Problem 100 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)\) be a non-negative continuous function such that the area bounded by the curve \(y=f(x), x\)-axis and the ordinates \(x=\frac{\pi}{4}\) and \(x=\beta>\frac{\pi}{4}\) is \(\left(\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta\right) .\) Then, \(\left(\frac{\pi}{2}\right) f\) is (A) \(\left(1-\frac{\pi}{4}-\sqrt{2}\right)\) (B) \(\left(1-\frac{\pi}{4}+\sqrt{2}\right)\) (C) \(\left(\frac{\pi}{4}+\sqrt{2}-1\right)\) (D) \(\left(\frac{\pi}{4}-\sqrt{2}+1\right)\)

Step-by-Step Solution

Answer

(B) \(1 - \frac{\pi}{4} + \sqrt{2}\).

Step 1: Understand the Problem

We need to find the value of \(f\left(\frac{\pi}{2}\right)\) given an integral expression for the area under \(y = f(x)\) from \(x = \frac{\pi}{4}\) to \(x = \beta\). The given area is equal to \((\beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta)\).

Step 2: Set Up the Integral Expression

The area under the curve from \(x = \frac{\pi}{4}\) to \(x = \beta\) is given by \(\int_{\frac{\pi}{4}}^{\beta} f(x) \, dx\). This integral is set equal to the given expression: \(\int_{\frac{\pi}{4}}^{\beta} f(x) \, dx = \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta\).

Step 3: Differentiate Both Sides with Respect to \(\beta\)

Differentiate both sides of the equation with respect to \(\beta\). The Fundamental Theorem of Calculus tells us that the left side becomes \(f(\beta)\) when differentiated. For the right-hand side, use the product and chain rules: differentials of \(\beta \sin \beta\), \(\frac{\pi}{4} \cos \beta\), and \(\sqrt{2} \beta\) are \(\sin \beta + \beta \cos \beta\), \(-\frac{\pi}{4} \sin \beta\), and \(\sqrt{2}\).

Step 4: Solve for \(f(\beta)\)

Equating the derivatives from Step 3, we find \(f(\beta) = \sin \beta + \beta \cos \beta - \frac{\pi}{4} \sin \beta + \sqrt{2}\). Simplifying, this gives \(f(\beta) = \left(1 - \frac{\pi}{4}\right) \sin \beta + \beta \cos \beta + \sqrt{2}\).

Step 5: Substitute \(\beta = \frac{\pi}{2}\) to Find \(f\left(\frac{\pi}{2}\right)\)

Substitute \(\beta = \frac{\pi}{2}\) into the expression for \(f(\beta)\): \(f\left(\frac{\pi}{2}\right) = \left(1 - \frac{\pi}{4}\right) \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) + \sqrt{2}\).

Step 6: Simplify Result to Find \(f\left(\frac{\pi}{2}\right)\)

Calculate \(\sin\left(\frac{\pi}{2}\right) = 1\) and \(\cos\left(\frac{\pi}{2}\right) = 0\). Thus, \(f\left(\frac{\pi}{2}\right) = \left(1 - \frac{\pi}{4}\right)\cdot 1 + 0 + \sqrt{2}\). Simplifying, \(f\left(\frac{\pi}{2}\right) = 1 - \frac{\pi}{4} + \sqrt{2}\).

Step 7: Match with Options

Compare \(f\left(\frac{\pi}{2}\right)\) with the given options. We find it matches option (B): \(1 - \frac{\pi}{4} + \sqrt{2}\).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of Integral Calculus. This theorem connects differentiation with integration, showing they are essentially inverse operations. It has two main parts:

* **First Part:** It tells us that if a function is continuous over an interval, then its integral can be computed using an antiderivative of that function. In simpler terms, for a function \(f(x)\), the integral from \(a\) to \(b\) is \(F(b) - F(a)\), where \(F(x)\) is an antiderivative of \(f(x)\).

* **Second Part:** It states that the derivative of an integral of a function is the original function itself. More formally, if \F(x) = \int_{a}^{x} f(t) \, dt\ then \F'(x) = f(x)\.

In our exercise, this theorem was used in step 3 to differentiate the integral, which led us to finding \(f(\beta)\). This reliance on the fundamental theorem simplifies problems where we are given an integral and need to extract a function value.

Continuous Function

A continuous function is a mathematical concept where a function \(f(x)\) has no abrupt changes, jumps, or breaks in its graph. It is crucial for the application of calculus and the fundamental theorem. Here are some features:

* **No Jumps:** The function doesn't suddenly break off and continue somewhere else. Think of a seamless curve.
* **Defined Everywhere:** For it to be continuous over an interval, it must be defined at every point within that interval.
* **Limit Matches Value:** At any point \(c\), the limit of the function as \(x\) approaches \(c\) should equal \(f(c)\).

In this exercise, \(f(x)\) is given as a non-negative, continuous function. This ensures that the integral \int_{rac{\pi}{4}}^{\beta} f(x) \, dx\ can be properly evaluated, and \(f(x)\) is smooth and predictable between these bounds.

Definite Integral

Definite Integrals are a vital concept, allowing us to compute the area under a curve for a specific interval. Here's what makes them special:

* **Bounds:** They have set limits on integration, defining the region over which you calculate the area, such as from \(a\) to \(b\) in \int_{a}^{b} f(x) \, dx\.
* **Numerical Result:** Unlike indefinite integrals, a definite integral results in a number representing the total accumulated value between the bounds.
* **Area Interpretation:** In real-world contexts, the definite integral can be interpreted as the total area under the curve of the function, above the x-axis.

In the original problem, we set up a definite integral from \frac{\pi}{4}\ to \beta\, representing the area under \(f(x)\). Solving this gave us the expression for the area, linking it back to \(f(x)\) through differentiation, as per the fundamental theorem of calculus.