Maharashtra Board Class 12 Maths Sample Paper Set 3 with Solutions

Maharashtra State Board Class 12th Maths Sample Paper Set 3 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Maths Model Paper Set 3 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. The negation of inverse of ~p → q is _____.
(A) p ∧ q
(B) ~p ∧ ~q
(C) ~p ∧ q
(D) ~q → ~p (2)
Answer:
(A) p ∧ q

Explanation:
Inverse of ~p → q ≡ p → ~q.
∴ Negation of inverse of ~p → q
≡ ~(p → q)
≡ p ∧ q …… [~(p→q) ≡ p ∧ ~q]

ii. In ∆ABC, if c² + a² – b² = ac, then ∠B = ______
(A) \(\frac{\pi}{4}\)
(B) \(\frac{\pi}{3}\)
(C) \(\frac{\pi}{2}\)
(D) \(\frac{\pi}{6}\)
Answer:
(B) \(\frac{\pi}{3}\)

Explanation:
c² + a² – b² = ac ……(i)[Given]
By cosine rule, we have b² = c² + a² – 2ca cos B
∴ cos B = \(\frac{c^2+a^2-b^2}{2 a c}\) = \(\frac{a c}{2 a c}\) ……[From (i)]
∴ cos B = \(\frac{1}{2}\)
∴ ∠B = \(\frac{\pi}{3}\)

iii. The value of \(\hat{\mathrm{i}}\) ⋅ (\(\hat{\mathrm{j}}\) × \(\hat{\mathrm{k}}\) + \(\hat{\mathrm{j}}\) ⋅ (\(\hat{\mathrm{i}}\) × \(\hat{\mathrm{k}}\)) + k ⋅ (\(\hat{\mathrm{i}}\) × \(\hat{\mathrm{j}}\)) is
(A) 0
(B) -i
(C) 1
(D) 3 (2)
Answer:
(C) 1

Explanation:
\(\hat{\mathrm{i}}\) × (\(\hat{\mathrm{j}}\) × \(\hat{\mathrm{k}}\)) + \(\hat{\mathrm{k}}\) × (\(\hat{\mathrm{i}}\) × \(\hat{\mathrm{j}}\))
= \(\hat{\mathrm{i}}\) × (\(\hat{\mathrm{i}}\)) + \(\hat{\mathrm{j}}\) × (-\(\hat{\mathrm{j}}\)) + \(\hat{\mathrm{k}}\) × (\(\hat{\mathrm{k}}\))
= 1 – 1 + 1
= 1

iv. If the line \(\frac{x+1}{2}\) = \(\frac{y-\mathrm{m}}{3}\) = \(\frac{z-4}{6}\) lies in the plane 3x – 14y + 6z + 49 = 0, then the value of m is
(A) 5
(B) 3
(C) 2
(D) -5 (2)
Answer:
(A) 5
Explanation:
If line \(\frac{x+1}{2}\) = \(\frac{y-m}{3}\) = \(\frac{z-4}{6}\) lies in the plane, then (-1, m, 4) should satisfy the equation of a plane 3x – 14y + 6z + 49 = 0
∴ 3(-1) – 14m + 6(4) + 49 = 0
∴ -3 – 14m + 24 + 49 = 0
∴ 14m = 70 ⇒ m = 5

v. If y = sec (tan^-1 x) then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at x = 1, is equal to
(A) \(\frac{1}{2}\)
(B) 1
(C) \(\frac{1}{\sqrt{2}}\)
(D) \(\sqrt{2}\) (2)
Answer:
(C) \(\frac{1}{\sqrt{2}}\)

Explanation:
y = sec(tan^-1 x) = sec(sec^-1\(\sqrt{x^2+1}\)) = \(\sqrt{x^2+1}\)
⇒ \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = \(\frac{1}{2 \sqrt{x^2+1}}\)(2x) = \(\frac{x}{\sqrt{x^2+1}}\)
\(\left.\frac{d y}{d x}\right|_{x=1}\) = \(\frac{1}{\sqrt{2}}\)

vi. \(\int \frac{1}{x+x^5}\) dx = f(x) + c, then \(\int \frac{x^4}{x+x^5}\) dx =
(A) log | x | – f(x) + c
(B) f(x) + log | x | + c
(C) f(x) – log|x| + c
(D) \(\frac{1}{5}\)x⁵f(x) + c (2)
Answer:
(A) log |x| – f(x) + c

