Maharashtra Board Class 12 Maths Sample Paper Set 6 with Solutions

Maharashtra State Board Class 12th Maths Sample Paper Set 6 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Maths Model Paper Set 6 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 most appropriate answer from the given alternatives for each question: [16 Marks]

i. If(p ∧ ~ q) → (~ p ∨ r) is a false statement, then the respective truth values of p, q and r are
(A) T, F, F
(B) F, T, T
(C) T, T, T
(D) F, F, F
Answer:
(A) T, F, F

Explanation:
Since, (p ∧ ~q) → (~ p ∨ r) ≡ F
∴ p ∧ ~ q ≡ T and ~ p ∨ r ≡ F
∴ p ≡ T, ~ q ≡ T and ~ p ≡ F, r ≡ F
∴ p ≡ T, q ≡ F, r ≡ F

ii. The principal solutions of cot x = \(\sqrt{3}\) are (2)
(A) \(\frac{\pi}{3}, \frac{7 \pi}{6}\)
(B) \(\frac{\pi}{6}, \frac{7 \pi}{6}\)
(C) \(\frac{\pi}{3}, \frac{4 \pi}{3}\)
(D) \(\frac{3 \pi}{4}, \frac{7 \pi}{4}\)
Answer:
(B) \(\frac{\pi}{6}\), \(\frac{7 \pi}{6}\)

Explanation:
cot x = \(\sqrt{3}\)
∴ cot x = cot \(\left(\frac{\pi}{6}\right)\) and cot x = cot (π + \(\frac{\pi}{6}\)) = cot \(\left(\frac{7 \pi}{6}\right)\)
such that 0 < \(\frac{\pi}{6}\) < 2π and 0 < \(\frac{7 \pi}{6}\) < 2π
∴ the required principal solutions are x = \(\frac{\pi}{6}\) and x = \(\frac{7 \pi}{6}\)

iii. If the sum of slopes of the pair of lines represented by 4x² + 2bxy – 7y² = 0 is equal to the product of the slopes, then the value of h is
(A) -6
(B) -2
(C) -4
(D) 4
Answer:
(B) -2

Explanation:
Given equation of pair of lines is
4x² + 2hxy – 7y² = 0
∴ A = 4, H = h, B = -7
Now, m₁ + m₂ = \(-\frac{2 H}{B}\) = \(\frac{2 h}{7}\) and m₁m₂ = \(\frac{A}{B}\) = \(\frac{4}{-7}\)
Given that, m₁ + m₂ = m₁m₂
∴ \(\frac{2 h}{7}\) = \(\frac{4}{-7}\), ∴ h = -2

iv. If θ be the angle between any two vectors \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\), then \(|\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}|\) = \(|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|\), when θ is equal to
(A) 0
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{2}\)
(D) π
Answer:
(B) \(\frac{\pi}{4}\)

Explanation:
\(|\bar{a} \cdot \bar{b}|\) = \(|\bar{a} \times \bar{b}|\)
∴ \(|\overline{\mathrm{a}}||\overline{\mathrm{b}}|\) cos θ = \(|\bar{a}||\bar{b}|\) sin θ
∴ tan θ = 1
∴ θ = \(\frac{\pi^c}{4}\)

v. Let f (x) and g (x) be differentiable for 0 < x < 1 such that f (0) = 0, g (0) = 0, f (1) = 6. Let there exist a real number c in (0, 1) such that f ‘(c) = 2g’ (c), then the value of g (1) must be (A) 1
(B) 3
(C) 2.5
(D) -1
Answer:
(B) 3

Explanation:
By LMVT, ∃ c ∈ (0, 1) such that
f'(c) = \(\frac{f(1)-f(0)}{1-0}\) and g'(c) = \(\frac{g(1)-g(0)}{1-0}\)
∴ f'(c) = 6 – 0 = 6 and g'(c) = g(1) – 0 = g(1)
Given that f'(c) = 2g'(c)
∴ 6 = 2g(1)
∴ g(1) = 3

vi. \(\int \frac{\cos 2 x-1}{\cos 2 x+1} d x\) =
(A) tan x – x + c
(B) x + tan x + c
(C) x – tan x + c
(D) – x – cot x + c (2)
Answer:
(C) x – tan x + c

