Problem 102 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\begin{tabular}{l} \hline \multicolumn{1}{c} { Column-I } \\ \hline I. The function \(y\) defined by the equa- \\ 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 \(y(x)\) \\ 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 \\ \(x\), where \(f(x), g(x)\) (C) 4 \\ ferentiable 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 \\ 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 \end{tabular}

Step-by-Step Solution

Answer

I: k=1, II: k=-2, III: k=24, IV: 2 values.

Step 1: Finding k in Problem I

The given function is in implicit form: \[ x y - \log y = 1. \]Taking derivatives with respect to \( x \), through implicit differentiation, leads to:\[ y + x y' - \frac{1}{y} y' = 0. \]Thus:\[ y = - (x - \frac{1}{y}) y'. \]Assuming the relation \( x(y y'' + y'^{2}) - y'' + k y y' = 0 \) holds true, substitute the expression for \( y \) derived from implicit differentiation. After rearranging and simplifying, the terms become:\[ x y'^{2} = ky y'. \]For these terms to cancel for all \( y \), the coefficient comparison yields that:\[ k = 1. \]

Step 2: Calculating k in Problem II

Given the function in terms of parameters \( x = \sin(t) \) and \( y = ae^{t\sqrt{2}} + be^{t\sqrt{2}} \), with the differential equation:\[ (1 - x^2) y'' - x y' = k y. \]Start by finding \( y' \) and \( y'' \) using parameter \( t \):\[ y' = \sqrt{2}(ae^{t\sqrt{2}} + be^{t\sqrt{2}})\cdot \cos(t), \]\[ y'' = 2(ae^{t\sqrt{2}} + be^{t\sqrt{2}}) \cdot (\cos^2(t) - \sin^2(t)). \]By rearranging the equation and matching coefficients, obtain:\[ k = -2. \]

Step 3: Finding k in Problem III

Given the function \( F(x) = f(x)g(x)h(x) \) and the condition at \( x_0 \): \[ F'(x_0) = 21 F(x_0), \]use the product rule for differentiation:\[ F'(x_0) = f'(x_0)g(x_0)h(x_0) + f(x_0)g'(x_0)h(x_0) + f(x_0)g(x_0)h'(x_0). \]Substitute the given derivatives:\[ 21 f(x_0)g(x_0)h(x_0) = 4 f(x_0)g(x_0)h(x_0) - 7 f(x_0)g(x_0)h(x_0) + k f(x_0)g(x_0)h(x_0). \]Thus:\[ 21 = 4 - 7 + k, \]leading to:\[ k = 24. \]

Step 4: Determining number of n for Problem IV

The function \( f(x) = x^n \) differentiates to:\[ f'(x) = nx^{n-1}. \]The condition \( f'(a+b) = f'(a) + f'(b) \) implies:\[ n(a+b)^{n-1} = na^{n-1} + nb^{n-1}. \]This is only true if \( n = 0 \) or \( n = 1 \). Verifying:- If \( n = 0 \), both sides become 0.- If \( n = 1 \), both sides become 1.Thus, the number of values of \( n \) is 2.

Key Concepts

Implicit Differentiation

Implicit differentiation is a handy technique when dealing with equations where one variable is not isolated from others. Instead of solving for one variable, you differentiate both sides of the equation as they are. For instance, consider an equation like \( x y - \log y = 1 \). To differentiate this explicitly in terms of \( x \), you'd assume \( y \) is a function of \( x \), denoted as \( y(x) \). Then, differentiate both sides of the equation using the product rule and the chain rule.

Here's how it works:

• For the term \( xy \), apply the product rule: the derivative is \( y + x y' \).
• The derivative of \( \log y \) requires the chain rule, giving \( -\frac{1}{y} y' \).
• Set the entire derivative equal to zero (as the derivative of a constant is zero).

This process allows you to solve for \( y' \), representing the rate of change of \( y \) with respect to \( x \), without needing to solve explicitly for \( y \).

Differential Equations

Differential equations involve relationships between a function and its derivatives. They have profound applications, from modeling population dynamics to predicting heat flow. In our problem II, the differential equation \((1-x^2) y'' - xy' = ky\) appears. This is a second-order linear differential equation because it involves the second derivative of \( y \) with respect to \( x \).

To solve such equations, often begin by letting \( y \) and its derivatives be functions expressed in terms of a third variable, like \( t \). For instance, if \( x = \sin t \) and \( y = ae^{t\sqrt{2}} + be^{t\sqrt{2}} \), compute \( y' \) and \( y'' \) by differentiating with respect to \( t \), thus simplifying the problem.

• First, find the derivatives: \( y' = \sqrt{2}(ae^{t\sqrt{2}} + be^{t\sqrt{2}})\cdot \cos(t) \).
• Then, \( y'' = 2(ae^{t\sqrt{2}} + be^{t\sqrt{2}}) \cdot (\cos^2(t) - \sin^2(t)) \).

Substituting these into the original equation allows for the determination of constants such as \( k \), which is derived by comparing coefficients.

Product Rule of Differentiation

The product rule is an essential tool in calculus, especially when differentiating expressions where two functions are multiplied together. The rule states that if you have two differentiable functions, \( u(x) \) and \( v(x) \), then the derivative of their product is given by \( (uv)' = u'v + uv' \).

In Problem III, where we have a composite function \( F(x) = f(x)g(x)h(x) \), the product rule extends naturally:

• The derivative \( F'(x) \) involves summing the derivative of each factor times the remaining functions.
• At a particular point \( x_0 \), simplify using given values like \( f'(x_0), g'(x_0), \) and \( h'(x_0) \).

This approach simplifies the problem to solving a straightforward algebraic equation to find constants like \( k \). This method ensures full exploitation of existing derivative values to find unknowns efficiently.

Exponential Functions

Exponential functions, like \( e^x \), are fundamental in many areas, characterized by their constant growth rate relative to their value. In our problems, they appear in expressions such as \( ae^{t\sqrt{2}} + be^{t\sqrt{2}} \). Here, \( e^{t\sqrt{2}} \) is an exponential function with a rate determined by \( \sqrt{2}t \).

Exponential functions differentiate in a simple and elegant manner: the derivative of \( e^{u(x)} \) is \( e^{u(x)} \frac{du}{dx} \), leveraging the chain rule. Thus, they make solving differential equations, like those found in Problem II, more manageable by providing clear, consistent results.

• Differentiating \( e^{t\sqrt{2}} \) gives \( \sqrt{2}e^{t\sqrt{2}} \).
• These features aid in transforming complex differential equations into simpler forms.

By understanding exponential functions' properties, you can effectively manipulate them to fit the structure required by a differential equation, ultimately aiding in discovering solutions and constants such as \( k \).