Problem 109 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(f(x)=\sin (\sin x)\) and \(f^{\prime \prime}(x)+\tan x f^{\prime}(x)+g(x)=0\), then \(g(x)\) is: (a) \(\cos ^{2} x \cos (\sin x)\) (b) \(\sin ^{2} x \cos (\cos x)\) (c) \(\sin ^{2} x \sin (\cos x)\) (d) \(\cos ^{2} x \sin (\sin x)\)

Step-by-Step Solution

Answer

The function \( g(x) \) is (d) \( \cos^2 x \sin(\sin x) \).

Step 1: Understand the given expressions

The function given is \( f(x) = \sin(\sin x) \). The differential equation involving \( f''(x) \) and its derivative is \( f''(x) + \tan(x) f'(x) + g(x) = 0 \), where we need to find the expression for \( g(x) \).

Step 2: Find the first derivative \( f'(x) \)

Use the chain rule to differentiate \( \sin(\sin x) \). \[ f'(x) = \cos(\sin x) \cdot \cos x \].

Step 3: Find the second derivative \( f''(x) \)

Apply the product rule and chain rule to differentiate \( f'(x) = \cos(\sin x) \cdot \cos x \):\[ f''(x) = [-(\sin(\sin x) \cdot \cos x) \cdot \cos x] + [\cos(\sin x)(-\sin x)] \].Simplifying this gives:\[ f''(x) = -\sin(\sin x) \cdot \cos^2 x - \cos(\sin x) \cdot \sin x \].

Step 4: Substitute derivatives into the given equation

Substitute \( f''(x) \) and \( f'(x) \) into the equation:\[ -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x + \tan x (\cos(\sin x) \cos x) + g(x) = 0 \].

Step 5: Simplify and solve for \( g(x) \)

Combining terms:\[ -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x + \frac{\sin x}{\cos x} \cdot \cos(\sin x) \cos x + g(x) = 0 \]simplifies to:\[ -\sin(\sin x) \cos^2 x + g(x) = 0 \].Thus, \( g(x) = \sin(\sin x) \cos^2 x \).

Step 6: Match \( g(x) \) with given options

Comparing with the provided answers: \( g(x) = \sin(\sin x) \cos^2 x \) matches option (d) \( \cos^2 x \sin(\sin x) \).

Key Concepts

Differentiation

Differentiation is one of the fundamental concepts in calculus. It involves finding the derivative of a function, which represents the rate of change of a quantity. Imagine traveling on a road trip; differentiation helps us understand how our speed changes over time. Just as knowing whether you're speeding up or slowing down is crucial for safe driving, differentiation helps us understand how functions change.
In mathematical terms, the derivative of a function at a point is the slope of the tangent line at that point on the graph. To differentiate a function, we apply specific rules and techniques, notably the chain rule and the product rule. Simplifying differentiation helps us solve complex problems in physics, engineering, and economics by breaking them down into manageable parts.

Chain Rule

The chain rule is a powerful technique used in calculus to differentiate composite functions. A composite function is one that is made by combining two or more functions, like nesting dolls inside each other.
For example, in solving the given exercise, our function is nested as \( f(x) = \sin(\sin x) \), meaning it has layers of functions: the sine function applied to another sine function. The chain rule helps us differentiate this by moving through each layer one step at a time:

• First, find the derivative of the outside function at the outer layer.
• Then, multiply it by the derivative of the inner function, moving one layer inward.

This sequential approach allows us to accurately reach the derivative of complex, nested functions.

Product Rule

The product rule is used when we need to differentiate a product of two or more functions. It's particularly helpful when functions are multiplied together, which is common in many calculus problems.
For instance, in the exercise, we applied the product rule to \( f'(x) = \cos(\sin x) \cdot \cos x \). The product rule states that if you have two functions multiplying each other, \( u(x) \) and \( v(x) \), the derivative of their product is given by:

• \( (uv)' = u'v + uv' \)

This means you find the derivative of the first function, multiply it by the second function, and then add the product of the first function with the derivative of the second function. This reciprocal process allows for accurate differentiation of products of functions.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are essential in calculus, and they depict relationships between angles and sides of triangles. They're used in various applications, from modeling wave patterns to solving geometry problems.
In the original exercise, these trigonometric functions play a significant role. Understanding how to differentiate them is crucial:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).

These identities form the backbone of solving many calculus problems involving trigonometric functions. Knowing how they interact within composite and product functions—as demonstrated in the problem—helps tackle complex problems involving waves, cycles, and oscillations.