_-(1)\ne f'_+(1),$$ $f$ is **not differentiable at $x=1$**. ## 5. Final set Therefore, the set of all points where $f$ is not differentiable is $$S=\{0,1\}.$$ So the correct option is **A**.

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