Explanation:

vii. The area bounded by the curve y = x³, the X-axis and the lines x = – 2 and x = 1 is
(A) – 9 sq. units
(B) \(-\frac{15}{4}\)
(C) \(\frac{15}{4}\) sq. units
(D) \(\frac{17}{4}\) sq. units
Answer:
(D) \(\frac{17}{4}\) sq. unit

Explanation:

viii. The probability of a shooter hitting a target is \(\frac{3}{4}\)
How many minimum number of times must he fire so that the probability of hitting the target at least once is more than 0⋅99?
(A) 2
(B) 3
(C) 4
(D) 5 (2)
Answer:
(C) 4

Explanation:
Given, p = \(\frac{3}{4}\) and P(X ≥ 1) > 0.99
∴ q = 1 – P = \(\frac{1}{4}\)
∴ P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0)
∴ 0.99 < 1 – P(X = 0)
∴ P(X = 0) < 1 – 0.99
∴ \({ }^n C_0\left(\frac{3}{4}\right)^0\left(\frac{1}{4}\right)\) < 0.01
∴ \(\left(\frac{1}{4}\right)^n\) < \(\frac{1}{100}\)
∴ 100 < (4)ⁿ
∴ Minimum value of n which satisfies the above inequality is n = 4.

Question 2.
Answer the following: [4 Marks]

i. If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}, determine the truth value of the following statement:
∃ x ∈ A such that x + 8 = 15. (1)
Answer:
For x = 7, x + 8 = 7 + 8 = 15
∴ x = 7 satisfies the equation x + 8 = 15.
∴ The qiven statement is true.
∴ Its truth value is T.

ii. Find the direction cosines of any normal to the XY plane. (1)
Answer:
Direction cosines of the normal to the XY plane are 0, 0, 1.

iii. If f(2) = 27 , f’ (2) = 40 and h = 0.01, then find the approximate value of f(2.01). (1)
Answer:
f(2.01) ≈ f (2) + hf'(2)
≈ 27 + (0.01) (40)
≈ 27.4

iv. If the p.d.f. of a continuous r.v.X is given by
f(x) = kx, 0 < x < 4
= 0 otherwise
then what is the value of k? (1)
Answer:
Since the function f(x) represents p.d.f of r.v.x, we get
\(\int_0^4 k x d x\) = 1
∴ \(\frac{k}{2}[16-0]\) = 1
∴ k = \(\frac{1}{8}\)

Section – B

Attempt any EIGHT of the following questions:

Question 3.
Using truth table, verify that ~(p ∨ q) ≡ ~p ∧ ~q (2)
Answer:

In the above truth tab’e, entries in columns 4 and 7 are identical.
∴ ~(p ∨ q) ≡ ~p ∧ ~q

Question 4.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) and X is a 2 × 2 matrix such that AX = I, then find X. (2)
Answer:

Question 5.
Find the cartesian co-ordinates of the point whose polar co-ordinates are \(\left(\frac{3}{4}, \frac{3 \pi}{4}\right)\) (2)
Answer:

Question 6.
If \(|\overline{\mathrm{a}}|\) = \(|\overline{\mathrm{b}}|\) = 1, \(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}\) = 0 and \(|\overline{\mathrm{a}}|\) + \(|\overline{\mathrm{b}}|\) + \(|\overline{\mathrm{c}}|\) = 0, then find |\(|\overline{\mathrm{c}}|\)| (2)
Answer:

Question 7.
If \(|\overline{\mathrm{c}}|\) = 3\(|\overline{\mathrm{a}}|\) – 2\(|\overline{\mathrm{c}}|\) = 0, then prove that \(\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]\) = 0 (2)
Answer:

