Let f be a differentiable function on R such that f(2)=1, f'(2)=4. Let lim x->0 ( f(2+x) )^(3/x) = e^alpha. Then the number of times the curve y=4x^3-4x^2-4(alpha-7)x-alpha meets x-axis is :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q16 – JEE-MAIN 2025

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Home Maths Limits and DerivativesQuestion #276

Let

f

be a differentiable function on

R

such that

f (2) = 1

f^{'} (2) = 4

. Let

lim_{x \to 0} {(f (2 + x))}^{3 / x} = e^{α}

. Then the number of times the curve

y = 4 x^{3} - 4 x^{2} - 4 (α - 7) x - α

meets x-axis is :

Clue 1 of 3: Limit Evaluation

Recognize the indeterminate form of the limit and recall the standard formula for $1^{\infty}$ type limits.

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Step 1: Evaluate the limit to find

α

✦ Active

The given limit is $lim_{x \to 0} {(f (2 + x))}^{3 / x} = e^{α}$ . Since $f (2) = 1$ , this is of the indeterminate form $1^{\infty}$ . We use the property that if $lim_{x \to a} g (x)^{h (x)}$ is of the form $1^{\infty}$ , then the limit is $e^{lim_{x \to a} h (x) (g (x) - 1)}$ . Here, $g (x) = f (2 + x)$ and $h (x) = \frac{3}{x}$ .

e^{α} = e^{lim_{x \to 0} \frac{3}{x} (f (2 + x) - 1)}

The exponent is $lim_{x \to 0} \frac{3 (f (2 + x) - 1)}{x}$ . This is a $\frac{0}{0}$ form, so we apply L'Hopital's Rule:

lim_{x \to 0} \frac{3 f^{'} (2 + x)}{1} = 3 f^{'} (2)

Given $f^{'} (2) = 4$ , the exponent is $3 \times 4 = 12$ . Therefore, $e^{α} = e^{12}$ , which implies $α = 12$ .

Step 2: Formulate the polynomial equation○ Expand

Substitute $α = 12$ into the given curve equation $y = 4 x^{3} - 4 x^{2} - 4 (α - 7) x - α$ :

y = 4 x^{3} - 4 x^{2} - 4 (12 - 7) x - 12

\Rightarrow y = 4 x^{3} - 4 x^{2} - 4 (5) x - 12

\Rightarrow y = 4 x^{3} - 4 x^{2} - 20 x - 12

To find where the curve meets the x-axis, we set $y = 0$ :

4 x^{3} - 4 x^{2} - 20 x - 12 = 0

Divide the entire equation by 4:

x^{3} - x^{2} - 5 x - 3 = 0

Step 3: Find the distinct real roots of the polynomial○ Expand

Let $P (x) = x^{3} - x^{2} - 5 x - 3$ . We test integer divisors of the constant term -3 (i.e., $\pm 1, \pm 3$ ). By inspection, $P (- 1) = (- 1)^{3} - (- 1)^{2} - 5 (- 1) - 3 = - 1 - 1 + 5 - 3 = 0$ . Thus, $x = - 1$ is a root, and $(x + 1)$ is a factor of $P (x)$ . We perform polynomial division or synthetic division:

(x + 1) (x^{2} - 2 x - 3) = 0

Now, factor the quadratic term $x^{2} - 2 x - 3$ :

(x + 1) (x - 3) (x + 1) = 0

(x + 1)^{2} (x - 3) = 0

The roots are $x = - 1$ (with multiplicity 2) and $x = 3$ (with multiplicity 1). The distinct real roots are $x = - 1$ and $x = 3$ . Therefore, the curve meets the x-axis at 2 distinct points.

💡 Teacher's Secret Hint

Remember that a root with even multiplicity (like $x = - 1$ here) means the curve touches the x-axis, while a root with odd multiplicity (like $x = 3$ ) means it crosses. Both are considered 'meeting' the x-axis.

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Maths Question 16 – JEE-MAIN 2025

Clue 1 of 3: Limit Evaluation

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