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

Q: Wait, why is the limit $1^\infty$?

Great question! As $x \to 0$, the base $f(2+x)$ approaches $f(2)$, which is given as $1$. The exponent $3/x$ approaches $3/0$, which is $\infty$. Thus, we have the classic $1^\infty$ indeterminate form.

Q: Why do we use the formula instead of taking logs?

You *can* take logs, but the formula $e^{\lim h(x)(g(x)-1)}$ is a powerful, time-saving shortcut for JEE. It directly converts the $1^\infty$ form into an exponential limit, which is much cleaner to solve.

Q: Does a repeated root count as two intersections?

The question asks for the number of times the curve *meets* the x-axis. This refers to the number of distinct points of intersection. A repeated root (like $x=-1$ in our case) indicates that the curve is tangent to the x-axis at that point, which counts as one intersection point.

Question

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

Options:

(A) 2

(B) 1

(C) 0

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a function $f(x)$ with known properties at a specific point: $f(2) = 1$ and $f'(2) = 4$. Our goal is to evaluate the limit $\lim_{x 	o 0} (f(2 + x))^{3/x} = e^{\alpha}$ and subsequently analyze the roots of a related cubic equation.

### The Limit Challenge

As $x 	o 0$, the base $f(2+x)$ approaches $f(2) = 1$, while the exponent $3/x$ approaches $\infty$. This results in the indeterminate form $1^{\infty}$.

To resolve this, we utilize the standard limit identity:
$$\lim_{x 	o a} [g(x)]^{h(x)} = e^{\lim_{x 	o a} h(x)(g(x)-1)}$$

Applying this to our expression, we obtain:
$$e^{\lim_{x 	o 0} \frac{3}{x}(f(2+x)-1)}$$

### The Derivative Connection

We substitute $1$ with $f(2)$ in the exponent to reveal the definition of the derivative:
$$3 \cdot \lim_{x 	o 0} \frac{f(2+x) - f(2)}{x} = 3 \cdot f'(2)$$

Given that $f'(2) = 4$, the exponent evaluates to $3 	imes 4 = 12$. Therefore, we identify that $\alpha = 12$.

### The Polynomial Geometry

With $\alpha = 12$, we substitute this value into the cubic equation:
$$y = 4x^3 - 4x^2 - 4(12 - 7)x - 12$$

This simplifies to:
$$y = 4x^3 - 4x^2 - 20x - 12$$

To find the intersections with the $x$-axis, we set $y = 0$ and divide by $4$:
$$x^3 - x^2 - 5x - 3 = 0$$

### Solving for Roots

By testing $x = -1$, we observe that $(-1)^3 - (-1)^2 - 5(-1) - 3 = -1 - 1 + 5 - 3 = 0$. Thus, $(x+1)$ is a factor.

Performing polynomial division or synthetic division, we factor the equation as:
$$(x+1)(x^2 - 2x - 3) = 0$$

Further factoring the quadratic term yields:
$$(x+1)(x+1)(x-3) = 0 \implies (x+1)^2(x-3) = 0$$

### Final Calculation

The roots of the equation are $x = -1$ (a root of multiplicity 2) and $x = 3$.

Because the curve touches the $x$-axis at $x = -1$ and crosses it at $x = 3$, there are exactly **2** distinct points of intersection.

Select Answer:

Visualized Solution

Analyze the Limit Form

Apply $1^{\infty}$ Formula

Relate to Derivative $f^{'} (2)$

Calculate the Value of $α$

Form the Cubic Equation

Simplify the Polynomial

Find the First Root

Factorize Completely

Identify the Roots on Graph

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Limit Challenge

The Derivative Connection

The Polynomial Geometry

Solving for Roots

Final Calculation

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Select Answer:

Visualized Solution

Analyze the Limit Form

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1∞ Formula

Relate to Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(2)

Calculate the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α