Question 8.
Find the acute angle between lines x = -y, z = 0 and x = 0, z = 0. (2)
Answer:
Given equations of lines are x = -y, z = 0 i.e., \(\frac{x}{1}\) = \(\frac{y}{-1}\) = \(\frac{z}{0}\) and x = 0, z = 0 which represents Y-axis.
Direction ratios of above lines are
a₁ = 1, b₁ = -1, c₁ = 0 and a₂ = 0, b₂ = 1, c₂ = 0 (Direction ratio of Y-axis)
Angle between two lines is

Question 9.
Differentiate the following w. r. t. x. : cos^-1 (4 cos³ x – 3 cos x) (2)
Answer:
Let y = cos^-1 (4 cos³ x – 3 cos x)
= cos^-1 (cos 3x)
∴ y = 3x
Differentiating w. r. t. x., we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = \(\frac{d}{d x}\)(3x)
∴ \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = 3

Question 10.
Evaluate: \(\int \sqrt{1-\cos 2 x} d x\) (2)
Answer:

Question 11.
Evaluate : \(\int \frac{1}{1+x-x^2} \mathrm{~d} x\)
Answer:

Question 12.
Evaluate :
\(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2+\sin x}{2-\sin x}\right) \mathrm{d} x\) (2)
Answer:

Question 13.
Find the area of the region bounded by the following curve, X-axis and the given lines:
y² = 16x, x = 0, x = 4
Answer:
Given equation of the curve is
y² = 16x
∴ y = ±4\(\sqrt{x}\)
∴ y = 4\(\sqrt{x}\) ……[∵ In first quadrant, y > 0]

Question 14.
Solve the following differential equation.
log \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)\) = 2x + 3y
Answer:

Section – C

Attempt any EIGHT of the following questions: [24 Marks]

Question 15.
Examine whether the following statement pattern is a tautology or a contradiction or a contingency:
(~ p → q) ∧ (p ∧ r) (3)
Answer:

Truth values in the last column are not identical. Hence, ¡t is contingency.

Question 16.
Find the general solution of tan³ θ = 3tanθ.
Answer:
tan³ = 3 tan θ
∴ tan³θ – 3 tan θ = 0
∴ tan θ(tan²θ – 3) = 0
∴ tan θ = 0 or tan²θ = 3 = \((\sqrt{3})^2\)
∴ tan θ = o or tan² θ = (tan\(\frac{\pi}{3}\))²
Since tan θ = 0 implies θ = nπ and tan²θ = tan²α implies θ = nt ± α, n ∈ Z.
∴ θ = nπ or θ = nπ ± \(\frac{\pi}{3}\)
∴ The required general solution is θ = nπ or nπ ± \(\frac{\pi}{3}\), where n ∈ Z.

Question 17.
With usual notations prove that 2 {asin²\(\frac{C}{2}\) + csin²\(\frac{A}{2}\)} = a – b + c. (3)
Answer:
Consider L.H.S. = 2{asin²\(\frac{C}{2}\) + csin²\(\frac{A}{2}\)}.
= 2{a\(\left(\frac{1-\cos C}{2}\right)\) + c\(\left(\frac{1-\cos A}{2}\right)\)}
= a – a cos C + c – c cos A
= (a + c) – (a cos C + c cos A)
= a + c – b ……… [By projection rule]
= a – b + c
= R.H.S.
∴ 2{asin²\(\frac{C}{2}\) + csin²\(\frac{A}{2}\)} = a – b + c

Question 18.
If Q is the foot of the perpendicular from P(2, 4, 3) on the line joining the points A(1, 2, 4) and B(3; 4, 5), find coordinates of Q. (3)
Answer:
Let PQ be the perpendicular drawn from point P(2, 4, 3) to the line joining the points A(1, 2, 4), and B(3, 4, 5).
Let Q divides AB internally in the ratio λ : 1.
∴ Q ≡ \(\left(\frac{3 \lambda+1}{\lambda+1}, \frac{4 \lambda+2}{\lambda+1}, \frac{5 \lambda+4}{\lambda+1}\right)\) ………. (i)
Direction ratios of PQ are

Question 19.
Prove that: \(\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}}+\overline{\mathrm{c}} & \overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}
\end{array}\right]\) = 0 (3)
Answer:

