12th Maths Question Paper 2024 Maharashtra Board Pdf

Maharashtra State Board Class 12th Maths Question Paper 2024 with Solutions Answers Pdf Download.

Class 12 Maths Question Paper 2024 Maharashtra State Board with Solutions

Time: 3 Hrs.
Max. Marks: 80

General Instructions: The question paper is divided into FOUR sections.

1. Section A: Q.1 contains Eight multiple choice type of questions, each carrying Two marks.
   Q.2 contains Four very short answer type questions, each carrying one mark.
2. Section B: Q.3 to Q. 14 contain Twelve short answer type questions, each carrying Two marks. (Attempt any Eight)
3. Section C: Q.15 to Q.26 contain Twelve short answer type questions, each carrying Three marks. (Attempt any Eight)
4. Section D: Q. 27 to Q.34 contain Eight long answer type questions, each carrying Four marks. (Attempt any Five)
5. Use of log table is allowed. Use of calculator is not allowed.
6. Figures to the right indicate full marks.
7. Use of graph paper is not necessary. Only rough sketch of graph is expected.
8. For each multiple choice type of question, Only the first attempt will be considered for evaluation.
9. Start answer to each section on a new page.

Section – A

Question 1.
Select and write the correct answer for the following multiple choice type of questions: [16]

(i) The dual of statement t ∨ (p ∨ q) is ……….. (2)
(a) c ∧ (p ∨ q)
(b) c ∧ (p ∧ q)
(c) t ∧ (p ∧ q)
(d) t ∧ (p ∨ q)
Answer:
(b) c ∧ (p ∧ q)
To obtain the dual of a statement, we replace ‘∨’ with ‘∧’ and ‘t’ with ‘c’
∴ The dual of statement t ∨ (p ∨ q) is c ∧ (p ∧ q).

(ii) The principle solutions of the equation cos θ = \(\frac{1}{2}\) are ………. (2)
(a) \(\frac{\pi}{6}, \frac{5 \pi}{6}\)
(b) \(\frac{\pi}{6}, \frac{7 \pi}{6}\)
(c) \(\frac{\pi}{3}, \frac{5 \pi}{3}\)
(d) \(\frac{\pi}{3}, \frac{2 \pi}{6}\)
Answer:

(iii) If α, β, γ are direction angles of a line and α = 60°, β = 45°, then γ = _____. (2)
(a) 30° or 90°
(b) 45° or 60°
(c) 90° or 130°
(d) 60° or 120°
Answer:
(d) 60° or 120°
Given that ∝ = 60°, β = 45°, γ = ?
We know that,
cos² ∝ + cos² β + cos² γ = 1
[Properties of direction angle]
⇒ cos² (60°) + cos² (45°) + cos² γ = 1
⇒ \(\left(\frac{1}{2}\right)^2\) + \(\left(\frac{1}{\sqrt{2}}\right)^2\) + cos² γ = 1
⇒ \(\frac{1}{4}+\frac{1}{2}\) + cos² γ = 1
⇒ cos² γ = 1 – \(\frac{3}{4}\)
⇒ cos² γ = \(\frac{1}{4}\)
⇒ cos γ = ±\(\frac{1}{2}\)
⇒ cos γ = \(\frac{1}{2}\)
or cos γ = –\(\frac{1}{2}\)
cos γ = cos\(\frac{\pi}{3}\)
or cos γ = – cos\(\frac{\pi}{3}\)
⇒ γ = \(\frac{\pi}{3}\)
or cos γ = cos[π – \(\frac{\pi}{3}\)]
⇒ γ = π – \(\frac{\pi}{3}\)
γ = \(\frac{2 \pi}{3}\)
γ = \(\frac{\pi}{3}\) = 60°
or γ = \(\frac{2 \pi}{3}\) = 120°
Hence, γ = 60° or 120°.