Explanation:

vii. The solution of (x + y)²\(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = 1 is
(A) x = tan^-1 (x + y) + c
(B) ytan^-1 \(\left(\frac{x}{y}\right)\) = c
(C) y = tan^-1 (x + y) + c
(D) y + tan^-1 (x + y) = c (2)
Answer:
(C) y = tan^-1 (x + y) + c

Explanation:
\(\frac{d y}{d x}\) = \(\frac{1}{(x+y)^2}\) …….. (i)
Put x + y = u ……..(ii)
∴ 1 + \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\)
∴ \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) – 1 ……… (iii)
Substituting (ii) and (iii) in (i), we get
\(\frac{\mathrm{du}}{\mathrm{~d} x}\) – 1 = \(\frac{1}{u^2}\)
∴ \(\int \frac{u^2}{1+u^2}\)du = ∫dx
∴ \(\int\left(1-\frac{1}{1+u^2}\right)\) du = ∫dx
∴ u – tan^-1u = x + c
∴ x + y – tan^-1 (x + y) = x + c
∴ y = tan^-1 (x + y) + c

viii. A die is thrown 100 times. If getting an even number is considered a success, then the standard deviation of the number of successes is
(A) \(\sqrt{50}\)
(B) 5
(C) 25
(D) 10 (2)
Answer:
(B) 5

Explanation:
Here, n = 100
P(getting an even number) = P = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ Var(X) = npq = 100 × \(\frac{1}{2}\) × \(\frac{1}{2}\) = 25

Question 2.
Answer the following: [4 Marks]

i. Find the distance of the point (4, 3, 5) from the Y-axis. .. (1)
Answer:
The distance from (4, 3, 5) to the Y-axis is
\(\sqrt{x^2+z^2}\) = \(\sqrt{4^2+5^2}\) = \(\sqrt{16+25}\) = \(\sqrt{41}\) units

ii. Find the value of p if the equation
px² – 8xy + 3y + 14x + 2y – 8 = 0 represents a pair of perpendicular lines. (1)
Answer:
Here, a = p and b = 3
The given equation represents a pair of lines perpendicular to each other.
∴ a + b = 0
∴ p + 3 = 0
∴ p = -3

iii. If s = 60 + 2t – 10t² is the displacement of the particle at time t, then find the rate of change in displacement w.r.t. t. (1)
Answer:
s = 60 + 2t – 10t²
∴ Rate of change of displacement w.r.t. t is
\(\frac{d s}{d t}\) = 2 – 20t

iv. If r.v. X denote the number of prime numbers appear on the upper most face of a fair die when it rolled twice. Find the possible values of X. (1)
Answer:
x denote the number of prime numbers.
Since die is rolled twice
∴ Possible values of X are 0, 1, 2.

Section – B [16 Marks]

Attempt any EIGHT of the following questions:

Question 3.
Find the adjoint of the matrix. (2)
Answer:
Here,
a₁₁ = 1
∴ M₁₁ = -1 and A₁₁ = (-1)^{1 + 1} (-1) = -1
a₁₂ = 3
∴ M₁₂ = 4 and A₁₂ = (-1)^{1 + 2} (4) = -4
a₂₁ = 4
∴ M₂₁ = 3 and A₂₁ = (-1)^{2 + 1} (3) = -3
∴ a₂₂ = -1
∴ M₂₂ = 1 and A₂₂ = (-1)^{2 + 2}(1) = 1
∴ The matrix of the co-factors is
\(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\) = \(\left[\begin{array}{cc}
-1 & -4 \\
-3 & 1
\end{array}\right]\)

Question 4.
Find the general solution of cosec θ = –\(\sqrt{2}\). (2)
Answer:
cosec θ = –\(\sqrt{2}\)
sin θ = –\(\frac{1}{\sqrt{2}}\)
sin θ = – sin \(\left(\frac{\pi}{4}\right)\) = sin (π + \(\frac{\pi}{4}\)) = sin \(\left(\frac{5 \pi}{4}\right)\)
Since sin θ = sin α implies θ = nπ + (-1)ⁿ ∝, n ∈ Z.
∴ The required qeneral solution is θ = nπ + (-1)ⁿ \(\left(\frac{5 \pi}{4}\right)\), where n ∈ Z.

