Question

Let \(f(0)=0\) and \(f^{\prime}(x)=\frac{1}{\sqrt{\left(1-x^{2}\right)}}\) for \(-1

Short answer

Question: Prove that for the function f(x) whose derivative is \(f^{\prime}(x)=\frac{1}{\sqrt{1-x^{2}}}\), the expression \(f(x) + f(a) = f\left(x \sqrt{1-a^{2}} +a \sqrt{1-x^{2}}\right)\) holds.

Step-by-step solution

Finding the function f(x) from its derivative

Given that \(f^{\prime}(x)=\frac{1}{\sqrt{1-x^{2}}}\), we need to find the function f(x) by integrating f'(x) with respect to x. Using the substitution method, let \(x = \sin{u}\), we will integrate: $$f(x) = f(0) + \int_{0}^{x} \frac{1}{\sqrt{1-t^{2}}} dt$$ Since \(x = \sin{u}\), we get \(dx = \cos{u} du\) and \(t = \sin{u}\). Now, substituting these values we get: $$f(x) = 0 + \int_{0}^{\arcsin{x}} \frac{\cos{u}}{\sqrt{1-\sin^{2}{u}}} du$$ Here, the integral becomes: $$f(x) = \int_{0}^{\arcsin{x}} du = \arcsin{x}$$ Therefore, f(x) = arcsin(x).

Verifying the given expression

Now that we have the function f(x), we need to prove that: $$f(x) + f(a) = f\left(x \sqrt{1-a^{2}} +a \sqrt{1-x^{2}}\right)$$ We know that \(f(x) = \arcsin{x}\) and \(f(a) = \arcsin{a}\). So, we can write the given expression as: $$\arcsin{x} + \arcsin{a} = \arcsin{\left(x \sqrt{1-a^{2}} +a \sqrt{1-x^{2}}\right)}$$ Let \(\arcsin{x} = \alpha\) and \(\arcsin{a} = \beta\). Then we can write the given expression as: $$\alpha + \beta = \arcsin{\left(\sin\alpha \cos\beta + \cos\alpha \sin\beta\right)}$$ Since \(\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta\), we can rewrite the given expression as: $$\alpha + \beta = \arcsin{\sin(\alpha + \beta)}$$ It is known that if \(-\frac{\pi}{2} \le \alpha,\beta \le \frac{\pi}{2}\), then \(\arcsin\sin(x) = x\) for \(-\pi \le x \le \pi\). Therefore, the given expression holds, and we have successfully proven that: $$f(x) + f(a) = f\left(x \sqrt{1-a^{2}} +a \sqrt{1-x^{2}}\right)$$

Key concepts

Integration Techniques

Integration is an essential concept in calculus that allows us to find the total accumulation of quantities, such as area, volume, and other physical quantities. In Integral Calculus for IIT JEE, understanding different integration techniques is crucial for solving complex problems.

One common technique is substitution, which simplifies an integral by transforming it into a more manageable form. Just like in the original problem, by substituting with a trigonometric function, we turned a difficult integral into one that is easy to solve. Another technique is integration by parts, where the integrand is split into two parts and integrated separately, often used for products of functions. Furthermore, there is partial fractions, which helps in integrating rational functions by decomposing them into simpler fractions. Mastering these techniques opens the door to solving a wide range of integrals encountered in calculus.

Inverse Trigonometric Functions

Inverse trigonometric functions are another cornerstone of calculus, particularly in the context of Integral Calculus for IIT JEE. These functions, like arcsine, arccosine, and arctangent, provide angles when given the sine, cosine, or tangent values respectively.

In the given problem, the function f(x) involved an integral that resulted in the arcsine function, represented by \(\arcsin{x}\). Understanding these functions is crucial because they often appear in integrals involving root expressions, as seen in the original exercise. These functions also come with their specific ranges and domains, which are important when dealing with their derivatives or during the integration process. To handle such functions effectively, one must be familiar with their properties, identities, and the intuition behind their graphical representations.

Proofs in Calculus

Proofs are the backbone of mathematics, ensuring that the principles and formulas we use are founded on solid logic. Proofs in calculus test our understanding of concepts and solidify our knowledge base.

In the exercise, we were required to verify a relationship between the function values. The proof used standard trigonometric identities to simplify and prove the given equation. The ability to perform such proofs relies on a deep understanding of functions, their properties, and the rules of calculus. Proving identities or relationships can also involve strategies such as direct substitution, contradiction, or constructing a logical sequence of equalities, all of which serve to enhance problem-solving skills within calculus.

Derivative and Integral Relationships

The connection between derivatives and integrals lies at the heart of calculus. These two operations are inversely related through the Fundamental Theorem of Calculus. Derivatives represent the rate of change, while integrals can be understood as the accumulation of quantities.

In the example provided, this relationship is explored by finding the function \(f(x)\) from the given derivative \(f'(x)\), showcasing how integration is essentially the 'reverse' of differentiation. Moreover, the problem also illustrates that the derivative of inverse trigonometric functions can lead back to the original functions when integrated. Understanding this relationship enables students to transition smoothly between the physical interpretation of these mathematical operations and their symbolic representations, enhancing their comprehension of calculus as a whole.