Let the function $f(x) = (x^2 - 1)|x^2 - ax + 2| + \cos|x| $ be not differentiable at the two points $ x = \alpha = 2 $ and $ x = \beta $. Then the distance of the point $(\alpha, \beta)$ from the line $12x + 5y + 10 = 0$ is equal to:

Question

Let $f(x)=|\log_e x|-|x-1|+5$. Statement 1: $f(x)$ is differentiable for all $x\in(0,\infty)$
Statement 2: $f(x)$ is increasing in $(1,\infty)$
Statement 3: $f(x)$ is decreasing in $(0,1)$
Which of the following is correct?

Answer 1

Step 1: Identifying non-differentiable points. The function $\cos|x|$ is universally differentiable. Consequently, the focus shifts to identifying points where $|x^2 - ax + 2|$ is non-differentiable. This occurs when the expression inside the absolute value is zero. Set the inner expression to zero: \[x^2 - ax + 2 = 0\] Given that one root is $\alpha = 2$, substitute this value into the equation: \[4 - 2a + 2 = 0 \implies a = 3\] With $a = 3$, the quadratic equation becomes $x^2 - 3x + 2 = 0$, and its roots are $\alpha = 2$ and $\beta = 1$.

Step 2: Finding the distance from the line. The coordinates of the point are $(\alpha, \beta) = (2, 1)$. Apply the point-to-line distance formula to find the distance $d$ from the point $(2, 1)$ to the line $12x + 5y + 10 = 0$: \[d = \frac{|12(2) + 5(1) + 10|}{\sqrt{12^2 + 5^2}} = \frac{|24 + 5 + 10|}{\sqrt{144 + 25}} = \frac{39}{\sqrt{169}} = \frac{39}{13} = 3\]

Let the function \(f(x) = (x^2 - 1)|x^2 - ax + 2| + \cos|x| \) be not differentiable at the two points \( x = \alpha = 2 \) and \( x = \beta \). Then the distance of the point \((\alpha, \beta)\) from the line \(12x + 5y + 10 = 0\) is equal to:

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The Correct Option is A

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