Question 5.
Find k, if the sum of the slopes of the lines represented by x² + kxy – 3y² = 0 is twice their product. (2)
Answer:
Given equation of the lines is x² + kxy – 3y² = 0.
Comparing with ax² + 2hxy + by² = 0, we get
a = 1, 2h = k, b = -3.
Let m₁ and m₂ be the slopes of the lines represented by x² + kxy – 3y² = 0.
∴ m₁ + m₂ = –\(\frac{2 h}{b}\) = \(\frac{\mathrm{k}}{3}\) and m₁m₂ = \(\frac{a}{b}\) = –\(\frac{1}{3}\)
According to the given condition,
m₁ + m₂ = 2(m₁m₂)
∴ \(\frac{k}{3}\) = 2(-\(\frac{1}{3}\))
∴ k = -2

Question 6.
Find \(\overline{\mathrm{a}}\) • (\(\overline{\mathrm{b}}\) × \(\overline{\mathrm{c}}\)),if \(\overline{\mathrm{a}}\) = 3\(\hat{\mathrm{i}}\) – \(\hat{\mathrm{j}}\) + 4\(\hat{\mathrm{k}}\), \(\overline{\mathrm{b}}\) = 2\(\hat{\mathrm{i}}\) + 3\(\hat{\mathrm{j}}\) – \(\hat{\mathrm{k}}\) and \(\overline{\mathrm{c}}\) = -5\(\hat{\mathrm{i}}\) + 2\(\hat{\mathrm{j}}\) + 3\(\hat{\mathrm{k}}\). (2)
Answer:
\(\bar{a}\) = 3\(\hat{\mathbf{i}}\) – \(\hat{j}\) + 4\(\hat{k}\), \(\bar{b}\) = 2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\), \(\bar{c}\) = -5\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) + 3\(\hat{\mathbf{k}}\)
\(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) = \(\left|\begin{array}{ccc}
3 & -1 & 4 \\
2 & 3 & -1 \\
-5 & 2 & 3
\end{array}\right|\)
= 3(9 + 2) + 1(6 – 5) + 4(4 + 15)
= 3(11) + 1 (1) + 4(19)
= 33 + 1 + 76
∴ \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) = 110

Question 7.
Find the vector equation of the line passing through points having position vectors 3\(\hat{\mathrm{i}}\) + 4\(\hat{\mathrm{j}}\) – 7\(\hat{\mathrm{k}}\) and 6\(\hat{\mathrm{i}}\) – \(\hat{\mathrm{j}}\) + \(\hat{\mathrm{k}}\). (2)
Answer:
Let \(\bar{a}\) = 3\(\hat{i}\) + 4\(\hat{j}\) – 7\(\hat{k}\) , \(\bar{b}\) = 6\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) be the position vectors of the points by which the line is passing through.
The vector equation of a line passing through the points having postion vectors \(\bar{a}\) and \(\bar{b}\) is given by \(\overline{\mathrm{r}}\) = \(\bar{a}\) + λ(\(\bar{b}\) – \(\bar{a}\))
∴ \(\bar{r}\) = (3\(\hat{i}\) + 4\(\hat{j}\) – 7\(\hat{k}\)) + λ[(6\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) – (3\(\hat{i}\) + 4\(\hat{j}\) – 7\(\hat{k}\))
∴ \(\bar{r}\) = (3\(\hat{i}\) + 4\(\hat{j}\) – 7\(\hat{k}\)) + λ(3\(\hat{i}\) – 5\(\hat{j}\) + 8\(\hat{k}\))

Question 8.
Find the cartesian equation of the plane passing through A(7, 8, 6) and parallel to the plane \(\overline{\mathrm{r}}\) – (6\(\hat{\mathrm{i}}\) + 8\(\hat{\mathrm{j}}\) + 7\(\hat{\mathrm{k}}\)) = 0. (2)
Answer:
The plane passes through the point A(7, 8, 6).
∴ x₁ = 7, y₁ = 8, z₁ = 6.
Since plane is parallel to the plane
\(\overline{\mathrm{r}}\) × (6\(\hat{i}\) + 8\(\hat{j}\) + 7\(\hat{k}\)) = 0, direction ratios of normal vector will be a = 6, b = 8, c = 7.
Equation of a plane in cartesian form is
a(x x₁) + b(y y₁) + c(z z₁) = 0
∴ 6(x – 7) + 8(y – 8) + 7(z – 6) = 0
∴ 6x – 42 + 8y – 64 + 7z – 42 = 0
∴ 6x + 8y + 7z = 42 + 42 + 64
∴ 6x + 8y + 7z = 148

Question 9.
Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\), if xe^y + ye^x = 1. (2)
Answer:
xe^y + ye^x = 1
Differentiating w.r.t. x, we get

Question 10.
Test whether the following function is increasing or decreasing.
f(x) = x³ – 6x² + 12x – 16, x ∈ R (2)
Answer:
f (x) = x³ – 6x² + 12x – 16, x ∈ R
∴ f'(x) = 3x² – 12x + 12
= 3(x² – 4x + 4) = 3(x – 2)²
3(x – 2)² is always positive for x ≠ 2
∴ f ‘(x) ≥ 0 for all x ∈ R.
Hence, f(x) is an increasing function for all x ∈ R.