Question 20.
Find the Cartesian equation of the plane \(\overline{\mathrm{r}}\) = λ(\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)) + μ(\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) + 3\(\hat{\mathbf{k}}\)) (3)
Answer:
The equation \(\overline{\mathrm{r}}\) = \(\overline{\mathrm{a}}\) + \(\lambda \bar{b}\) + \(\mu \bar{c}\) represents a plane passing through a point having position vector \(\bar{a}\) and parallel to vectors \(\bar{b}\) and \(\bar{c}\).
Here, \(\bar{a}\) = 0\(\hat{i}\) + 0\(\hat{j}\) + 0\(\hat{k}\), \(\bar{b}\) = \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\), \(\bar{c}\) = \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\)
The given plane is perpendicular to the vector \(\bar{n}\) = \(\overline{\mathrm{b}} \times \overline{\mathrm{c}}\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & -1 \\
1 & 2 & 3
\end{array}\right|\)
= \(\hat{i}\) (3 – (-2)) – \(\hat{j}\) (3 – (-1)) + \(\hat{k}\) (2 – 1)
= 5\(\hat{i}\) – 4\(\hat{j}\) + \(\hat{k}\)
Vector equation of the plane in scalar product form is \(\bar{r} \cdot \bar{n}\) = \(\bar{a} \cdot \bar{n}\)

Question 21.
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then prove that the composite function y = f[g(x)] is a differentiable function of x and \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = \(\frac{\mathrm{d} y}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{du} x}\). (3)
Answer:
Let δx be a small increment in the value of x.
Since u is a function of x, there should be a corresponding increment 8u in the value of u.
Also y is a function of u.
∴ There should be a corresponding increment δy in the value of y.

Here, R.H.S. of (ii) exists and is finite.
Hence, L.H.S. of (ii) should also exists arid be finite.
∴ \(\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}\) = \(\frac{d y}{d x}\) exists and is finite.
\(\frac{d y}{d x}\) = \(\frac{d y}{d u} \times \frac{d u}{d x}\)

Question 22.
Find the values of x for which f(x) = 2x³ – 15x² – 144x – 7 is strictly decreasing (3)
Answer:
f(x) = 2x³ – 15x² – 144x – 7
∴ f’(x) = 6x² – 30x – 144
= 6(x² – 5x – 24)
= 6(x + 3)(x – 8)
f(x) is strictly decreasing, if f ‘(x) < 0
∴ 6(x + 3)(x – 8) < 0
∴ (x + 3)(x – 8) < 0
ab < 0 ⇔ a > 0 and b < 0 or a < 0 and b > 0
∴ Either (x + 3) > 0 and (x – 8 ) < 0 or
(x + 3) < 0 and (x – 8) > 0

Case 1: x + 3 < 0 and x – 8 > 0
∴ x < -3 and x > 8,
which is not possible.

Case 2: x + 3 > 0 and x – 8 < 0 ∴ x > -3 and x < 8
Thus, f(x) is strictly decreasing function for -3 < x < 8 ,i.e., x ∈(-3, 8).

Question 23.
If each side of an equilateral triangle increases at the rate of \(\sqrt{2}\) cm/ sec, find the rate of increase of its area when its side is of length 3 cm. (3)
Answer:
Let a be the length of each side of an equilateral triangle and A be its area.

Question 24.
Obtain the differential equation by eliminating arbitrary constants A and B. (3)
y = A cos (log x) + B sin (log x)
Answer:
y = A cos (log x) + Bsin (log x) ………… (i)
Here, A and B are arbitrary constants.
Differentiating (i) wr.t. x, we get

Question 25.
Reduce each of the following differential to the variable separable form and hence solve.
cos² (x – 2y) = 1 – 2\(\frac{\mathrm{d} y}{\mathrm{~d} x}\) (3)
Answer:
cos²(x – 2y) = 1 – 2\(\frac{d y}{d x}\) ……….(i)
Put x – 2y = u ………(ii)
Differentiating w.r.t. x, we get
1 – 2\(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) …….(iii)
Substituting (ii) and (iii) in (i), we get
cos²u = \(\frac{d u}{d x}\)
∴ dx = \(\frac{d u}{\cos ^2 u}\)
∴ dx = sec²u du
Integrating on both sides, we get
∫dx = ∫sec²u du
∴ x = tan u + c
∴ x = tan (x – 2y) + c

Question 26.
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is \(\frac{1}{100}\). Find the probability that he will win a prize at least once. (3)
Answer:
Let X denote the number of times a person wins a prize.

