Geometric Progression is an important topic in the Mathematics section in MHT CET exam. Practising this topic will increase your score overall and make your conceptual grip on MHT CET exam stronger.
This article gives you a full set of MHT CET PYQs for Geometric Progression with explanations for effective preparation. Practice of MHT CET Mathematics PYQs including Geometric Progression questions regularly will improve accuracy, speed, and confidence in the MHT CET 2026 exam.
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MHT CET PYQs for Geometric Progression with Solutions
1.
The function \( f(x) = \tan^{-1} (\sin x + \cos x) \) is an increasing function in:
- \( \left( 0, \frac{\pi}{2} \right) \)
- \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
- \( \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \)
- \( \left( -\frac{\pi}{2}, \frac{\pi}{4} \right) \)
2.
Evaluate \( i^2 + i^3 + \dots + i^{4000} \):- \( 1 \)
- \( 0 \)
- \( i \)
- \( -i \)
3.
The absolute maximum value of the function \( f(x) = 2x^3 - 3x^2 - 36x + 9 \) defined on \( [-3, 3] \) is- \( 36 \)
- \( 53 \)
- \( 63 \)
- \( 72 \)
4.
Calculate the area of the intersection between the disk \(x^2 + y^2 \leq 1\) and the region defined by \(y^2 \leq 1 - x\).
- \( \frac{\pi}{2} - \frac{2}{3} \)
- \( \frac{\pi}{2} + \frac{2}{3} \)
- \( \frac{\pi}{2} + \frac{4}{3} \)
- \( \frac{\pi}{2} - \frac{4}{3} \)
5.
The tangent at the point \( (x_1, y_1) \) on the curve \( y = x^3 + 3x^2 + 5 \) passes through the origin. Then \( (x_1, y_1) \) does NOT lie on the curve:- \( {x^2} + \frac{y^2}{81} = 2 \)
- \( \frac{y^2}{9} - x^2 = 8 \)
- \( y = 4x^2 + 5 \)
- \( \frac{x}{3} - y^2 = 2 \)
6.
Let the following system of equations: $$ kx + y + z = 1, \quad x + ky + z = k, \quad x + y + kz = k^2 $$ have no solution. Find $|k|$:- \( 0 \)
- \( 1 \)
- \( 2 \)
- \( 3 \)
7.
Evaluate the following limit: \[ \lim_{\theta \to -\frac{\pi}{4}} \frac{\cos\theta + \sin\theta}{\theta + \frac{\pi}{4}}. \]- \( 0 \)
- \( 1 \)
- \( \sqrt{2} \)
- \( \frac{1}{\sqrt{2}} \)
8.
The equation of a circle with center \( (5, 4) \) and touching the \( Y \)-axis is:- \( x^2 + y^2 - 10x - 8y - 16 = 0 \)
- \( x^2 + y^2 - 10x - 8y - 16 = 0 \)
- \( x^2 + y^2 + 10x + 8y + 16 = 0 \)
- \( x^2 + y^2 - 10x - 8y + 16 = 0 \)
9.
If \( x = \frac{1 - t^2}{1 + t^2} \) and \( y = \frac{2t}{1 + t^2} \), then \( \frac{dy}{dx} \) is equal to:- \( -\frac{y}{x} \)
- \( \frac{y}{x} \)
- \( -\frac{x}{y} \)
- \( \frac{x}{y} \)
10.
Two players A and B are alternately throwing a coin and a die together. A player who first throws a head and a 6 wins the game. If A starts the game, then the probability that B wins the game is:- \( \frac{12}{23} \)
- \( \frac{11}{23} \)
- \( \frac{5}{119} \)
- \( \frac{12}{119} \)
11.
Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k}, \, \vec{b} = \hat{i} - \hat{j} + \hat{k}, \, \text{and} \, \vec{c} = \hat{i} + \hat{j} - \hat{k} \). A vector in the plane of \( \vec{a} \) and \( \vec{b} \) whose projection on \( \vec{c} \) is \( \frac{1}{\sqrt{3}} \), is:- \( 4\hat{i} - \hat{j} + 4\hat{k} \)
- \( 3\hat{i} + \hat{j} - 3\hat{k} \)
- \( 2\hat{i} + \hat{j} - 2\hat{k} \)
- \( 4\hat{i} + \hat{j} - 4\hat{k} \)
12.
Let \( f(x) = \frac{2 - \sqrt{x + 4}}{\sin 2x}, \, x \neq 0 \). In order that \( f(x) \) is continuous at \( x = 0 \), \( f(0) \) is to be defined as:
- \( -\frac{1}{8} \)
- \( \frac{1}{2} \)
- \( 1 \)
- \( \frac{1}{8} \)
13.
The range of the function \( f(x) = \sin^{-1}(x - \sqrt{x}) \) is equal to?
- \( \left[ \sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right] \)
- \( \left[ \sin^{-1}\left(\frac{1}{2}\right), \frac{\pi}{2} \right] \)
- \( \left[ -\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2} \right] \)
- \( \left[ -\sin^{-1}\left(\frac{1}{2}\right), \frac{\pi}{2} \right] \)
14.
If \( A = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} \), then \( A^{50} \) is:
\( \begin{bmatrix} 1 & 0 \\ 0 & 50 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 25 \\ 0 & 1 \end{bmatrix} \)
15.
The solution of the differential equation \( \sqrt{1 - y^2} \, dx + x \, dy - \sin^{-1} y \, dy = 0 \) is:- \( x = \sin^{-1} y - 1 + ce^{-\sin^{-1} y} \)
- \( y = x\sqrt{1 - y^2} + \sin^{-1} y + c \)
- \( x = 1 + \sin^{-1} y + ce^{\sin^{-1} y} \)
- \( y = \sin^{-1} y - 1 + x\sqrt{1 - y^2} + c \)
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