Problem 64 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Show that the equation \(x=a \sin x+b\), where \(00\), has at least one positive root which does not exceed \(a+b\).

Step-by-Step Solution

Answer

Based on the properties of the sine function and the defined values of variables \(a\) and \(b\), as well as the application of the Intermediate Value Theorem, it can be shown that the given equation \(x=a \sin x+b\) has at least one positive root that does not exceed \(a+b\).

Step 1 Understand properties of the sine function

Firstly, it's important to understand that the sine function fluctuates between -1 and 1, i.e., \(-1 \leq \sin x \leq 1\). Therefore, the result of \(a \sin x\) will be from \(-a\) to \(a\). Adding \(b\) to it, we get \(-a+b\leq a \sin x + b \leq a + b\).

Step 2 Apply the conditions on a and b

Given that \(00\), it means that the lower limit \(-a+b\) is positive and \(a + b\) is greater than \(b\) which is also positive. Therefore, the range of \(a \sin x + b\) is in the positive domain.

Step 3 Apply Intermediate Value Theorem (IVT)

The intermediate value theorem (IVT) states that if a function \(f\) is continuous on the real line and \(d\) is between \(f(a)\) and \(f(b)\), then there exists \(c\) between \(a\) and \(b\) such that \(f(c) = d\). According to the conditions, the function \(f(x) = a \sin x + b - x\) is continuous on \([0, a+b]\) and \(f(0) = b > 0\) while \(f(a+b) = a \sin(a+b) + b - (a+b) \leq a + b - (a+b) = 0\). So, by the IVT, the function \(f(x)\) has at least one positive root on \([0, a+b]\), which means that equation \(x = a \sin x + b\) has at least one positive solution.