(iv) The perpendicular distance of the plane \(\bar{r}\).(3\(\hat{i}\) +4\(\hat{j}\) +12\(\hat{k}\)) = 78, from the origin is ____. (2)
(a) 4
(b) 5
(c) 6
(d) 8
Answer:
(c) 6

(v) The slope of the tangent to the curve x = sin θ and u = cos 2θ at θ = \(\frac{\pi}{6}\) is ____.
(a) \(-2 \sqrt{3}\)
(b) \(\frac{-2}{\sqrt{3}}\)
(c) -2
(d) \(-\frac{1}{2}\)
Answer:
(c) -2
Given, x = sin θ and y = cos 2θ
On differentiate w.r.t θ, we get

(vi) If \(\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} x^3 \cdot \sin ^4 x d x\) = k then k = _____. (2)
(a) 1
(b) 2
(c) 4
(d) 0
Answer:
(d) 0
Let f(x) = x³ . sin⁴x
∴ f(- x) = [-x]³ [sin(-x)]⁴
∴ = -x³(-sin x)⁴
∴ = -f(x).
∴ f(x) is on odd function.
⇒ \(\int_{-\pi / 4}^{\pi / 4}\)x³.sin⁴ x dx = 0
[∵ F(x) is an odd number]
k = 0

(vii) The integrating factor of linear differential equation x\(\frac{d y}{d x}\) + 2y = x² log x is ___. (2)
(a) x
(b) \(\frac{1}{x}\)
(c) x²
(d) \(\frac{1}{x^2}\)
Answer:
(c) x²
\(\frac{d y}{d x}\) + 2y = x²log x
Given. \(\frac{d y}{d x}+\frac{2}{x} y\) = x log x
Here p = \(\frac{2}{x}\)
Thus integrating factor is.
I.F. = \(e^{\int \mathrm{Pd} d x}\) = \(e^{\int \frac{2}{x} d x}\)
= \(e^{2 \log x}\)
= x²

(viii) If the mean and variance of a binomial distribution are 18 and 12 respectively, then the value of n is ___. (2)
(a) 36
(b) 54
(c) 18
(d) 27
Answer:
(b) 54
Mean = np = 18 …….. (i)
variance = npq = 12 …….. (ii)
Equation (ii) divide by equation (i), we get
So, \(\frac{n p q}{n p}\) = \(\frac{12}{18}\)
∴ q = \(\frac{2}{3}\)
Now, p = 1 – q = 1 – \(\frac{2}{3}\)
∴ p = \(\frac{1}{3}\).
Put the value of p in equation (i).
np = 18
⇒ n[latex]\frac{1}{3}[/latex] = 18
⇒ n = 54

Question 2.
Answer the following questions: [4]

(i) Write the compound statement ‘Nagpur is in Maharashtra and Chennai is in Tamilnadu symbolically. (1)
Answer:
Let p : Nagpur is in Maharashtra
q : Chennai is in Tamilnadu
∴ The symbolic form of the given statement is [p ∧ q]

(ii) If the vectors 2\(\hat{i}\) – 3\(\hat{j}\) + 4\(\hat{k}\) and p\(\hat{i}\) + 6\(\hat{j}\) – 8\(\hat{k}\) are collinear, then find the value of p. (1)
Answer:
The given vectors are 2\(\hat{i}\) – 3\(\hat{j}\) + 4\(\hat{k}\) and p\(\hat{i}\) + 6\(\hat{j}\) – 8\(\hat{k}\)
According to the question,
∴ The given vectors are collinear

(iii) Evaluate : ∫\(\frac{1}{x^2+25}\)dx
Answer:

(iv) A particle is moving along X-axis. Its acceleration at time t is proportional to its velocity at that time. Find the differential equation of the motion of the particle.
Answer:
Let x be the distance travelled by the partide at time t.
Now,
the velocity is given by \(\frac{d x}{d t}\) and acceleration is given by \(\frac{d^2 x}{d t^2}\)
According to the question.
\(\frac{d^2 x}{d t^2}\) \(\frac{d x}{d t}\) ∝
\(\frac{d^2 x}{d t^2}\) = k.\(\frac{d x}{d t}\)
where k is a constant of proportionalitly.
This is the required differential equation.

Section – B

Attempt any EIGHT of the following questions:

Question 3.
Construct the truth table for the statement pattern: (2)
[(p → q ∧ q] → p
Answer:
Construct the truth table for the statement attern [(p → q) ∧ q] → p

Question 4.
Check whether the matrix \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) is invertible or not. (2)
Answer:
Let A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\)
then, |A| = \(\left|\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right|\)
= cos²θ – (-sin²θ)
= cos²θ + sin²θ
= 1 ≠ 0
∴ A is a non-singular matrix.
Hence, matrix A is invertible.

