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,

cos^{2} ∝ + cos^{2} β + cos^{2} γ = 1

[Properties of direction angle]

⇒ cos^{2} (60°) + cos^{2} (45°) + cos^{2} γ = 1

⇒ \(\left(\frac{1}{2}\right)^2\) + \(\left(\frac{1}{\sqrt{2}}\right)^2\) + cos^{2} γ = 1

⇒ \(\frac{1}{4}+\frac{1}{2}\) + cos^{2} γ = 1

⇒ cos^{2} γ = 1 – \(\frac{3}{4}\)

⇒ cos^{2} 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) = x^{3} . sin^{4}x

∴ f(- x) = [-x]^{3} [sin(-x)]^{4}

∴ = -x^{3}(-sin x)^{4}

∴ = -f(x).

∴ f(x) is on odd function.

⇒ \(\int_{-\pi / 4}^{\pi / 4}\)x^{3}.sin^{4} 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 = x^{2} log x is ___. (2)

(a) x

(b) \(\frac{1}{x}\)

(c) x^{2}

(d) \(\frac{1}{x^2}\)

Answer:

(c) x^{2}

\(\frac{d y}{d x}\) + 2y = x^{2}log 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. = \(e^{\int \mathrm{Pd} d x}\) = \(e^{\int \frac{2}{x} d x}\)

= \(e^{2 \log x}\)

= x^{2}

(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|\)

= cos^{2}θ – (-sin^{2}θ)

= cos^{2}θ + sin^{2}θ

= 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 x^{2} + kxy – 3y^{2} = 0 is twice their product. (2)

Answer:

On comparing the given equation x^{2} + kxy – 3y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0

we get, a = 1, 2h = k and b = -3

Let m_{1} and m_{2} be the stopes of the lines represented by x^{2} + kxy – 3y^{2} = 0

∴ m_{1} + m_{2} = \(\frac{-2 h}{b}\) = \(\frac{-k}{-3}\) = \(\frac{k}{3}\)

and m_{1}m_{2} = \(\frac{a}{b}\) = \(\frac{1}{-3}\) = \(-\frac{1}{3}\)

According to the question,

m_{1} + m_{2} = 2[m_{1}m_{2}]

∴ \(\frac{k}{3}\) = -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 = x^{2}, and the lines x = 1, x = 2 and y = 0. (2)

Answer:

The given equation of the curve is y = x^{2}

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 S_{1}

q : The switch S_{2}

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 – x^{2})\(\frac{d^2 y}{d x^2}\) – 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^{-1}x

⇒ 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) = kx^{2} (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 ax^{2} + 2hxy + by^{2} = 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 m_{1} and m_{2} are the slopes of the ones represented by the equation ax^{2} + 2hxy + by^{2} = 0

then m_{1} + m_{2} = \(\frac{-2 h}{b}\) and m_{1}m_{2} = \(\frac{a}{b}\)

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, (m_{1} – m_{2})^{2} = (m_{1} + m_{2})^{2} – 4m_{1}m_{2}

Case 2 : If one of the tines is parallel to the y-axis then one of the slope m_{1}. m_{2} 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 = at^{2}, 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 x^{2} + 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:

= At^{2} + 9A + Bt^{2} + Bt + Ct + C

1 = (A + B)t^{2} + (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.