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JEE Mathematics 2026 Question with Solution

If f(x)={a∣x∣+x2−2(sin⁡∣x∣)(cos⁡∣x∣)x,x≠0b,x=0f(x) = \begin{cases} \frac{a|x| + x^2 - 2(\sin|x|)(\cos|x|)}{x} & , x \neq 0\\ b & , x = 0 \end{cases} is continuous at x=0x=0, then a+ba+b is equal to

  • A

    00

  • B

    11

  • C

    44

  • D

    22

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

f(x)={a∣x∣+x2−2(sin⁡∣x∣)(cos⁡∣x∣)x,x≠0 b,x=0f(x) = \begin{cases} \frac{a|x| + x^2 - 2(\sin|x|)(\cos|x|)}{x}, & x \neq 0 \\\ b, & x = 0 \end{cases}

Find: The value of a+ba+b such that f(x)f(x) is continuous at x=0x=0.

Using 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta = \sin 2\theta, for x≠0x \neq 0 we get

f(x)=a∣x∣+x2−sin⁡(2∣x∣)xf(x) = \frac{a|x| + x^2 - \sin(2|x|)}{x}

For the right hand limit, when x→0+x \to 0^+, we have ∣x∣=x|x|=x. Therefore,

lim⁡x→0+ax+x2−sin⁡(2x)x=a+0−2=a−2\lim_{x \to 0^+} \frac{ax + x^2 - \sin(2x)}{x} = a + 0 - 2 = a-2

For the left hand limit, when x→0−x \to 0^-, we have ∣x∣=−x|x|=-x. Therefore,

lim⁡x→0−−ax+x2−sin⁡(−2x)x\lim_{x \to 0^-} \frac{-ax + x^2 - \sin(-2x)}{x}

Since sin⁡(−2x)=−sin⁡(2x)\sin(-2x)=-\sin(2x),

lim⁡x→0−−ax+x2+sin⁡(2x)x=−a+0+2=−a+2\lim_{x \to 0^-} \frac{-ax + x^2 + \sin(2x)}{x} = -a + 0 + 2 = -a+2

For continuity at x=0x=0, the right hand limit and left hand limit must be equal.

a−2=−a+2a-2 = -a+2

So,

2a=42a=4 a=2a=2

Now the common limit is

a−2=2−2=0a-2 = 2-2 = 0

Since f(0)=bf(0)=b and the function is continuous at x=0x=0, we must have

b=0b=0

Thus,

a+b=2+0=2a+b = 2+0 = 2

Therefore, the correct option is D.

Direct continuity check

Given: The function is continuous at x=0x=0.

Find: a+ba+b.

Near x=0x=0, use sin⁡(2∣x∣)∼2∣x∣\sin(2|x|) \sim 2|x|. Then for x≠0x \neq 0,

f(x)=a∣x∣+x2−sin⁡(2∣x∣)x∼a∣x∣+x2−2∣x∣xf(x) = \frac{a|x| + x^2 - \sin(2|x|)}{x} \sim \frac{a|x| + x^2 - 2|x|}{x}

For x>0x>0, this becomes

(a−2)x+x2x→a−2\frac{(a-2)x + x^2}{x} \to a-2

For x<0x<0, this becomes

(2−a)(−x)+x2x→−a+2\frac{(2-a)(-x) + x^2}{x} \to -a+2

Equating both gives

a−2=−a+2⇒a=2a-2=-a+2 \Rightarrow a=2

Then the limit at x=0x=0 is 00, so by continuity b=0b=0. Hence a+b=2a+b=2, and the correct option is D.

Common mistakes

  • Taking ∣x∣=x|x|=x on both sides of 00 is incorrect. For x<0x<0, ∣x∣=−x|x|=-x. Always evaluate the right hand limit and left hand limit separately when absolute value is present.

  • Not using the identity 2sin⁡∣x∣cos⁡∣x∣=sin⁡(2∣x∣)2\sin|x|\cos|x| = \sin(2|x|) makes the limit harder to simplify. First rewrite the trigonometric term into a standard form before taking limits.

  • Equating the common limit directly to bb before ensuring the left and right limits are equal is incomplete. First check RHL=LHL\text{RHL} = \text{LHL}, then use continuity to set that common value equal to bb.

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