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Limits Continuity and Differentiability question

2024 · 27 Jan · Shift 1 · Q39
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  5. /2024 · 27 Jan · Shift 1 · Q39

Limits Continuity and Differentiability question

2024 · 27 Jan · Shift 1 · Q39

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Consider the function. f(x)={a(7x−12−x2)b∣x2−7x+12∣,x3 b,x=3,f(x)=\left\{\begin{array}{cc} \frac{\mathrm{a}\left(7 x-12-x^2\right)}{\mathrm{b}\left|x^2-7 x+12\right|} & , x3 \\\\ \mathrm{~b} & , x=3, \end{array}\right.f(x)=⎩⎨⎧​b∣x2−7x+12∣a(7x−12−x2)​ b​,x3,x=3,​ where [x][x][x] denotes the greatest integer less than or equal to xxx. If S\mathrm{S}S denotes the set of all ordered pairs (a, b) such that f(x)f(x)f(x) is continuous at x=3x=3x=3, then the number of elements in S\mathrm{S}S is :
  1. A
    Infinitely many
  2. B
    4
  3. C
    2
  4. D
    1
View written solutionFree

Correct answer: 0

  1. Interpret the function carefully

The given function is

f(x)={a(7x−12−x2)b ∣x2−7x+12∣,x≠3b,x=3f(x)= \begin{cases} \dfrac{a(7x-12-x^2)}{b\,|x^2-7x+12|}, & x\ne 3 \\ b, & x=3 \end{cases}f(x)=⎩⎨⎧​b∣x2−7x+12∣a(7x−12−x2)​,b,​x=3x=3​

Also, the statement says [x][x][x] denotes the greatest integer function, but it does not actually appear in the displayed formula. So we solve using the function as written.

We need all ordered pairs (a,b)(a,b)(a,b) such that f(x)f(x)f(x) is continuous at x=3x=3x=3.


  1. Simplify the expression for x≠3x\ne 3x=3

First factor:

x2−7x+12=(x−3)(x−4)x^2-7x+12=(x-3)(x-4)x2−7x+12=(x−3)(x−4)

Also,

7x−12−x2=−(x2−7x+12)=−(x−3)(x−4)7x-12-x^2=-(x^2-7x+12)=-(x-3)(x-4)7x−12−x2=−(x2−7x+12)=−(x−3)(x−4)

Hence for x≠3x\ne 3x=3,

f(x)=a [−(x2−7x+12)]b∣x2−7x+12∣=−ab⋅x2−7x+12∣x2−7x+12∣f(x)=\frac{a\,[-(x^2-7x+12)]}{b|x^2-7x+12|} = -\frac{a}{b}\cdot \frac{x^2-7x+12}{|x^2-7x+12|}f(x)=b∣x2−7x+12∣a[−(x2−7x+12)]​=−ba​⋅∣x2−7x+12∣x2−7x+12​

So for x≠3x\ne 3x=3,

f(x)=−ab sgn⁡(x2−7x+12)f(x)= -\frac{a}{b}\,\operatorname{sgn}(x^2-7x+12)f(x)=−ba​sgn(x2−7x+12)

where sgn⁡(t)=t∣t∣\operatorname{sgn}(t)=\dfrac{t}{|t|}sgn(t)=∣t∣t​ for t≠0t\ne 0t=0.


  1. Check the sign near x=3x=3x=3

Near x=3x=3x=3,

x2−7x+12=(x−3)(x−4)x^2-7x+12=(x-3)(x-4)x2−7x+12=(x−3)(x−4)
  • If x→3−x\to 3^-x→3−, then x−3<0x-3<0x−3<0 and x−4<0x-4<0x−4<0, so

    (x−3)(x−4)>0(x-3)(x-4)>0(x−3)(x−4)>0

    Therefore,

    f(x)=−ab(x<3 near 3)f(x)= -\frac{a}{b} \quad (x<3 \text{ near } 3)f(x)=−ba​(x<3 near 3)
  • If x→3+x\to 3^+x→3+, then x−3>0x-3>0x−3>0 and x−4<0x-4<0x−4<0, so

    (x−3)(x−4)<0(x-3)(x-4)<0(x−3)(x−4)<0

    Therefore,

    f(x)=+ab(x>3 near 3)f(x)= +\frac{a}{b} \quad (x>3 \text{ near } 3)f(x)=+ba​(x>3 near 3)

Thus,

lim⁡x→3−f(x)=−ab,lim⁡x→3+f(x)=ab\lim_{x\to 3^-} f(x)= -\frac{a}{b}, \qquad \lim_{x\to 3^+} f(x)= \frac{a}{b}x→3−lim​f(x)=−ba​,x→3+lim​f(x)=ba​
  1. Condition for continuity at x=3x=3x=3

For continuity at x=3x=3x=3, the left-hand and right-hand limits must be equal, and equal to f(3)=bf(3)=bf(3)=b.

So first,

−ab=ab-\frac{a}{b}=\frac{a}{b}−ba​=ba​

This gives

ab=0\frac{a}{b}=0ba​=0

Since the expression for x≠3x\ne 3x=3 has denominator bbb, we must have

b≠0b\ne 0b=0

Hence,

a=0a=0a=0

Then both one-sided limits become 000. For continuity, we also need

f(3)=b=0f(3)=b=0f(3)=b=0

So this requires

b=0b=0b=0

But this contradicts the requirement b≠0b\ne 0b=0 for the formula when x≠3x\ne 3x=3 to be defined.

Therefore, no ordered pair (a,b)(a,b)(a,b) exists such that f(x)f(x)f(x) is continuous at x=3x=3x=3.

So,

S=∅S=\varnothingS=∅

and the number of elements is

000
  1. Compare with the options

The correct count should be 000, but this is not among the options.

So the stored answer D: 1\boxed{\text{D: }1}D: 1​ does not agree with the mathematics of the displayed function.

It is likely there is a typo in the question statement or formula.


  1. Final conclusion

The number of ordered pairs (a,b)(a,b)(a,b) is

0\boxed{0}0​

Hence the stored answer DDD is not correct for the function as written.

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