Section – D

Attempt any FIVE of the following questions:

Question 27.
Find the inverse of \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) by adjoint method.
Answer:

Question 28.
If θ is the measure of acute angle between the pair of lines given by ax² + 2hxy + by² = 0, then prove that tan θ = \(\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|\), a + b ≠ 0.
Hence, find the acute angle between the lines x² + xy = 0.
Answer:
Let m₁ and m₂ be the slopes of the lines represented by the equation ax² + 2hxy + by² = 0

Comparing equation x² + xy = 0 with ax² + 2hxy + by² = 0, we get
a = 1, h = \(\frac{1}{2}\) and b = 0.
Let θ be the acute angle between them.
∴ tan θ = \(\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|\) = \(\left|\frac{2 \sqrt{\frac{1}{4}-0}}{1+0}\right|\) = 1
∴ θ = \(\frac{\pi}{4}\)

Question 29.
Find the co-ordinates of the foot of the perpendicular drawn from the point 2\(\hat{\mathrm{i}}\) – \(\hat{\mathrm{j}}\) + 5\(\hat{\mathrm{k}}\) to the line \(\overline{\mathrm{r}}\) = (11\(\hat{\mathrm{i}}\) – 2\(\hat{\mathrm{j}}\) – 8\(\hat{\mathrm{k}}\)) + λ(10\(\hat{\mathrm{i}}\) – 4\(\hat{\mathrm{j}}\) – 11\(\hat{\mathrm{k}}\)). Also find the length of the perpendicular.
Answer:

Question 30.
Solve the following L.P.P.
Maximize z = 5x₁ + 6x₂ subject to 2x₁ + 3x₂ ≤ 18, 2x₁ + x₂ ≤ 12, x₁ ≥ 0, x₂ ≥ 0
Answer:
To find the graphical solution, construct the table as follows:


The shaded portion OBEC is the feasible region.
Whose vertices are O (0, 0), B (0, 6), E, C (6, 0)
E is the point of intersection of the lines
2x₁ + x₂ = 12 …….. (i)
and 2x₁ + 3x₂ = 18 …….(ii)
∴ By (i) – (ii), we get

Here, the objective function is Z = 5x₁ + 6x₂
Now, we will find maximum value of Z as follows:

∴ Z has maximum value 40.5 at E(4.5, 3)
∴ Z is maximum, when x₁ = 4.5, x₂ = 3

Question 31.
If x = a sin t – b cos t, y = a cos t + b sin t, show that \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) = \(-\frac{x^2+y^2}{y^3}\) (4)
Answer:
x = asint – b cost, y = acos t + b sint
Squaring and adding, we get
x² + y² = (a sin t – b cos t)² + (a cos t + b sin t)²
= a² (sin²t + cos²t) + b² (cos²t + sin²t)
= a²(1) + b²(1)
∴ x² + y² = a² + b²
Differentiating w.r.t. x, we get

Question 32.
Prove that:

Answer:
Put x = a tanθ
∴ tanθ = \(\frac{x}{a}\)
Differentiating (i) w.r.t. θ, we get
dx = a sec² θ dθ

Question 33.
Prove that: \(\int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x\) = \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-x) \mathrm{d} x\).
Hence, evaluate \(\int_0^{\frac{\pi}{2}} \log (\tan x) d x\).
Answer:

Question 34.
The p. d. f. of r. v. X is given by
f(x) = \(\frac{\mathrm{k}}{\sqrt{x}}\), for 0 < x < 4 and
= 0, otherwise.
Determine k. Determine c. d. f. of X and hence find P (X ≤ 2) and P (X ≤ 1).
Answer:

Maharashtra Board Class 12 Maths Previous Year Question Papers