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Functions question

2023 · 11 Apr · Shift 2 · Q36
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Functions question

2023 · 11 Apr · Shift 2 · Q36

JEE MainMathematicsFunctionsMCQ+4 / −1
The domain of the function f(x)=1[x]2−3[x]−10f(x)=\frac{1}{\sqrt{[x]^{2}-3[x]-10}}f(x)=[x]2−3[x]−10​1​ is : ( where [x][\mathrm{x}][x] denotes the greatest integer less than or equal to xxx )
  1. A
    (−∞,−2)∪[6,∞)(-\infty,-2) \cup[6, \infty)(−∞,−2)∪[6,∞)
  2. B
    (−∞,−3]∪[6,∞)(-\infty,-3] \cup[6, \infty)(−∞,−3]∪[6,∞)
  3. C
    (−∞,−2)∪(5,∞)(-\infty,-2) \cup(5, \infty)(−∞,−2)∪(5,∞)
  4. D
    (−∞,−3]∪(5,∞)(-\infty,-3] \cup(5, \infty)(−∞,−3]∪(5,∞)
View written solutionFree

Correct answer: A

  1. We need the function
f(x)=1[x]2−3[x]−10f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}f(x)=[x]2−3[x]−10​1​

to be defined.

Since the square root is in the denominator, the expression inside the root must be strictly positive:

[x]2−3[x]−10>0.[x]^2-3[x]-10>0.[x]2−3[x]−10>0.
  1. Let
[x]=n,[x]=n,[x]=n,

where nnn is an integer. Then we solve

n2−3n−10>0.n^2-3n-10>0.n2−3n−10>0.

Factorizing,

n2−3n−10=(n−5)(n+2).n^2-3n-10=(n-5)(n+2).n2−3n−10=(n−5)(n+2).

So the inequality becomes

(n−5)(n+2)>0.(n-5)(n+2)>0.(n−5)(n+2)>0.
  1. This product is positive when both factors are positive or both are negative:
  • n>5n>5n>5, or
  • n<−2n<-2n<−2.

Since nnn is an integer, this means

n≥6orn≤−3.n\ge 6 \quad \text{or} \quad n\le -3.n≥6orn≤−3.
  1. Now translate back in terms of xxx.

  • If [x]≥6[x]\ge 6[x]≥6, then x∈[6,∞).x\in [6,\infty).x∈[6,∞).

  • If [x]≤−3[x]\le -3[x]≤−3, then all real numbers with greatest integer at most −3-3−3 are x<−2.x< -2.x<−2. Indeed, for example, x=−2.5x=-2.5x=−2.5 gives [x]=−3[x]=-3[x]=−3, but x=−2x=-2x=−2 gives [x]=−2[x]=-2[x]=−2, which is not allowed. Hence this part is (−∞,−2).(-\infty,-2).(−∞,−2).

  1. Therefore the domain is
(−∞,−2)∪[6,∞).(-\infty,-2)\cup[6,\infty).(−∞,−2)∪[6,∞).
  1. Checking options:
  • A: (−∞,−2)∪[6,∞)(-\infty,-2)\cup[6,\infty)(−∞,−2)∪[6,∞) ✅
  • B: includes x=−2x=-2x=−2, but at x=−2x=-2x=−2, [x]=−2[x]=-2[x]=−2 and (−2)2−3(−2)−10=4+6−10=0,(-2)^2-3(-2)-10=4+6-10=0,(−2)2−3(−2)−10=4+6−10=0, denominator becomes 000, so not allowed. ❌
  • C: includes (5,6)(5,6)(5,6), but for 5<x<65<x<65<x<6, [x]=5[x]=5[x]=5 and 25−15−10=0,25-15-10=0,25−15−10=0, not allowed. ❌
  • D: has both above mistakes. ❌

Hence the correct option is A.

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