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Limits Continuity and Differentiability question

2025 · 24 Jan · Shift 1 · Q27
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  5. /2025 · 24 Jan · Shift 1 · Q27

Limits Continuity and Differentiability question

2025 · 24 Jan · Shift 1 · Q27

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f:R−{0}→Rf: \mathbb{R}-\{0\} \rightarrow \mathbb{R}f:R−{0}→R be a function such that f(x)−6f(1x)=353x−52f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}f(x)−6f(x1​)=3x35​−25​. If the lim⁡x→0(1αx+f(x))=β;α,β∈R\lim\limits _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta ; \alpha, \beta \in \mathbb{R}x→0lim​(αx1​+f(x))=β;α,β∈R, then α+2β\alpha+2 \betaα+2β is equal to
  1. A
    6
  2. B
    5
  3. C
    3
  4. D
    4
View written solutionFree

Correct answer: D

  1. Given functional equation

We have

f(x)−6f(1x)=353x−52,x≠0.f(x)-6f\left(\frac{1}{x}\right)=\frac{35}{3x}-\frac{5}{2}, \qquad x\ne 0.f(x)−6f(x1​)=3x35​−25​,x=0.

We need to find α,β\alpha,\betaα,β such that

lim⁡x→0(1αx+f(x))=β.\lim_{x\to 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta.x→0lim​(αx1​+f(x))=β.
  1. Write another equation by replacing xxx with 1x\frac{1}{x}x1​

Substitute x↦1xx\mapsto \frac{1}{x}x↦x1​ in the given relation:

f\left(\frac{1}{x}\right)-6f(x)=\frac{35}{3/(1/x)}-\frac{5}{2}= rac{35x}{3}-\frac{5}{2}.

So,

f(1x)−6f(x)=35x3−52.f\left(\frac{1}{x}\right)-6f(x)=\frac{35x}{3}-\frac{5}{2}. f(x1​)−6f(x)=335x​−25​.
  1. Solve the two linear equations

Let

A=f(x),B=f(1x).A=f(x), \qquad B=f\left(\frac{1}{x}\right).A=f(x),B=f(x1​).

Then

A−6B=353x−52...(1)A-6B=\frac{35}{3x}-\frac{5}{2} \quad \text{...(1)}A−6B=3x35​−25​...(1)

and

B−6A=35x3−52....(2)B-6A=\frac{35x}{3}-\frac{5}{2}. \quad \text{...(2)}B−6A=335x​−25​....(2)

From (2),

B=6A+35x3−52.B=6A+\frac{35x}{3}-\frac{5}{2}.B=6A+335x​−25​.

Substitute into (1):

A−6(6A+35x3−52)=353x−52.A-6\left(6A+\frac{35x}{3}-\frac{5}{2}\right)=\frac{35}{3x}-\frac{5}{2}.A−6(6A+335x​−25​)=3x35​−25​. A−36A−70x+15=353x−52.A-36A-70x+15=\frac{35}{3x}-\frac{5}{2}.A−36A−70x+15=3x35​−25​. −35A=353x−52+70x−15.-35A=\frac{35}{3x}-\frac{5}{2}+70x-15.−35A=3x35​−25​+70x−15. −35A=353x+70x−352.-35A=\frac{35}{3x}+70x-\frac{35}{2}.−35A=3x35​+70x−235​.

Hence,

A=f(x)=−13x−2x+12.A=f(x)=-\frac{1}{3x}-2x+\frac{1}{2}.A=f(x)=−3x1​−2x+21​.

So,

f(x)=−13x−2x+12.f(x)=-\frac{1}{3x}-2x+\frac{1}{2}.f(x)=−3x1​−2x+21​.
  1. Use the limit condition

Now,

1αx+f(x)=1αx−13x−2x+12.\frac{1}{\alpha x}+f(x)=\frac{1}{\alpha x}-\frac{1}{3x}-2x+\frac{1}{2}.αx1​+f(x)=αx1​−3x1​−2x+21​.

For the limit as x→0x\to 0x→0 to exist and be finite, the coefficient of 1x\frac{1}{x}x1​ must be zero:

1α−13=0.\frac{1}{\alpha}-\frac{1}{3}=0.α1​−31​=0.

Thus,

1α=13  ⟹  α=3.\frac{1}{\alpha}=\frac{1}{3} \implies \alpha=3.α1​=31​⟹α=3.

Then,

lim⁡x→0(1αx+f(x))=lim⁡x→0(−2x+12)=12.\lim_{x\to 0}\left(\frac{1}{\alpha x}+f(x)\right)=\lim_{x\to 0}\left(-2x+\frac{1}{2}\right)=\frac{1}{2}.x→0lim​(αx1​+f(x))=x→0lim​(−2x+21​)=21​.

So,

β=12.\beta=\frac{1}{2}.β=21​.
  1. Compute α+2β\alpha+2\betaα+2β
α+2β=3+2(12)=3+1=4.\alpha+2\beta=3+2\left(\frac{1}{2}\right)=3+1=4.α+2β=3+2(21​)=3+1=4.
  1. Check options

The correct option is

4\boxed{4}4​

which is Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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