MHT CET 2024 15 May Shift 1 question paper is available here . MHT CET 2024 question paper comprises comprises 150 MCQs carrying a total weightage of 200 marks.
MHT CET 2024 Question Paper for PCM is divided into three subjects- Physics, Chemistry and Mathematics. The Physics and Chemistry and Mathematics section consists of 50 questions for each section (10 questions from Class 11 and 40 questions from Class 12th syllabus).
MHT CET 2024 15 May Shift 1 Question Paper PDF Download
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MHT CET 2024 15 May Shift 1 Detailed Solutions
First section is of Mathematics:-
Question 1:
The statement (p ∧ (~q)) ∨ ((~p) ∧ q) ∨ ((~p) ∧ (~q)) is equivalent to:
(a) (~p) ∨ (~q)
(b) p ∨ (~q)
(c) p ∨ q
(d) (~p) ∨ q
Question 2:
If B = [[3, a, -1], [1, 3, 1], [-1, 1, 3]] is the adjoint of a 3×3 matrix A and |A| = 4, then a is equal to:
(a) 1
(b) 2
(c) -1
(d) 0
Question 3:
If A = [[0, 1, 2], [1, 2, 3], [3, 1, 1]], then A⁻¹ is:
(a) [[1/2, -1/2, 1/2], [-4, 3, -1], [5/2, -3/2, 1/2]]
(b) [[1, 0, -1], [2, -1, 3], [0, 1, 2]]
(c) [[0, 1, 0], [1, -1, 2], [4, 0, -1]]
(d) None of these
Question 4:
The value of sin(cot⁻¹(x)) is:
(a) 1/√(1 + x²)
(b) √(1 + x²)
(c) 1/(x√(1 + x²))
(d) x√(1 + x²)
View Solution
Let θ = cot⁻¹(x). Then, cot(θ) = x.
1. Represent θ in a right triangle: cot(θ) = adjacent / opposite = x / 1.
2. Hypotenuse = √(x² + 1).
3. sin(θ) = opposite / hypotenuse = 1 / √(x² + 1).
Thus, sin(cot⁻¹(x)) = 1 / √(1 + x²).
Question 5:
If y = 5cos(x) - 3sin(x), then d²y/dx² + y equals:
View Solution
We are given:
dy/dx = -5sin(x) - 3cos(x).
Step 1: Compute the second derivative.
Differentiate dy/dx with respect to x:
d²y/dx² = -5cos(x) + 3sin(x).
Step 2: Add y to the second derivative.
y = 5cos(x) - 3sin(x). Add y to d²y/dx²:
d²y/dx² + y = (-5cos(x) + 3sin(x)) + (5cos(x) - 3sin(x)).
Step 3: Simplify the expression.
Combine terms:
d²y/dx² + y = 0.
Question 6:
A wire of length 20 units is divided into two parts such that the product of one part and the cube of the other part is maximum. The product of these parts is:
View Solution
Let x be one part and y be the other part.
Step 1: Write the condition.
x + y = 20, so y = 20 - x.
We aim to maximize f(x) = (20 - x)x³ = 20x³ - x⁴.
Step 2: Differentiate to find critical points.
f'(x) = 60x² - 4x³. Solve f'(x) = 0:
4x²(15 - x) = 0, so x = 0 or x = 15.
Step 3: Evaluate the second derivative.
f''(x) = 120x - 12x². At x = 15:
f''(15) = -900 < 0, so f(x) is maximum at x = 15.
Step 4: Compute the product.
y = 20 - x = 5. Product = 15 × 5 = 75.
Question 7:
The integral of sec²/³(x) csc⁴/³(x) dx from π/6 to π/3 is equal to:
View Solution
Step 1: Rewrite the integral:
I = ∫ sec²/³(x) csc⁴/³(x) dx from π/6 to π/3.
Step 2: Use substitution t = tan(x). Adjust limits:
x = π/6 → t = 1/√3; x = π/3 → t = √3.
Step 3: Simplify using substitution.
I = ∫ t^(2/3)/(1 + t²)^(2/3) dt. Further simplify and solve:
I = 3^(7/6) - 3^(5/6).
Question 8:
The distribution function F(X) of a discrete random variable X is given. Then P[X = 4] + P[X = 5]:
| X | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| F(X) | 0.2 | 0.37 | 0.48 | 0.62 | 0.85 | 1 |
Question 9:
If the lines (x - k)/2 = (y + 1)/3 = (z - 1)/4 and (x - 3)/1 = (y - 9/2)/2 = z/1 intersect, then the value of k is:
Question 2:
From the options given below, which one of the following is a primary amine?
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