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Limits Continuity and Differentiability question

2025 · 4 Apr · Shift 2 · Q45
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Limits Continuity and Differentiability question

2025 · 4 Apr · Shift 2 · Q45

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let fff be a differentiable function on R\mathbf{R}R such that f(2)=1,f′(2)=4f(2)=1, f^{\prime}(2)=4f(2)=1,f′(2)=4. Let lim⁡x→0(f(2+x))3/x=eα\lim \limits_{x \rightarrow 0}(f(2+x))^{3 / x}=\mathrm{e}^\alphax→0lim​(f(2+x))3/x=eα. Then the number of times the curve y=4x3−4x2−4(α−7)x−αy=4 x^3-4 x^2-4(\alpha-7) x-\alphay=4x3−4x2−4(α−7)x−α meets xxx-axis is :
  1. A
    3
  2. B
    1
  3. C
    2
  4. D
    0
View written solutionFree

Correct answer: C

  1. Find α\alphaα using the given limit

We are given

lim⁡x→0(f(2+x))3/x=eα,\lim_{x\to 0}(f(2+x))^{3/x}=e^\alpha,x→0lim​(f(2+x))3/x=eα,

with

f(2)=1,f′(2)=4.f(2)=1,\qquad f'(2)=4.f(2)=1,f′(2)=4.

Take natural logarithm. Let

L=lim⁡x→0(f(2+x))3/x.L=\lim_{x\to 0}(f(2+x))^{3/x}.L=x→0lim​(f(2+x))3/x.

Then

ln⁡L=lim⁡x→03xln⁡(f(2+x)).\ln L=\lim_{x\to 0}\frac{3}{x}\ln(f(2+x)).lnL=x→0lim​x3​ln(f(2+x)).

Since f(2)=1f(2)=1f(2)=1, we have f(2+x)→1f(2+x)\to 1f(2+x)→1 as x→0x\to 0x→0, so this is of the form 0/00/00/0.

Using the standard expansion near x=0x=0x=0:

f(2+x)=f(2)+xf′(2)+o(x)=1+4x+o(x).f(2+x)=f(2)+xf'(2)+o(x)=1+4x+o(x).f(2+x)=f(2)+xf′(2)+o(x)=1+4x+o(x).

Hence

ln⁡(f(2+x))=ln⁡(1+4x+o(x))=4x+o(x).\ln(f(2+x))=\ln(1+4x+o(x))=4x+o(x).ln(f(2+x))=ln(1+4x+o(x))=4x+o(x).

Therefore,

ln⁡L=lim⁡x→03x(4x+o(x))=12.\ln L=\lim_{x\to 0}\frac{3}{x}(4x+o(x))=12.lnL=x→0lim​x3​(4x+o(x))=12.

So

L=e12.L=e^{12}.L=e12.

Given L=eαL=e^\alphaL=eα, we get

α=12.\alpha=12.α=12.
  1. Form the cubic curve

The curve is

y=4x3−4x2−4(α−7)x−α.y=4x^3-4x^2-4(\alpha-7)x-\alpha.y=4x3−4x2−4(α−7)x−α.

Substitute α=12\alpha=12α=12:

y=4x3−4x2−4(12−7)x−12y=4x^3-4x^2-4(12-7)x-12y=4x3−4x2−4(12−7)x−12 y=4x3−4x2−20x−12.y=4x^3-4x^2-20x-12.y=4x3−4x2−20x−12.

Factor out 444:

y=4(x3−x2−5x−3).y=4(x^3-x^2-5x-3).y=4(x3−x2−5x−3).

So we need the number of real roots of

x3−x2−5x−3=0.x^3-x^2-5x-3=0.x3−x2−5x−3=0.
  1. Factor the cubic

Try rational roots ±1,±3\pm 1,\pm 3±1,±3.

Check x=−1x=-1x=−1:

(−1)3−(−1)2−5(−1)−3=−1−1+5−3=0.(-1)^3-(-1)^2-5(-1)-3=-1-1+5-3=0.(−1)3−(−1)2−5(−1)−3=−1−1+5−3=0.

So (x+1)(x+1)(x+1) is a factor.

Divide:

x3−x2−5x−3=(x+1)(x2−2x−3).x^3-x^2-5x-3=(x+1)(x^2-2x-3).x3−x2−5x−3=(x+1)(x2−2x−3).

Now factor the quadratic:

x2−2x−3=(x−3)(x+1).x^2-2x-3=(x-3)(x+1).x2−2x−3=(x−3)(x+1).

Thus

x3−x2−5x−3=(x+1)2(x−3).x^3-x^2-5x-3=(x+1)^2(x-3).x3−x2−5x−3=(x+1)2(x−3).

Hence

y=4(x+1)2(x−3).y=4(x+1)^2(x-3).y=4(x+1)2(x−3).
  1. Count how many times the curve meets the xxx-axis

The xxx-intercepts are at

x=−1(double root),x=3(simple root).x=-1 \quad (\text{double root}), \qquad x=3 \quad (\text{simple root}).x=−1(double root),x=3(simple root).

A double root means the curve touches the xxx-axis there, and a simple root means it crosses the xxx-axis there.

So the curve meets the xxx-axis at 2 distinct points.

  1. Final answer

Therefore, the number of times the curve meets the xxx-axis is

2.\boxed{2}.2​.

So the correct option is C.

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