Gate CS 2014 - Set 1 - 2 Marks Question

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This educational video features a lecture by Yash Jain Sir from Knowledge Gate Educator, focusing on a calculus problem from the GATE CS 2014 exam. The problem asks to identify a true statement about a continuous function f(x) defined on the interval [0, 2], given specific values: f(0) = -1, f(2) = -1, and f(1) = 1. The instructor systematically analyzes the four multiple-choice options, ultimately proving that Option A is correct using the Intermediate Value Theorem. He constructs an auxiliary function to demonstrate the existence of a specific point y within the interval (0, 1) that satisfies the condition f(y) = f(y+1).

Chapters

1. 0:00 – 2:00 00:00-02:00

   The video begins with the instructor presenting the problem statement clearly on the screen. The text reads: 'Q. A function is continuous in the interval [0, 2]. It is known that f(0) = f(2) = -1 and f(1) = 1. Which one of the following statements must be true. (GATE CS 2014) (SET-1) (2 MARKS)'. He points to the options A, B, C, and D. He writes 'f(x) [0, 2]' on the whiteboard to establish the domain of the function. He emphasizes the continuity condition and the specific function values provided in the question.

2. 2:00 – 5:00 02:00-05:00

   The instructor begins analyzing the options, specifically targeting Option A: 'There exist a y in the interval (0, 1), such that f(y) = f(y+1)'. He decides to test this statement by defining a new function, which he labels g(y). He writes the equation 'g(y) = f(y) - f(y+1)' on the board. He explains that since f(x) is continuous, the function g(y) must also be continuous. He writes 'f(x) -> con' and 'g(x) -> con' to denote continuity. He notes that the domain for y is (0, 1), which implies y+1 is in (1, 2), keeping the function within the original domain [0, 2].

3. 5:00 – 6:26 05:00-06:26

   The instructor proceeds to evaluate the function g(y) at the boundaries of the interval (0, 1). He calculates g(0) by substituting y=0 into the equation: 'g(0) = f(0) - f(1) = -1 - 1 = -2'. Next, he calculates g(1): 'g(1) = f(1) - f(2) = 1 - (-1) = +2'. He draws a coordinate system with the y-axis representing the function value and the horizontal axis representing y. He marks -2 at y=0 and +2 at y=1. He draws a curve connecting these points, showing it must cross the x-axis (where g(y)=0) somewhere between 0 and 1. He concludes that since g(0) is negative and g(1) is positive, by the Intermediate Value Theorem, there exists a y in (0, 1) such that g(y) = 0. This implies f(y) - f(y+1) = 0, or f(y) = f(y+1), confirming Option A is the correct answer.

The lecture effectively demonstrates a problem-solving strategy for calculus questions involving continuous functions. By defining an auxiliary function g(y) = f(y) - f(y+1), the instructor transforms the problem of finding a point where two function values are equal into a root-finding problem. The calculation of g(0) = -2 and g(1) = +2 provides the necessary sign change to apply the Intermediate Value Theorem. This guarantees the existence of a solution y in the interval (0, 1), validating the statement in Option A. The visual aids, including the board writing and graph sketching, reinforce the logical steps taken to reach the conclusion.