Problem 102 - A Complete Resource Book in Mathematics for JEE Main

Question

Column-I Column-II I. The function \(y\) defined by the equa- (A) 24 tion \(x y-\log y=1\) satisfies \(x\left(y y^{\prime \prime}+\right.\) \(\left.y^{\prime 2}\right)-y^{\prime \prime}+k y y^{\prime}=0 .\) The value of \(k\) is II. If the function \(y(x)\) (B) 2 represented by \(x=\sin t, y=\) \(a e^{t \sqrt{2}}+b e^{t \sqrt{2}}, t \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) satisfies the equation \(\left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}\) \(=k y\), then \(k\) is equal to III. Let \(F(x)=f(x) g(x) h(x)\) for all real (C) 4 \(x\), where \(f(x), g(x)\) and \(h(x)\) are differentiable functions. At some point \(x_{0}\), if \(F^{\prime}\left(x_{0}\right)=21 F\left(x_{0}\right), f^{\prime}\left(x_{0}\right)=4\) \(f\left(x_{0}\right), g^{\prime}\left(x_{0}\right)=-7 g\left(x_{0}\right)\) and \(h^{\prime}\left(x_{0}\right)=\) \(k h\left(x_{0}\right)\) then \(k\) is equal to IV. Let \(f(x)=x^{n}, n\) being a non-negative (D) 3 integer. The number of values of \(n\) for which the equality \(f^{\prime}(a+b)\) \(=f^{\prime}(a)+f^{\prime}(b)\) is valid for all \(a, b\) \(>0\), is

Step-by-Step Solution

Answer

I. k = 0, II. k = 4, III. k = 2, IV. n has 1 value.

Step 1: Analyze the first function's expression

Given the function's expression as \( xy - \log y = 1 \), we need to find the value of \( k \) in the differential equation \( x(yy'' + (y')^2) - y'' + ky y' = 0 \). First, compute the derivatives \( y' \) and \( y'' \) using implicit differentiation of the given equation, then substitute them into the differential equation to solve for \( k \).

Step 2: Analyze the second function's parameters

Given the solution \( x=\sin t, y=a e^{t \sqrt{2}}+b e^{-t \sqrt{2}} \), substitute these into the differential equation \( (1-x^2)y'' - xy' = ky \) and compute \( y' \) and \( y'' \). Replace back into the equation to balance terms and find \( k \). This should yield \( k = 4 \).

Step 3: Differentiate the composite function for the third equation

For \( F(x) = f(x)g(x)h(x) \), use the given condition \( F'(x_0) = 21F(x_0) \) and differentiate \( F(x) \) using the product rule as \( F'(x) = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) \). Plug in the values \( f'(x_0) = 4f(x_0), g'(x_0) = -7g(x_0) \) to solve for \( h'(x_0) = k h(x_0) \) and equate to find \( k = 2 \).

Step 4: Evaluate for valid integers in the fourth expression

For \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \). The condition \( f'(a+b) = f'(a) + f'(b) \) implies \( n(a+b)^{n-1} = na^{n-1} + nb^{n-1} \). Check for which values of \( n \), this holds true. Solve \( (a+b)^{n-1} = a^{n-1} + b^{n-1} \), focusing on simple non-negative integer values, it holds for \( n = 1 \) only. Therefore, \( n \) has only one possible value.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when it's difficult or impossible to solve for a variable before differentiating. In the context of this exercise, you see an equation in which the variables \( x \) and \( y \) are interrelated by the expression \( xy - \log y = 1 \). This equation doesn’t easily solve for \( y \) as a function of \( x \). Therefore, we use implicit differentiation. When using implicit differentiation:

• Differentiate both sides of the equation with respect to \( x \).
• Treat \( y \) as a function of \( x \), meaning the derivative of \( y \) will be \( y' \).
• Apply chain rule wherever necessary, especially when differentiating terms involving \( y \).

For the given expression, differentiate \( xy \) using the product rule and consider \( \log y \) differentiation involving the chain rule. This will allow you to solve for derivatives like \( y' \) and \( y'' \) which are key to finding the value of \( k \) in the associated differential equation.

Product Rule

When differentiating functions that are products of other functions, we use the product rule. This method is essential when we encounter an expression like \( F(x) = f(x)g(x)h(x) \), as in our third example. The product rule states:

• If you have a product \( u(x)v(x) \), then its derivative is \( u'(x)v(x) + u(x)v'(x) \).

For three functions, \( f(x)g(x)h(x) \), the derivative \( F'(x) \) becomes:

• Use the derivative: \( F'(x) = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) \).

In the exercise, we apply this rule directly within the given constraints: \( F'(x_0) = 21F(x_0) \) and provided values of \( f'(x_0), g'(x_0), h'(x_0) \). This leads us to solve for \( h'(x_0) \) by balancing the equation to find the constant \( k \).

Functional Equations

Functional equations involve finding unknown functions that satisfy a given equation for all values in its domain. In the context of the exercise, we have a specific function \( f(x) = x^n \) and need to check under what conditions the derivative function equation \( f'(a+b) = f'(a) + f'(b) \) holds.
Firstly, the derivative of the function \( f(x) = x^n \) is \( f'(x) = nx^{n-1} \). The equation \( f'(a+b) = f'(a) + f'(b) \) leads us to solve:

• \( n(a+b)^{n-1} = na^{n-1} + nb^{n-1} \)

We check simple integer values to find that this holds true only for \( n = 1 \) because, in such cases, the calculation simplifies perfectly as it mirrors the identity property of addition. This shows that \( n \) has only one feasible value satisfying the given conditions.