GATE CS 2014 – Set 1 – 1 Mark Question
Duration: 7 min
This video lesson is available to enrolled students.
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The video is an educational lecture by Yash Jain Sir from Knowledge Gate Educator, focusing on a calculus problem from the GATE CS 2014 exam. The problem defines a function f(theta) as a 3x3 determinant involving trigonometric terms. The instructor analyzes the properties of this function over the interval (pi/6, pi/3) to determine the truth of two statements regarding its derivative f'(theta). The core of the solution relies on evaluating the function at the endpoints of the interval and applying Rolle's Theorem, followed by determining if the function is constant or not to address the second statement.
Chapters
0:00 – 2:00 00:00-02:00
The video opens with a 'CALCULUS' title card before displaying a mathematical problem on a digital whiteboard. The problem defines a function f(theta) as a determinant with sin theta, cos theta, and tan theta in the first row, and fixed values for pi/6 and pi/3 in the subsequent rows. The variable theta is restricted to the interval (pi/6, pi/3). The instructor, Yash Jain Sir, asks the viewer to identify true statements regarding the derivative f'(theta). Statement (I) claims there exists a theta where f'(theta) = 0, and Statement (II) claims there exists a theta where f'(theta) != 0. He notes this is a 1-mark question from GATE CS 2014. He writes the interval endpoints as a and b and explicitly writes 'Rolle's theorem' on the board, signaling the method to be used.
2:00 – 5:00 02:00-05:00
The instructor begins solving by testing the boundary conditions required for Rolle's Theorem. He substitutes theta = pi/6 into the function. He points out that the first row becomes sin(pi/6), cos(pi/6), tan(pi/6), which is exactly the same as the second row. He explains the determinant property that if two rows are identical, the value is zero, so f(pi/6) = 0. He repeats this logic for theta = pi/3, where the first row matches the third row, resulting in f(pi/3) = 0. He writes f(a) = f(b) = 0 on the board. To visualize this, he draws a graph showing a curve starting and ending at the same height, illustrating the geometric interpretation of Rolle's Theorem. He also draws a horizontal line labeled 'constant' and writes f'(theta) = 0, considering the possibility that the function might be constant throughout the interval.
5:00 – 7:06 05:00-07:06
The instructor finalizes the solution by addressing both statements. He confirms that since f(a) = f(b) = 0 and the function is continuous and differentiable, Rolle's Theorem guarantees at least one point c where f'(c) = 0, validating Statement (I). He then refutes the idea that the function is constant. He explains that expanding the determinant yields a non-zero expression involving sin theta, cos theta, and tan theta, meaning the function varies. Because the function is not constant, its derivative cannot be zero everywhere, which implies there must be points where f'(theta) != 0, validating Statement (II). He circles the option 'Both I and II' and marks it as the correct answer. The video concludes with a black screen displaying 'THANKS FOR WATCHING' in white text.
The lecture demonstrates a classic application of Rolle's Theorem in the context of determinants. By evaluating the function at the boundaries, the instructor establishes the necessary condition f(a) = f(b) to prove the existence of a critical point. The second part of the problem requires recognizing that satisfying the boundary condition does not imply the function is constant, thereby ensuring the derivative is non-zero at other points. This dual analysis confirms both statements are true.