Q.22. Let and . If the number of points where g is not continuous and is not differentiable are an

Q: What is the correct answer for JEE Main Mathematics Q22?

The correct answer is 4. Sol. First, analyze $f(x)$. For $x < 0$, $f(x) = \frac{e^x-1}{x}$. This function is continuous and differentiable for $x < 0$. As $x \to 0^-$, $\lim_{x \to 0^-} \frac{e^x-1}{x} = 1$ (standard limit). For $x \ge 0$, $f(x) = x^2-5x+6 = (x-2)(x-3)$. This is a polynomial, so it's continuous and differentiable for $x \ge 0$. At $x=0$: $f(0) = 0^2-5(0)+6 = 6$. $\lim_{x \to 0^-} f(x) = 1$. Since $\lim_{x

Q: Q.22. Let and . If the number of points where g is not continuous and is not differentiable are and respectively, then is equal to ________.

The correct answer is 4. Sol. First, analyze $f(x)$. For $x < 0$, $f(x) = \frac{e^x-1}{x}$. This function is continuous and differentiable for $x < 0$. As $x \to 0^-$, $\lim_{x \to 0^-} \frac{e^x-1}{x} = 1$ (standard limit). For $x \ge 0$, $f(x) = x^2-5x+6 = (x-2)(x-3)$. This is a polynomial, so it's continuous and differentiable for $x \ge 0$. At $x=0$: $f(0) = 0^2-5(0)+6 = 6$. $\lim_{x \to 0^-} f(x) = 1$. Since $\lim_{x

Question

Q.22. Let  and . If the number of points where g is not continuous and is not differentiable are  and  respectively, then  is equal to ________.

FirstInTest · Accepted Answer

Sol. First, analyze $f(x)$.
For $x < 0$, $f(x) = \frac{e^x-1}{x}$. This function is continuous and differentiable for $x < 0$.
As $x 	o 0^-$, $\lim_{x 	o 0^-} \frac{e^x-1}{x} = 1$ (standard limit).
For $x \ge 0$, $f(x) = x^2-5x+6 = (x-2)(x-3)$. This is a polynomial, so it's continuous and differentiable for $x \ge 0$.
At $x=0$:
$f(0) = 0^2-5(0)+6 = 6$.
$\lim_{x 	o 0^-} f(x) = 1$.
Since $\lim_{x 	o 0^-} f(x) 
e f(0)$, $f(x)$ is discontinuous at $x=0$.
Now, analyze $f(|x|)$.
$f(|x|) = \begin{cases} \frac{e^{|x|}-1}{|x|} & , |x| < 0 \ |x|^2-5|x|+6 & , |x| \ge 0 \end{cases}$.
Since $|x| \ge 0$ for all $x$, the first case is never met. So $f(|x|) = |x|^2-5|x|+6 = x^2-5|x|+6$.
This function is continuous everywhere. It is not differentiable at $x=0$ because of $|x|$.
Now, analyze $|f(x)|$.
$f(x) = \begin{cases} \frac{e^x-1}{x} & , x < 0 \ (x-2)(x-3) & , x \ge 0 \end{cases}$.
For $x < 0$, $e^x-1 < 0$ and $x < 0$, so $\frac{e^x-1}{x} > 0$. Thus $|f(x)| = f(x)$ for $x < 0$.
For $x \ge 0$, $f(x) = (x-2)(x-3)$.
$f(x) \ge 0$ for $x \in [0, 2] \cup [3, \infty)$.
$f(x) < 0$ for $x \in (2, 3)$.
So $|f(x)| = \begin{cases} f(x) & , x < 0 \ f(x) & , x \in [0, 2] \cup [3, \infty) \ -f(x) & , x \in (2, 3) \end{cases}$.
$|f(x)|$ is continuous everywhere except where $f(x)$ is discontinuous. $f(x)$ is discontinuous at $x=0$.
So $|f(x)|$ is discontinuous at $x=0$.
$|f(x)|$ is not differentiable at $x=0$ (due to discontinuity), and at $x=2, x=3$ where $f(x)$ changes sign and $f'(x) 
e 0$.
Now, $g(x) = f(|x|) + |f(x)|$.
Points of discontinuity for $g(x)$:
$f(|x|)$ is continuous everywhere.
$|f(x)|$ is discontinuous at $x=0$.
So $g(x)$ is discontinuous at $x=0$. Thus $\alpha = 1$ (one point of discontinuity).
Points of non-differentiability for $g(x)$:
$f(|x|) = x^2-5|x|+6$ is not differentiable at $x=0$.
$|f(x)|$ is not differentiable at $x=0$ (discontinuity), $x=2$, $x=3$.
So $g(x)$ is not differentiable at $x=0, 2, 3$.
Thus $\beta = 3$ (three points of non-differentiability).
Therefore, $\alpha + \beta = 1 + 3 = 4$.

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