Let the function f(x) = (x^2-1) |x^2-ax+2| + cos|x| be not differentiable at the two points x=α=2 and x=β.

Question

Let the function \(f(x)=\left(x^{2}-1\right)\left|x^{2}-a x+2\right|+\cos |x|\) 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
 

Answer 1

Correct option is (1) 3  

\(f(x)=\left(x^2-1\right)\left|x^2-a x+2\right|+\cos |x|\)

Notice that \(\cos (-x)=\cos x=\cos |x|\) which means \(\cos |x|\) is differentiable

everywhere in \(x \in R\)

\(\Rightarrow f(x)\) can be non differentiable where \(\left|x^2-a x+2\right|\)

= 0

\( \Rightarrow 4-2 a+2=0 \Rightarrow a=3 \)

\(\Rightarrow\left(x^2-3 x+2\right)=0 \quad \Rightarrow x=1,2 \)

\( \beta=1\)  

but f(x) is differentiable at x = 1.

so it should be bonus.

distance of \((\alpha, \beta)\) from line

\(12 x+5 y+10=0 \)

\(\Rightarrow \frac{|2(12)+5(1)+10|}{13}=\frac{39}{13}=3\)