Question ID: #550
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:
- (1) 3
- (2) 4
- (3) 2
- (4) 5
Solution:
$f(x) = (x^2 – 1)|x^2 – ax + 2| + \cos|x|$
The term $\cos|x|$ is differentiable everywhere. The non-differentiability arises from the modulus term.
Let $g(x) = x^2 – ax + 2$. The points of non-differentiability occur where $g(x) = 0$.
Given that $x = \alpha = 2$ is one such point:
$(2)^2 – a(2) + 2 = 0 \Rightarrow 6 – 2a = 0 \Rightarrow a = 3$.
So, $x^2 – 3x + 2 = 0 \Rightarrow (x-1)(x-2) = 0$.
The roots are $x=2$ and $x=1$.
However, at $x=1$, the coefficient $(x^2 – 1)$ becomes 0.
$f(x) = (x-1)(x+1)|(x-1)(x-2)| + \cos|x|$.
Due to the factor $(x-1)$ outside the modulus, the function becomes differentiable at $x=1$.
Thus, there is only one point of non-differentiability ($x=2$). The question states there are **two** points ($\alpha$ and $\beta$), which makes the data inconsistent/incorrect.
Ans. (Bonus)
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