Question 5.
In ∆ABC, if a = 18, b = 24 and c = 30 then find the value of sin\(\left(\frac{A}{2}\right)\). (2)
Answer:
Given : a = 18, b = 24 and c = 30

Question 6.
Find k, if the sum of the slopes of the lines represented by x² + kxy – 3y² = 0 is twice their product. (2)
Answer:
On comparing the given equation x² + kxy – 3y² = 0 with ax² + 2hxy + by² = 0
we get, a = 1, 2h = k and b = -3
Let m₁ and m₂ be the stopes of the lines represented by x² + kxy – 3y² = 0
∴ m₁ + m₂ = \(\frac{-2 h}{b}\) = \(\frac{-k}{-3}\) = \(\frac{k}{3}\)
and m₁m₂ = \(\frac{a}{b}\) = \(\frac{1}{-3}\) = \(-\frac{1}{3}\)
According to the question,
m₁ + m₂ = 2[m₁m₂]
∴ \(\frac{k}{3}\) = -2\(\left[\frac{-1}{3}\right]\)
∴ k = -2

Question 7.
If \(\bar{a}, \bar{b}, \bar{c}\) are the position vectors of the points A, B, C respectively and 5\(\bar{a}\) – 3\(\bar{b}\) – 2\(\bar{c}\) = \(\bar{0}\), then find the ratio in which the point C divides the line segment BA. (2)
Answer:
The given vector is 5\(\bar{a}\) – 3\(\bar{b}\) – 2\(\bar{c}\) = 0
∴ 2\(\bar{c}\) = 5\(\bar{a}\) – 3\(\bar{b}\)
∴ \(\bar{c}\) = \(\frac{5 \bar{a}-3 \bar{b}}{2}\)
∴ \(\bar{c}\) = \(\frac{5 \bar{a}-3 \bar{b}}{5-3}\)
∴ The point c divides the line segment BA externally in ratio 5 : 3.

Question 8.
Find the vector equation of the line passing through the point having position vector 4\(\hat{i}\) – \(\hat{j}\) + 2\(\hat{k}\) and parallel to the vector -2\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\). (2)
Answer:
The vector equation of a line passing through a point with position vector \(\bar{a}\) and parallel to \(\bar{b}\) is \(\bar{r}\) = \(\bar{a}\) + λ\(\bar{b}\).
∴ The vector equation of the line passing through the point having position vector 4\(\hat{i}\) – \(\hat{j}\) + 2\(\hat{k}\) and parallel to the vector -2\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) is r
= [4\(\hat{i}\) – \(\hat{j}\) + 2\(\hat{k}\)] + λ[-2\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)]
⇒ \(\bar{r}\) = (4 – 2λ)\(\hat{i}\) + \(\hat{j}\)(-1 – λ)\(\hat{k}\)(2 + λ)

Question 9.
Find \(\frac{d y}{d x}\), if y = (log x)^x. (2)
Answer:
y = (log x)^x
log y = log(log x)^x
∴ log y = xlog(log x)
Now, differentiate w.r.t. x, we get

Question 10.
Evaluate: ∫log x dx (2)
Answer:
I = ∫log x . 1 dx
Let u = log x and v = 1 dx
From integrating by ports

Question 11.
Evaluate: \(\int_0^{\frac{\pi}{2}} \cos ^2 x d x\) (2)
Answer:

Question 12.
Find the area of the region bounded by the curve y = x², and the lines x = 1, x = 2 and y = 0. (2)
Answer:
The given equation of the curve is y = x²

Question 13.
Solve: 1 + \(\frac{d y}{d x}\) = cosec(x + y) ; put x + y = u (2)
Answer:
Given, (x + u) = u ……….. (i)
On differentiate equation (i) w.r.t. x, we get
∴ 1 + \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\)
∴ Given differential equation becomes —
\(\frac{d u}{d x}\) = cosec u

⇒ sin u du = dx
On integrating both sides, we get
∴ ∫sinu . du = ∫dx
⇒ -cos u = x + c
⇒ x + cos u + c = 0
∴ x + cos[x + y] + c = 0. (From equation (i))

