JEE Main Maths Mock Test -1
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Jee Main Maths Mock Test -1
Total Questions: 20
Total Marks: 80
Duration: 48 Minutes
- Correct Answer : 4 Marks
- Wrong Answer: -1 Mark
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Question 1 of 20
1. Question
If $A = {1, 2, 3, 4}$ and $B = {2, 3, 5}$, then $A \cup B$ is
Correct
$A \cup B$ is the union of both sets, which includes all distinct elements from both sets.
Incorrect
$A \cup B$ is the union of both sets, which includes all distinct elements from both sets.
Unattempted
$A \cup B$ is the union of both sets, which includes all distinct elements from both sets.
-
Question 2 of 20
2. Question
If $z = 3 + 4i$, find $|z|$.
Correct
$|z| = \sqrt{3^2 + 4^2} = 5$
Incorrect
$|z| = \sqrt{3^2 + 4^2} = 5$
Unattempted
$|z| = \sqrt{3^2 + 4^2} = 5$
-
Question 3 of 20
3. Question
The determinant of $\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$ is
Correct
$\text{Det} = (2 \cdot 5) – (3 \cdot 4) = 10 – 12 = -7$
Incorrect
$\text{Det} = (2 \cdot 5) – (3 \cdot 4) = 10 – 12 = -7$
Unattempted
$\text{Det} = (2 \cdot 5) – (3 \cdot 4) = 10 – 12 = -7$
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Question 4 of 20
4. Question
How many 3-letter words can be formed using the letters A, B, C without repetition?
Correct
Permutations without repetition: $3! = 6$
Incorrect
Permutations without repetition: $3! = 6$
Unattempted
Permutations without repetition: $3! = 6$
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Question 5 of 20
5. Question
The general term of a binomial expansion $(a+b)^n$ is
Correct
The formula for the binomial expansion term is $T_k = C(n, k) a^{n-k} b^k$
Incorrect
The formula for the binomial expansion term is $T_k = C(n, k) a^{n-k} b^k$
Unattempted
The formula for the binomial expansion term is $T_k = C(n, k) a^{n-k} b^k$
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Question 6 of 20
6. Question
The sum of an infinite geometric series with the first term 3 and common ratio 1/2 is
Correct
$S = \frac{a}{1-r} = \frac{3}{1-1/2} = 6$
Incorrect
$S = \frac{a}{1-r} = \frac{3}{1-1/2} = 6$
Unattempted
$S = \frac{a}{1-r} = \frac{3}{1-1/2} = 6$
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Question 7 of 20
7. Question
Find the slope of the line passing through points (1, 2) and (3, 6).
Correct
$m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{6-2}{3-1} = 2$
Incorrect
$m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{6-2}{3-1} = 2$
Unattempted
$m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{6-2}{3-1} = 2$
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Question 8 of 20
8. Question
Evaluate $\lim_{x \to 2} \frac{x^2 – 4}{x-2}$.
Correct
Using direct substitution: $\lim_{x \to 2} \frac{x^2 – 4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = 4$
Incorrect
Using direct substitution: $\lim_{x \to 2} \frac{x^2 – 4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = 4$
Unattempted
Using direct substitution: $\lim_{x \to 2} \frac{x^2 – 4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = 4$
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Question 9 of 20
9. Question
The derivative of $f(x) = \sin x + \cos x$ is
Correct
$f‘(x) = \cos x – \sin x$
Incorrect
$f‘(x) = \cos x – \sin x$
Unattempted
$f‘(x) = \cos x – \sin x$
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Question 10 of 20
10. Question
The value of $\int_0^\pi \sin x dx$ is
Correct
$\int_0^\pi \sin x dx = [-\cos x]_0^\pi = 2$
Incorrect
$\int_0^\pi \sin x dx = [-\cos x]_0^\pi = 2$
Unattempted
$\int_0^\pi \sin x dx = [-\cos x]_0^\pi = 2$
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Question 11 of 20
11. Question
The area of a triangle with vertices at (0, 0), (4, 0), and (4, 3) is
Correct
Using the formula for the area of a right triangle: $\frac{1}{2} \times 4 \times 3 = 6$
Incorrect
Using the formula for the area of a right triangle: $\frac{1}{2} \times 4 \times 3 = 6$
Unattempted
Using the formula for the area of a right triangle: $\frac{1}{2} \times 4 \times 3 = 6$
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Question 12 of 20
12. Question
Find the value of $a$ if the roots of the equation $x^2 – 4x + a = 0$ are equal.
