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MCQMediumJEE 2025Differentiability

JEE Mathematics 2025 Question with Solution

Let the function f(x)=(x2−1)∣x2−ax+2∣+cos⁡∣x∣f(x) = (x^2 - 1)|x^2 - ax + 2| + \cos|x| be not differentiable at the two points x=α=2x = \alpha = 2 and x=βx = \beta. Then the distance of the point (α,β)(\alpha, \beta) from the line 12x+5y+10=012x + 5y + 10 = 0 is equal to:

  • A

    33

  • B

    44

  • C

    22

  • D

    55

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: f(x)=(x2−1)∣x2−ax+2∣+cos⁡∣x∣f(x) = (x^2 - 1)|x^2 - ax + 2| + \cos|x| is not differentiable at two points x=α=2x=\alpha=2 and x=βx=\beta.

Find: The distance of the point (α,β)(\alpha,\beta) from the line 12x+5y+10=012x + 5y + 10 = 0.

For absolute value functions, points of non-differentiability occur where the expression inside the absolute value becomes zero. Also, cos⁡∣x∣=cos⁡x\cos|x| = \cos x, so it is differentiable for all real xx. Therefore, we only need to consider

x2−ax+2=0x^2 - ax + 2 = 0

Since one root is given as α=2\alpha = 2, substitute x=2x=2:

4−2a+2=04 - 2a + 2 = 0 6−2a=06 - 2a = 0 a=3a = 3

Now the quadratic becomes

x2−3x+2=0x^2 - 3x + 2 = 0

Factoring,

(x−1)(x−2)=0(x-1)(x-2)=0

So the other root is

β=1\beta = 1

Hence the point is

(α,β)=(2,1)(\alpha,\beta) = (2,1)

Using the distance formula from the line 12x+5y+10=012x + 5y + 10 = 0,

d=∣12(2)+5(1)+10∣122+52d = \frac{|12(2) + 5(1) + 10|}{\sqrt{12^2 + 5^2}} d=∣24+5+10∣144+25d = \frac{|24+5+10|}{\sqrt{144+25}} d=39169=3913=3d = \frac{39}{\sqrt{169}} = \frac{39}{13} = 3

Therefore, the distance is 33, so the correct option is A.

Common mistakes

  • Mistake: Treating cos⁡∣x∣\cos|x| as non-differentiable at x=0x=0 because of the modulus. Why it is wrong: cos⁡∣x∣=cos⁡x\cos|x| = \cos x since cosine is an even function, so this term is differentiable everywhere. Do instead: check non-differentiability only from the factor ∣x2−ax+2∣|x^2-ax+2|.

  • Mistake: Forgetting that the non-differentiable points come from the roots of the expression inside the modulus. Why it is wrong: ∣g(x)∣|g(x)| can fail to be differentiable where g(x)=0g(x)=0. Do instead: first solve x2−ax+2=0x^2-ax+2=0 and use the given root α=2\alpha=2 to find aa.

  • Mistake: Using the point-to-line distance formula incorrectly by not taking absolute value in the numerator. Why it is wrong: distance is always non-negative. Do instead: use d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} carefully with the point (2,1)(2,1) and line 12x+5y+10=012x+5y+10=0.

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