Problem 44 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The interval in which \(\lambda\) should be if \(f(x)=\sin ^{3} x+\lambda\) \(\sin ^{2} x(-\pi / 2

Step-by-Step Solution

Answer

(C) \((-\frac{3}{2}, \frac{3}{2})\)

Step 1: Differentiate the Function

Differentiate \( f(x) = \sin^3 x + \lambda \sin^2 x \) with respect to \( x \) to find critical points. The derivative is \[ f'(x) = 3\sin^2 x \cos x + 2\lambda \sin x \cos x. \]

Step 2: Factor the Derivative

Factor the derivative to simplify finding critical points: \[ f'(x) = \cos x (3\sin^2 x + 2\lambda \sin x). \]

Step 3: Set the Derivative to Zero

Solve the equation \( \cos x (3\sin^2 x + 2\lambda \sin x) = 0 \) for critical points. This gives us two equations: 1. \( \cos x = 0 \) 2. \( 3\sin^2 x + 2\lambda \sin x = 0. \)

Step 4: Solve First Equation

Solve \( \cos x = 0 \) within the interval \( (-\pi/2, \pi/2) \). This gives \( x = \pm \pi/2 \), but these points are outside the open interval or are endpoints, so they do not count as interior critical points.

Step 5: Solve Second Equation

Solve \( 3\sin^2 x + 2\lambda \sin x = 0 \). Factor as \( \sin x (3\sin x + 2 \lambda) = 0 \), giving us critical points \( \sin x = 0 \) or \( 3\sin x + 2 \lambda = 0. \)

Step 6: Find Values for Second Equation

\[ \sin x = 0 \] has no solutions in \((-\pi/2, \pi/2)\). For \[ 3\sin x + 2 \lambda = 0 \Rightarrow \sin x = -\frac{2 \lambda}{3} \]. \( \sin x \) must be within \((-1,1)\). Therefore, \[ -1 < -\frac{2 \lambda}{3} < 1 \].

Step 7: Solve Inequality for \( \lambda \)

Solve the inequality: \[ -1 < -\frac{2 \lambda}{3} < 1. \]The equivalent inequality is \[ \frac{-3}{2} < \lambda < \frac{3}{2}. \] This corresponds to keeping \( \lambda \) strictly between \(-1.5\) and \(1.5\).

Step 8: Determine the Correct Interval

The solution for \( \lambda \) results in two intervals. Option (C), \( \left(-\frac{3}{2}, \frac{3}{2}\right) \), is precisely the desired interval for **\( \lambda \)** such that the function has exactly one maximum and one minimum.

Key Concepts

Interval Analysis

Interval analysis is crucial in understanding where a function behaves in a certain way within a specified range. In this exercise, we are particularly interested in analyzing the function

• within the interval \((-\pi/2, \pi/2)\). Knowledge of intervals helps in determining the critical points where the function reaches a maximum or minimum value.

We start by identifying critical points by setting the derivative of the function to zero, as these points typically indicate where a function might have extreme values, such as maxima or minima. Critical points are only considered if they lie within the specified open interval. By evaluating the function across different intervals, we can ascertain the values of parameters, like \( \lambda \), that ensure the function has the desired number of maxima or minima. This kind of analysis is essential in mathematical optimization tasks.

Maxima and Minima

Maxima and minima are the peaks and valleys in a function's graph. Finding these helps in understanding how a function behaves. In the problem given, we want to determine where the function

• has exactly one maximum and one minimum.

First, calculate the derivative, because the points where the derivative is zero are often the candidate points for being maxima or minima. For the function\(f(x) = \sin^3 x + \lambda \sin^2 x\), the derivative is \(f'(x) = \cos x (3\sin^2 x + 2\lambda \sin x)\).After solving for \(\cos x = 0\) and finding no solutions within the interval \((-\pi/2, \pi/2)\), focus shifts to solving \(3\sin^2 x + 2\lambda \sin x = 0\). Finding values of \(\lambda\) which ensure one maximum and one minimum involves ensuring the inequality \(-1 < -\frac{2 \lambda}{3} < 1\) holds true. This results in the interval \((-\frac{3}{2}, \frac{3}{2})\).Using derivative tests can help confirm whether these critical points are indeed maxima or minima by checking the sign of the derivative around these points.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, play a vital role in this exercise. The function \(f(x) = \sin^3 x + \lambda \sin^2 x\)depends heavily on these trigonometric terms, influencing its critical points.Trig functions are periodic, meaning they repeat values in a regular pattern. This characteristic means they have well-known derivative behaviors which can help locate these critical points easily. For instance, the derivative of \(\sin x\) is \(\cos x\), and solving for critical points involves tackling these same trigonometric entities.In the context of this function:

• The zeroes of the cosine and sine functions determine critical points.
• Sine has values ranging from -1 to 1, which affects how we solve \(3\sin^2 x + 2\lambda \sin x = 0\). Ensuring \(\lambda\) satisfies the derived inequality narrows its obtainable range and hence determines the number of maxima and minima within the given interval.