Question 14.
If two coins are tossed simultaneously, write the probability distribution of the number of heads. (2)
Answer:
When 2 coins are tossed then the sampLe space is
S = {HH, HT, TH, TT}
Let X be the number of heads in 2 tosses of a coins.
∴ X[HH]= 2, X[HT] = 1, X[TH] = 1, X[TT] = 0
So, X can take the value of 0, 1 and 2.
It is known that,
P[HH] = P[HT] = P[TH] = P[TT] = \(\frac{1}{4}\)
P[X = 0] = P[TT] = \(\frac{1}{4}\)
P[X = 1] = P[TH] + P[HT] = \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{1}{2}\)
P(X = 2) = P[HH] = \(\frac{1}{4}\)
Thus, the required probobility distribution is as follows:

Section – C

Attempt any EIGHT of the following questions: [24]

Question 15.
Express the following switching circuit in the symbolic form of logic. Construct the switching table: (3)

Answer:
Let p : The switch S₁
q : The switch S₂
Therefore, the symbolic form of the circuit is
[p ∨ q] ∧ [~p ∧ ~q].
Switching table:

Therefore, irrespective to the entries of the last column the lamp will not glow.

Question 16.
Prove that: tan^-1\(\left(\frac{1}{2}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\) = \(\frac{\pi}{4}\) (3)
Answer:

Question 17.
In ∆ABC, prove that:
\(\frac{\cos \mathrm{A}}{a}\) + \(\frac{\cos \mathrm{B}}{b}\) + \(\frac{\cos \mathrm{C}}{c}\) = \(\frac{a^2+b^2+c^2}{2 a b c}\) (3)
Answer:

Question 18.
Prove by vector method, the angle subtended on a semicircle is a right angle. (3)
Answer:
Let segment AB be a diameter of o circle with centre C and P be any point on the circle other than A and B.
Then ∠APB is on angle subtended on a semicircle.


∴ ∠APB is a right angle.
Hence, the angle subtended on a semicircle is the right angle.
Hence Proved.

Question 19.
Find the shortest distance between the lines \(\bar{r}\) = (4\(\hat{i}\) – \(\hat{j}\)) + λ(\(\hat{i}\) + 2\(\hat{j}\) – 3\(\hat{k}\)) and \(\bar{r}\) = (\(\hat{i}\) – \(\hat{j}\) – 2\(\hat{k}\)) + μ(\(\hat{i}\) + 4\(\hat{j}\) – 5\(\hat{k}\))
Answer:
Equation of lines are,

Question 20.
Find the angle between the line \(\bar{r}\) = (\(\hat{i}\) + 2\(\hat{j}\) + \(\hat{k}\)) + λ(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) and the plane \(\bar{r}\).(2\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) = 8.
Answer:

Question 21.
If y = sin^-1 x, then show that: (1 – x²)\(\frac{d^2 y}{d x^2}\) – x.\(\frac{d y}{d x}\) = 0.
Answer:

Question 22.
Find the approximate value of tan^-1 (1.002). [Given: π = 3.1416] (3)
Answer:
Let, f(x) = tan^-1x
⇒ f'(x) = \(\frac{1}{1+x^2}\)
∴ x = 1.002 = 1 + 0.002 = [a + h]
Here, a = 1 and h = 0.002

Question 23.
Prove that: \(\int \frac{1}{a^2-x^2} d x\) = \(\frac{1}{2 a} \log \left(\frac{a+x}{a-x}\right)+c\) (3)
Answer:

Question 24.
Solve the differential equation:
x.\(\frac{d y}{d x}\) – y + x.sin\(\left(\frac{y}{x}\right)\) = 0.
Answer:

Question 25.
Find k, if
f(x) = kx² (1 – x), for 0 < x < 1,
= 0 otherwise
is the p.d.f of random variable X. (3)
Answer:
The p.d.f. of a continuous random variable must satisfies the following condition

Question 26.
A die is thrown 6 times, if ‘getting an odd number’.is success, find the probability of 5 successes. (3)
Answer:
The repeated tosses of a die are Bernoulli trails. Let X denote the number of success of getting odd numbers in an expêriment of 6 traits.
Probability of getting an odd number in a single throw of a die is P = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ q = 1 – P = \(\frac{1}{2}\)
X has a binomial distribution.

Section – D

Attempt any FIVE of the following questions: [20]

Question 27.
Solve the following system of equations by the method of reduction : x + y + z = 6, y + 3z = 11, x + z = 2y. (4)
Answer:

⇒ x + y + z = 6 ……… (i)
and y + 3z = 11 ……. (ii)
and -3y = -6 …… (iii)
Put the value of y in equation (ii)
2 + 3z = 11
3z = 9
⇒ z = 3
Put the value of y and z in equation (1)
x + 2 + 3 = 6
x + 5 = 6
x = 1
Therefore, x = 1, y = 2, z = 3.

