If G(x) = -√(25 - x²) then lim (x→1) (G(x) - G(1))/(x - 1) has the value (a) 1/24 (b) 1/5 (c) -√24 (

Q: Wait, why is the limit equal to the derivative? I thought limits were just about approaching a value.

That is a fantastic question! You are absolutely right that limits are about approaching a value. However, the definition of the derivative $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ is specifically designed to measure the instantaneous rate of change. When you see this specific form, it is the mathematical way of saying 'what is the slope of the tangent line at $x=a$?' It is the bridge between algebra and geometry.

Q: Can I just plug in $x=1$ directly into the limit expression? It gives $0/0$.

Exactly! If you plug in $x=1$ directly, you get $\frac{G(1) - G(1)}{1 - 1} = \frac{0}{0}$, which is an indeterminate form. This is the classic signal that you are looking at a derivative definition. Instead of trying to evaluate the limit directly, you should recognize the definition and compute the derivative function $G'(x)$ first, then evaluate it at $x=1$.

Q: Why is the answer 'none of these'? Did I mess up the derivative?

Do not panic! In JEE Advanced, 'none of these' is a perfectly valid and common answer. It tests your confidence. If you have carefully calculated $G'(1) = \frac{1}{\sqrt{24}}$ and none of the options match, trust your work. The options provided were likely distractors designed to catch students who confused the y-coordinate with the slope or made simple arithmetic errors.

Question

If $G(x) = -\sqrt{25 - x^2}$ then $\lim_{x 	o 1} \frac{G(x) - G(1)}{x - 1}$ has the value

Options:

(A) $1/24$

(B) $1/5$

(C) $-\sqrt{24}$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Bowl

Imagine you are standing in a coordinate plane. You are given the function $G(x) = -\sqrt{25 - x^2}$.

At first glance, this might look like just another algebraic expression, but let's peel back the layers. If we square both sides, we get $G(x)^2 = 25 - x^2$, or $x^2 + G(x)^2 = 25$.

This is the equation of a circle with radius $5$ centered at the origin. However, because of the negative sign in front of the square root, we are restricted to the lower half of this circle. Imagine a bowl placed upside down below the x-axis, spanning from $x = -5$ to $x = 5$.

Our mission is to understand the behavior of this curve exactly at $x = 1$.

### The Secret Language of Limits

The problem asks us to evaluate the following limit:

$$\lim_{x 	o 1} \frac{G(x) - G(1)}{x - 1}$$

If you have spent enough time in the trenches of calculus, this expression should trigger an immediate recognition. This is the first-principle definition of the derivative at a point.

It is the mathematical embodiment of the slope of the tangent line to the curve at $x = 1$. We are not just calculating a limit; we are finding the steepness of our 'bowl' at a specific point. Our strategy is clear: find the derivative function $G'(x)$ and then evaluate it at $x = 1$.

### The Chain Rule: Unlocking the Derivative

To differentiate $G(x) = -(25 - x^2)^{1/2}$, we must respect the hierarchy of the function. We have an 'outer shell' (the negative square root) and an 'inner core' ($25 - x^2$).

This is the perfect scenario for the Chain Rule. We differentiate the outer shell first, treating the inner core as a single variable, and then multiply by the derivative of that inner core.

The derivative of the outer shell, using the power rule, is $-\frac{1}{2}(25 - x^2)^{-1/2}$. Now, we multiply this by the derivative of the inner core, which is $\frac{d}{dx}(25 - x^2) = -2x$.

Putting it all together, we get:

$$G'(x) = -\frac{1}{2}(25 - x^2)^{-1/2} \cdot (-2x)$$

### The Elegance of Simplification

Now, let's watch the magic happen as we simplify. The negative sign from the outer derivative and the negative sign from the inner derivative cancel out, leaving us with a positive result.

The $2$ in the denominator of the outer derivative cancels perfectly with the $2$ in the inner derivative. We are left with:

$$G'(x) = \frac{x}{\sqrt{25 - x^2}}$$

This is the general formula for the slope of the tangent at any point $x$ on our semi-circle. It is elegant, simple, and powerful.

### The Final Evaluation

We are almost at the finish line. We need the slope at $x = 1$. Substituting $x = 1$ into our derivative formula, we get:

$$G'(1) = \frac{1}{\sqrt{25 - (1)^2}} = \frac{1}{\sqrt{24}}$$

This simplifies to $\frac{1}{2\sqrt{6}}$. Looking at our options, we see $\frac{1}{24}$, $\frac{1}{5}$, and $-\sqrt{24}$.

None of these match our result of $\frac{1}{\sqrt{24}}$. This is a moment where you must trust your mathematical journey.

You have followed the logic, applied the rules of calculus, and arrived at a precise answer. If it is not listed, the correct choice is **'none of these'**. Stand tall in your calculation; you have mastered the physics of the curve.

Animated Solution for Mathematics - Differentiation: If $G (x) = - 25 - x^{2}$ then $lim_{x \to 1} \frac{G ( x ) - G ( 1 )}{x - 1}$ has the value

Select Answer:

Visualized Solution

Visualizing $G (x) = - 25 - x^{2}$

The Derivative Definition

Chain Rule Strategy

Differentiating the Outer Function

Differentiating the Inner Function

Simplifying the Derivative

Evaluating at $x = 1$

Final Conclusion

The Sigma Insight: Techniques of Differentiation

Analyzing the Geometry of the Bowl

The Secret Language of Limits

The Chain Rule

The Elegance of Simplification

The Final Evaluation

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If G(x)=−25−x2​ then limx→1​x−1G(x)−G(1)​ has the value