JEE Main 2025 Jan 29 Shift 2, Mathematics Q5: Differentiability Of A Function

JEE Main2025Jan 29, Shift 2Mathematics

Let the function $f (x) = (x^{2} + 1) ∣ x^{2} - a x + 2∣ + cos ∣ x ∣$ 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 :

A
$3$
B
$4$
C
$2$
D
$5$

Solution

Since $f(x)$ is not differentiable at $x=\alpha=2$, the expression inside the modulus must vanish at $x=2$.

$2^{2} - 2 a + 2 = 0$

$\Rightarrow 6 - 2 a = 0$

$\Rightarrow a = 3$

Hence,

$x^{2} - a x + 2 = x^{2} - 3 x + 2 = (x - 1) (x - 2)$

Therefore,

$f (x) = (x^{2} + 1) ∣ (x - 1) (x - 2) ∣ + cos ∣ x ∣$

Since $x^{2} + 1 \neq = 0$ for all $x$ , the function is not differentiable at the zeros of $(x - 1) (x - 2)$ .

$∴ α = 2, β = 1$

The required point is $(α, β) = (2, 1)$ .

Distance from the line $12 x + 5 y + 10 = 0$ is

$\frac{∣12 ( 2 ) + 5 ( 1 ) + 10∣}{1 2 ^{2} + 5 ^{2}} = \frac{39}{13} = 3$

$3$

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q5: Differentiability Of A Function

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