Question 28.
Prove that the acute angle θ between the lines represented by the equation ax² + 2hxy + by² = 0 is tanθ = \(\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|\). Hence find the condition that the lines are coincident. (4)
Answer:
Case 1 : Let m₁ and m₂ are the slopes of the ones represented by the equation ax² + 2hxy + by² = 0
then m₁ + m₂ = \(\frac{-2 h}{b}\) and m₁m₂ = \(\frac{a}{b}\)
If θ is the acute angle between the lines,
then, tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
Now, (m₁ – m₂)² = (m₁ + m₂)² – 4m₁m₂

Case 2 : If one of the tines is parallel to the y-axis then one of the slope m₁. m₂ does not exist. As the line passes through the origin so one line parallel to the y-axis. Its equation is x = 0 and b = 0
The other line is ax + 2by = 0, where slope is

Question 29.
Find the volume of the parallelopiped whose vertices are A(3, 2, -1), B(-2, 2, -3), C(3, 5, -2) and D(-2, 5, 4). (4)
Answer:

Question 30.
Solve the following LP.P. by graphical method:
Maximize: z = 10x + 25y
Subject to : 0 ≤ x ≤ 3,
0 ≤ y ≤ 3,
x + y ≤ 5.
Also find the maximum value of z. (4)
Answer:
The given inequations are 0 ≤ x ≤ 3; 0 ≤ y ≤ 3; x + y ≤ 5.

The shaded region OABCD is the feasible region with the vertices O.A. B.C and D

The maximum of z = 10x + 25y is 95 at the point C[2, 3].

Question 31.
If x = f(t) and y = g(t) are differentiable functions of t, so that y is function of x and \(\frac{d x}{d t}\) ≠ 0 then prove that \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\). Hence, find \(\frac{d y}{d x}\), if x = at², y = 2at. (4)
Answer:
Given, x and y are differentiable functions of t. Let there be a small increment δt in the value of t. At the value of x and y taking δx, δy respectively.

Question 32.
A box with a square base is to have an open top. The surface area of box is 147 sq.cm. What should be its dimensions in order that the volume is largest? (4)
Answer:
Let x cm be side of square base and h cm be its height
Then x² + 4xh = 147

Hence, the volume of the box is largest, when the side of square bose is 7 cm and its height is 3.5 cm.

Question 33.
Evaluate: \(\int \frac{5 e^x}{\left(e^x+1\right)\left(e^{2 x}+9\right)} d x\) (4)
Answer:

= At² + 9A + Bt² + Bt + Ct + C
1 = (A + B)t² + (B + C)t + (9A + C).
9A + C = 1 ………. (i)
A + B = 0 ………. (ii)
B + C = 0 ………(iii)
⇒ C = -B
Put in equation (ii)
9A – B = 1
A + B = 0
On solving, we get
10A = 1
⇒ A = \(\frac{1}{10}\)
B = \(\frac{-1}{10}\)
and C = \(\frac{1}{10}\)
Put the value A, B and C in equation (i)

Question 34.
Prove that
\(\int_0^{2 a} f(x) d x\) = \(\int_0^a f(x) d x\) + \(\int_0^a f(2 a-x) d x\).
Hence show that:
\(\int_0^\pi \sin x d x\) = 2\(\int_0^{\frac{\pi}{2}} \sin x d x .\)
Answer:
Proof:
Since ‘a’ lies between 0’ and ‘2a’
We have.

Now, put x = 2a – t
∴ dx = -dt
When x = a
then t = a
When x = 2a
then t = 0

To show that:
we use the proven property by setting f(x) = sin x and
2a = π, means a = \(\frac{\pi}{2}\)
⇒ \(\int_0^\pi \sin x d x\) = \(\int_0^{\pi / 2} \sin x d x\) + \(\int_0^{\pi / 2} \sin (\pi-x) d x\)
Knowing the Trignometric identity sin(π – x) = sin x, the equation simplifies to:
\(\int_0^\pi \sin x d x\) = 2\(2 \int_0^{\pi / 2} \sin x d x\).
Hence Proved.

Maharashtra Board Class 12 Maths Previous Year Question Papers