Correct
For equal roots, discriminant must be 0: $b^2 – 4ac = 0$, giving $a = 4$
Incorrect
For equal roots, discriminant must be 0: $b^2 – 4ac = 0$, giving $a = 4$
Unattempted
For equal roots, discriminant must be 0: $b^2 – 4ac = 0$, giving $a = 4$
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Question 13 of 20
13. Question
If $\vec{a} = 2\hat{i} + 3\hat{j}$ and $\vec{b} = \hat{i} – \hat{j}$, then $\vec{a} \cdot \vec{b}$ is
Correct
$\vec{a} \cdot \vec{b} = 2 \times 1 + 3 \times (-1) = 2 – 3 = -1$
Incorrect
$\vec{a} \cdot \vec{b} = 2 \times 1 + 3 \times (-1) = 2 – 3 = -1$
Unattempted
$\vec{a} \cdot \vec{b} = 2 \times 1 + 3 \times (-1) = 2 – 3 = -1$
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Question 14 of 20
14. Question
The probability of getting a sum of 7 when two dice are rolled is
Correct
There are 6 favorable outcomes out of 36 total outcomes, so $\frac{6}{36} = \frac{1}{6}$
Incorrect
There are 6 favorable outcomes out of 36 total outcomes, so $\frac{6}{36} = \frac{1}{6}$
Unattempted
There are 6 favorable outcomes out of 36 total outcomes, so $\frac{6}{36} = \frac{1}{6}$
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Question 15 of 20
15. Question
If $\cos \theta = \frac{3}{5}$, find $\sin \theta$ in the first quadrant.
Correct
Using $\sin^2 \theta + \cos^2 \theta = 1$: $\sin \theta = \frac{4}{5}$
Incorrect
Using $\sin^2 \theta + \cos^2 \theta = 1$: $\sin \theta = \frac{4}{5}$
Unattempted
Using $\sin^2 \theta + \cos^2 \theta = 1$: $\sin \theta = \frac{4}{5}$
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Question 16 of 20
16. Question
Find the integral $\int x^2 dx$.
Correct
$\int x^2 dx = \frac{x^3}{3} + C$
Incorrect
$\int x^2 dx = \frac{x^3}{3} + C$
Unattempted
$\int x^2 dx = \frac{x^3}{3} + C$
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Question 17 of 20
17. Question
Solve the differential equation $\frac{dy}{dx} = 3x^2$.
Correct
Integrate both sides: $y = x^3 + C$
Incorrect
Integrate both sides: $y = x^3 + C$
Unattempted
Integrate both sides: $y = x^3 + C$
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Question 18 of 20
18. Question
Find the coordinates of the centroid of a triangle with vertices at (0, 0), (6, 0), and (0, 8).
Correct
Centroid formula: $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = (2, \frac{8}{3})$
Incorrect
Centroid formula: $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = (2, \frac{8}{3})$
Unattempted
Centroid formula: $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = (2, \frac{8}{3})$
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Question 19 of 20
19. Question
Find the value of $\tan^{-1}(1) + \tan^{-1}(1)$.
Correct
$\tan^{-1}(1) = \frac{\pi}{4}$, so $\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$
Incorrect
$\tan^{-1}(1) = \frac{\pi}{4}$, so $\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$
Unattempted
$\tan^{-1}(1) = \frac{\pi}{4}$, so $\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$
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Question 20 of 20
20. Question
The vector $\vec{v} = 5\hat{i} + 12\hat{j}$ has magnitude
Correct
$|\vec{v}| = \sqrt{5^2 + 12^2} = 13$
Incorrect
$|\vec{v}| = \sqrt{5^2 + 12^2} = 13$
Unattempted
$|\vec{v}| = \sqrt{5^2 + 12^2} = 13$