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

Correct answer: C. 3 f(x)=(x^2+1)|x^2-a x+2|+ |x| Notice that (-x)= x= |x| which means |x| is differentiable everywhere in x ∈ R aligned & f(x) can be non differentiable where |x^2-a x+2| & =0 aligned x^2-a x+2=0 4-2 a+2=0 a=3 (x^2-3 x+2)=0 x=1,2 β=1 distance of (α, β) from line aligned & 12 x+5 y+10=0 & (|2(12)+5(1)+10|)/(13)=(39)/(13)=3 aligned

JEE Main 202529 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Continuity and Differentiability

Let the function f(x)=(x^2+1)|x^2-a x+2|+ |x| be not differentiable...

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

12 x+5 y+10=0

is equal to :

Options

Correct answer

C. 3

Step-by-step solution

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

\begin{aligned} & \Rightarrow f(x) \text{ can be non differentiable where }\left|x^2-a x+2\right| \\ & =0 \end{aligned}

\Rightarrow x^2-a x+2=0 \Rightarrow 4-2 a+2=0 \Rightarrow a=3 \Rightarrow \quad\left(x^2-3 x+2\right)=0 \quad \Rightarrow \quad x=1,2 \beta=1

distance of

(\alpha, \beta)

from line

\begin{aligned} & 12 x+5 y+10=0 \\ & \Rightarrow \frac{|2(12)+5(1)+10|}{13}=\frac{39}{13}=3 \end{aligned}

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