Q.25. The number of points, at which the function , , is not differentiable, is

Q: What is the correct answer for JEE Main Q25?

The correct answer is 3. Sol. First, consider $g(x) = \max \{6x, 2 + 3x^2\}$. The points where $6x = 2 + 3x^2$ are $3x^2 - 6x + 2 = 0$. Using the quadratic formula, $x = \frac{6 \pm \sqrt{36 - 4(3)(2)}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$. These are two points where $g(x)$ might be non-differentiable. Let $h(x) = |x-1| \cos \left(\frac{\pi}{x-1}\right)$. This function is

Q: Q.25. The number of points, at which the function , , is not differentiable, is

The correct answer is 3. Sol. First, consider $g(x) = \max \{6x, 2 + 3x^2\}$. The points where $6x = 2 + 3x^2$ are $3x^2 - 6x + 2 = 0$. Using the quadratic formula, $x = \frac{6 \pm \sqrt{36 - 4(3)(2)}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$. These are two points where $g(x)$ might be non-differentiable. Let $h(x) = |x-1| \cos \left(\frac{\pi}{x-1}\right)$. This function is

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Q.25. The number of points, at which the function , , is not differentiable, is

FirstInTest · Accepted Answer

Sol. First, consider $g(x) = \max \{6x, 2 + 3x^2\}$. The points where $6x = 2 + 3x^2$ are $3x^2 - 6x + 2 = 0$. Using the quadratic formula, $x = \frac{6 \pm \sqrt{36 - 4(3)(2)}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$. These are two points where $g(x)$ might be non-differentiable. Let $h(x) = |x-1| \cos \left(\frac{\pi}{x-1}ight)$. This function is not defined at $x=1$. However, the problem states $x \in (-\pi, \pi)$, so $x=1$ is a point of discontinuity. For $x 
e 1$, $h(x)$ is differentiable. The function $f(x)$ is the sum of $g(x)$ and $h(x)$. The points of non-differentiability for $f(x)$ are the points where $g(x)$ is non-differentiable and $x=1$ (due to discontinuity). The points $1 \pm \frac{\sqrt{3}}{3}$ are within $(-\pi, \pi)$. So, there are 3 points of non-differentiability: $1 - \frac{\sqrt{3}}{3}$, $1 + \frac{\sqrt{3}}{3}$, and $1$.

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