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JEE Mathematics 2026 Question with Solution

Let A={x:∣x2−10∣≤6}A = \{x : |x^2 - 10| \le 6\} and B={x:∣x−2∣>1}B = \{x : |x - 2| > 1\}. Then

  • A

    A−B=[2,3]A - B = [2, 3]

  • B

    A∩B=[−4,−2]∪[3,4]A \cap B = [-4, -2] \cup [3, 4]

  • C

    B−A=(−∞,−4)∪(−2,1)∪(4,∞)B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)

  • D

    A∪B=(−∞,1]∪(2,∞)A \cup B = (-\infty, 1] \cup (2, \infty)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A={x:∣x2−10∣≤6}A = \{x : |x^2 - 10| \le 6\} and B={x:∣x−2∣>1}B = \{x : |x - 2| > 1\}.

Find: Which set relation is correct.

For set AA,

−6≤x2−10≤6-6 \le x^2 - 10 \le 6

So,

4≤x2≤164 \le x^2 \le 16

Hence,

2≤∣x∣≤42 \le |x| \le 4

Therefore,

A=[−4,−2]∪[2,4]A = [-4, -2] \cup [2, 4]

For set BB,

∣x−2∣>1|x - 2| > 1

This gives

x−2>1 or x−2<−1x - 2 > 1 \text{ or } x - 2 < -1

So,

x>3 or x<1x > 3 \text{ or } x < 1

Therefore,

B=(−∞,1)∪(3,∞)B = (-\infty, 1) \cup (3, \infty)

Now evaluate B−AB - A. This means elements of BB that are not in AA. The complement used in the solution is

Ac=(−∞,−4)∪(−2,2)∪(4,∞)A^c = (-\infty, -4) \cup (-2, 2) \cup (4, \infty)

Thus,

B∩Ac=((−∞,1)∪(3,∞))∩((−∞,−4)∪(−2,2)∪(4,∞))B \cap A^c = \left(( -\infty, 1) \cup (3, \infty)\right) \cap \left(( -\infty, -4) \cup (-2, 2) \cup (4, \infty)\right)

which gives

(−∞,−4)∪(−2,1)∪(4,∞)(-\infty, -4) \cup (-2, 1) \cup (4, \infty)

So,

B−A=(−∞,−4)∪(−2,1)∪(4,∞)B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)

Therefore, the correct option is C.

The solution also notes that option A is technically correct since

A−B=[2,3]A - B = [2, 3]

but the extracted solution concludes that option C is the correct representation.

Checking the options one by one

From the solved sets,

A=[−4,−2]∪[2,4],B=(−∞,1)∪(3,∞)A = [-4, -2] \cup [2, 4], \qquad B = (-\infty, 1) \cup (3, \infty)

Option A:

A−B=A∩BcA - B = A \cap B^c

Since

Bc=[1,3]B^c = [1, 3]

we get

A−B=([−4,−2]∪[2,4])∩[1,3]=[2,3]A - B = ([-4, -2] \cup [2, 4]) \cap [1, 3] = [2, 3]

So this statement is true.

Option B:

A∩B=([−4,−2]∪[2,4])∩((−∞,1)∪(3,∞))A \cap B = ([-4, -2] \cup [2, 4]) \cap ((-\infty, 1) \cup (3, \infty))

Hence,

A∩B=[−4,−2]∪(3,4]A \cap B = [-4, -2] \cup (3, 4]

This is not equal to [−4,−2]∪[3,4][-4, -2] \cup [3, 4], so option B is false because 3∉B3 \notin B.

Option C:

B−A=(−∞,−4)∪(−2,1)∪(4,∞)B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)

This matches the derived result, so option C is true.

Option D:

A∪B=((−∞,1)∪(3,∞))∪([−4,−2]∪[2,4])A \cup B = ((-\infty, 1) \cup (3, \infty)) \cup ([-4, -2] \cup [2, 4])

which simplifies to

(−∞,1)∪[2,∞)(-\infty, 1) \cup [2, \infty)

This is not equal to (−∞,1]∪(2,∞)(-\infty, 1] \cup (2, \infty), so option D is false.

Thus, from the provided solution, the selected answer is C, even though option A also evaluates correctly.

Common mistakes

  • A common mistake is solving ∣x2−10∣≤6|x^2 - 10| \le 6 as only x2−10≤6x^2 - 10 \le 6. This ignores the lower bound from the absolute value. Use −6≤x2−10≤6-6 \le x^2 - 10 \le 6 before solving.

  • Students often write ∣x−2∣>1|x - 2| > 1 as 1<x<31 < x < 3, which is the interval for ∣x−2∣<1|x - 2| < 1. For a greater-than absolute value inequality, the solution lies outside the interval, so x<1x < 1 or x>3x > 3.

  • Another mistake is including boundary points incorrectly. Since ∣x−2∣>1|x - 2| > 1 is a strict inequality, x=1x = 1 and x=3x = 3 are not in BB. This is why intervals for BB are open at those points.

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