Let f(x) = 2^x - x^2, x ∈ R. If m and n are respectively the number of points at which the curves y

Q: Wait, why can't we just solve $2^x = x^2$ by taking logarithms on both sides?

That is a great instinct! If you take the natural log, you get $x \ln 2 = 2 \ln x$. This is still a transcendental equation because it involves both $x$ and $\ln x$. In JEE Advanced, when you see a mix of exponential and polynomial terms, graphical analysis is often the most robust and intended path to find the number of roots.

Q: How do I know there aren't more roots for $f'(x) = 0$ hidden somewhere?

Excellent question! We can analyze the second derivative $f''(x) = 2^x (\ln 2)^2 - 2$. Setting $f''(x) = 0$ tells us where the function changes concavity. Since $f'(x)$ is continuous and we have identified its turning points and behavior at the boundaries, we can be confident that there are no other roots. The function $f'(x)$ only changes direction once.

Q: Why did we specifically check $x=2$ and $x=4$ for the roots of $f(x)$?

In many JEE problems, integer roots are often 'hidden' in plain sight. It is a standard strategy to test small integers like $0, 1, 2, 3, 4$ first. Once you find these, you can use the behavior of the functions (growth rates) to confirm if any other roots exist.

Question

Let $f(x) = 2^x - x^2, x \in \mathbb{R}$. If $m$ and $n$ are respectively the number of points at which the curves $y = f(x)$ and $y = f'(x)$ intersects the $x$-axis, then the value of $m + n$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex landscape of curves. You have the function $f(x) = 2^x - x^2$.

Your mission is to find the number of times this curve kisses the $x$-axis, and then do the same for its derivative. This is not just algebra; it is a story of growth rates and intersections.

### The Roots of $f(x)$

To find $m$, the number of roots of $f(x) = 0$, we set $2^x - x^2 = 0$, which rearranges beautifully to:

$$2^x = x^2$$

Here, we have a battle between an exponential function and a parabola. The exponential $y = 2^x$ grows incredibly fast, while the parabola $y = x^2$ is steady.

By testing small integers, we find that at $x = 2$, $2^2 = 4$ and $2^2 = 4$. They meet! At $x = 4$, $2^4 = 16$ and $4^2 = 16$. They meet again!

But what about the negative side? As $x 	o -\infty$, $2^x$ approaches $0$ while $x^2$ grows to $\infty$. Since $2^0 = 1$ and $0^2 = 0$, the exponential starts above the parabola at $x=0$ and must cross it exactly once as we move into the negative region.

Thus, we have three roots: **$m = 3$**.

### The Derivative $f'(x)$

Now, we turn our attention to the derivative. Differentiating $f(x) = 2^x - x^2$ gives us:

$$f'(x) = 2^x \ln 2 - 2x$$

We want to find $n$, the number of roots of $f'(x) = 0$, which is equivalent to:

$$2^x \ln 2 = 2x$$

This is a new, more subtle dance. We are comparing an exponential curve $y = 2^x \ln 2$ with a straight line $y = 2x$.

### Finding $n$

Let us test the behavior. At $x = 0$, $2^0 \ln 2 \approx 0.69$, which is greater than $2(0) = 0$.

At $x = 2$, $2^2 \ln 2 \approx 2.77$, which is less than $2(2) = 4$. Because the exponential is above the line at $x=0$ and below it at $x=2$, the Intermediate Value Theorem guarantees a root in $(0, 2)$.

As $x$ increases further, the exponential growth takes over. At $x = 4$, $2^4 \ln 2 \approx 11.09$, which is greater than $2(4) = 8$. Again, the curves must cross in the interval $(2, 4)$.

Thus, we have exactly two roots for the derivative: **$n = 2$**.

### The Final Sum

We have navigated the curves, identified the intersections, and verified the behavior of the derivative. We found $m = 3$ and $n = 2$.

The problem asks for the sum $m + n$. Adding these together, we get:

$$3 + 2 = 5$$

The beauty of this problem lies in how it forces us to combine algebraic testing with the deep, intuitive power of calculus and graphical analysis. You have mastered the transcendental! The final answer is **5**.

Animated Solution for Mathematics - Differentiation: Let $f (x) = 2^{x} - x^{2}, x \in R$ . If $m$ and $n$ are respectively the number of points at which the curves $y = f (x)$ and $y = f^{'} (x)$ intersects the $x$ -axis, then the value of $m + n$ is

Enter Numerical Value:

Visualized Solution

The Function $f (x)$

Finding $m$ : Roots of $f (x)$

Visualizing $2^{x} = x^{2}$

Positive Roots for $m$

Negative Root for $m$

Finding $n$ : The Derivative $f^{'} (x)$

Analyzing $f^{'} (x) = 0$

Identifying the Roots for $n$

Final Calculation $m + n$

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Roots of $f (x)$

The Derivative $f^{'} (x)$

Finding $n$

The Final Sum

Similar Questions

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Let $f : R - {0} \to R$ be a function satisfying $f (\frac{x}{y}) = \frac{f ( x )}{f ( y )}$ for all $x, y, f (y) \neq = 0$ . If $f^{'} (1) = 2024$ , then

Enter Numerical Value:

Visualized Solution

The Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m: Roots of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x=x2

Positive Roots for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Negative Root for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n: The Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)=0

Identifying the Roots for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Final Calculation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Roots of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

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

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