Problem 9 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

For all \(x \in(0,1)\) IIIT (Screening)-2000] (a) \(e^{x}<1+x\) (b) \(\log _{e}(1+x)x\) (d) \(\log _{e} x>x\)

Step-by-Step Solution

Answer

The correct inequality is (b) \( \log_e(1+x) < x \) for \( x \in (0, 1) \).

Step 1: Analyze Option (a)

Consider the function \( f(x) = e^x - (1 + x) \). Let's find the derivative to determine the behavior of the function: \( f'(x) = e^x - 1 \). For \( x \in (0, 1) \), \( f'(x) > 0 \), meaning \( f(x) \) is increasing. At \( x = 0 \), \( f(0) = e^0 - (1 + 0) = 0 \). Since \( f(x) \) is increasing and starts at 0, \( f(x) > 0 \) for \( x > 0 \), thus \( e^x < 1 + x \) is false for all \( x \in (0, 1) \).

Step 2: Analyze Option (b)

Consider the function \( g(x) = \log_e(1+x) - x \). Compute its derivative: \( g'(x) = \frac{1}{1+x} - 1 \). For \( x \in (0, 1) \), we have \( g'(x) < 0 \) because \( \frac{1}{1+x} < 1 \). Therefore, \( g(x) \) is decreasing. At \( x = 0 \), \( g(0) = \log_e(1+0) - 0 = 0 \). Since \( g(x) \) is decreasing and starts at 0, \( g(x) < 0 \) for \( x > 0 \), thus \( \log_e(1+x) < x \) is true for \( x \in (0, 1) \).

Step 3: Analyze Option (c)

Consider the inequality \( \sin x > x \) for \( x \in (0, 1) \). The function \( h(x) = \sin x - x \) is defined. Compute its derivative: \( h'(x) = \cos x - 1 \). For small values of \( x \), particularly when close to 0, \( \cos x \approx 1 \), therefore \( h'(x) \leq 0 \) implying \( \sin x \leq x \). Thus, \( \sin x > x \) isn't true for all \( x \in (0, 1) \).

Step 4: Analyze Option (d)

For the function \( i(x) = \log_e x - x \), calculate its derivative: \( i'(x) = \frac{1}{x} - 1 \). For \( x \in (0, 1) \), \( \frac{1}{x} > 1 \), so \( i'(x) > 0 \); thus, \( i(x) \) is increasing. Near \( x = 0 \), \( \log_e x \to -\infty \), making \( \log_e x < x \) meaning \( \log_e x > x \) is false for all \( x \in (0, 1) \).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions with a constant base raised to a variable exponent. In the example problem, we look at the inequality involving the exponential function: \( e^x < 1 + x \). When interpreting this, \( e^x \) is the exponential function where the base \( e \) (approximately 2.718) is raised to the power \( x \). Exponential functions have several key characteristics:

• They grow rapidly as \( x \) increases. This means that as you plug in larger numbers for \( x \), the value of \( e^x \) increases significantly.
• At \( x = 0 \), the value of \( e^x \) is 1, because \( e^0 = 1 \).
• The derivative of an exponential function \( f(x) = e^x \) is itself: \( f'(x) = e^x \), showing continuous growth.

In this problem, the behavior of \( e^x \) compared to \( 1 + x \) is examined using derivatives. The function \( f(x) = e^x - (1 + x) \) displays an increase in value because its derivative \( f'(x) = e^x - 1 \) is greater than zero for \( x \) between 0 and 1. This means \( e^x < 1 + x \) isn't true, showcasing the rapid early rise of exponential functions.

Logarithmic Functions

Logarithmic functions have a base that is raised to a certain power to yield a given number. They are the inverse functions of exponentials. In our example, one focus is on \( \log_e(1+x) < x \). Here's what you should understand about logarithmic functions:

• They grow slower than exponential and linear functions.
• At\( x = 0 \), \( \log_e(1+x) \) evaluates to 0 because \(\log_e(1) = 0 \).
• The derivative \( g'(x) = \frac{1}{1+x} \) for \( \log_e(1+x) \) indicates that the rate of change decreases as \( x \) increases.

The inequality shows that \( \log_e(1+x) \) is less than \( x \) because the function \( g(x) = \log_e(1+x) - x \) decreases. Since the derivative \( g'(x) = \frac{1}{1+x} - 1 < 0 \), the function continually decreases for \( x \) in the given interval, confirming the inequality.

Trigonometric Functions

Trigonometric functions relate angles of a triangle to the ratios of its sides. Here, the focus is on \( \sin x > x \) for \( x \) between 0 and 1. Consider these properties of sine functions:

• \( \sin x \) is smooth and periodic with a maximum value of 1.
• Close to zero, sine behaves almost linearly: \( \sin x \approx x \).
• The derivative of \( \sin x \) is \( \cos x \), reflecting how the sine wave peaks and dips.

For small values close to zero, \( \cos x \approx 1 \), meaning the increase rate \( h'(x) = \cos x - 1 \approx 0 \). As such, \( \sin x \leq x \) often holds since the sine function changes more slowly than the linear \( x \). The inequality \( \sin x > x \) is not supported by the analysis for the entire interval.

Derivatives

The concept of derivatives is fundamental in calculus, describing how a function changes as its input changes. For instance, when considering whether \( e^x < 1 + x \), we compute the derivative \( f'(x) \) to see if \( f(x) = e^x - (1+x) \) is increasing or not.

• The derivative \( f'(x) = e^x - 1 \) suggests growth because it is positive for \( x \) in (0, 1).
• Taking derivatives helps us identify whether functions increase or decrease.
• The chain rule, product rule, and quotient rule are useful techniques in differentiation.

When you differentiate \( i(x) = \log_e x - x \) to obtain \( i'(x) = \frac{1}{x} - 1 \), it shows that the function is increasing in the interval since \( \frac{1}{x} > 1 \) for \( x \) between 0 and 1. Understanding derivatives not only solves inequalities but also illuminates function behaviors in specific intervals.