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.
- 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. - Section B: Q.3 to Q. 14 contain Twelve short answer type questions, each carrying Two marks. (Attempt any Eight)
- Section C: Q.15 to Q.26 contain Twelve short answer type questions, each carrying Three marks. (Attempt any Eight)
- Section D: Q. 27 to Q.34 contain Eight long answer type questions, each carrying Four marks. (Attempt any Five)
- Use of log table is allowed. Use of calculator is not allowed.
- Figures to the right indicate full marks.
- Use of graph paper is not necessary. Only rough sketch of graph is expected.
- For each multiple choice type of question, Only the first attempt will be considered for evaluation.
- 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 = 22 \int_0^{\pi / 2} \sin x d x.
Hence Proved.