Question 11.
Evaluate: \(\int \sqrt{1+\sin 2 x} \mathrm{~d} x\) (2)
Answer:

Question 12.
Differentiate the following w. r. t. x: cot³ [log (x³)] (2)
Answer:

Question 13.
Integrate the given function w.r.t. x: x⁹ ⋅ sec²(x¹⁰) (2)
Answer:
Let I = ∫x⁹ . sec²(x¹⁰) dx
Put x¹⁰ = t
Differentiating w.r.t.x, we get
10x⁹ dx = dt
∴ x⁹ dx = \(\frac{1}{10}\) dt 10
∴ x⁹ dx = \(\frac{1}{10}\) ∫ sec²t. dt
= \(\frac{1}{10}\)tan t + c
∴ I = \(\frac{1}{10}\) tan t + c
∴ I = \(\frac{1}{10}\) tan (x¹⁰) + c
∴ I = \(\frac{1}{10}\) tan (x¹⁰) + c

Question 14.
Solve the differential equation.
sec²x⋅tan y⋅dx + sec²y⋅tan x⋅dy = 0 (2)
Answer:
sec² x • tan y dx + sec² y • tan x dy = 0
Dividing both sides by tan x tan y, we get
\(\frac{\sec ^2 x \tan y}{\tan x \tan y}\)dx + \(\frac{\sec ^2 y \tan x}{\tan x \tan y}\)dy = 0
Integrating on both sides, we get
\(\int \frac{\sec ^2 x}{\tan x}\)dx + \(\int \frac{\sec ^2 y}{\tan y}\)dy = 0
∴ log |tan x| + log |tan y| = log |c|
∴ log |tan x – tan y| = log |c|
∴ tan x tan y = c

Section – C

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

Question 15.
Express the following circuits in the symbolic form of logic and write the input-output table. (3)

Answer:
Let p : The switch S₁ is closed.
q : The Switch S₂ is closed.
~p : The switch S₁‘ is closed.
~q : The Switch S₂‘ is closed.
Symbolic form of the given circuit is [p ∨ (~p ∧ ~q)] ∨ (p ∧ q)
Input – output table:

Question 16.
Prove that: sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) – 3sin^-1\(\left(\frac{\sqrt{3}}{2}\right)\) = –\(\frac{3 \pi}{4}\) (3)
Answer:
Let sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) = x
∴ sin x = \(\frac{1}{\sqrt{2}}\) = sin \(\frac{\pi}{4}\)
The principal value branch of sin^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(-\frac{\pi}{2}\) ≤ \(\frac{\pi}{4}\) ≤ \(\frac{\pi}{2}\).
∴ x = \(\frac{\pi}{4}\)

Question 17.
Using the truth table, prove the following logical equivalence: (3)
p ↔ q ≡ ~ [(p ∨ q) ∧ ~ (p ∧ q)] (3)
Answer:

Question 18.
In ∆ABC, with usual notations prove that a² = b² + c² – 2bc cos A. (3)
Answer:
Take A as the origin. X – axis along AB and the line perpendicular to AB through A as the y – axis. The co-ordinates of A, B and C are (0, 0), (0, 0) and (b cos A, b sin A) respectively.
To prove that a² = b² + c² – 2bc cos A

L.H.S. = a² = BC²
= (c – b cos A)² + (0 – b sin A)² …[By distance formula]
= c² + b² cos² A – 2 bc cos A + b² sin² A
= c² + b² cos² A + b² sin² A – 2bc cos A
= c² + b² – 2 bc cos A = R.H.S.
∴ a² = b² + c² – 2bc cosA

Question 19.
Find the joint equation of the pair of lines through the origin and perpendicular to the lines given by 2x² – 3xy – 9y² = 0.
Answer:
Given equation of the lines is 2x² – 3xy – 9y² = 0
Comparing with ax² + 2hxy + by² = 0, we get
a = 2, 2h = -3, b = -9
Let m₁ and m₂ be the slopes of the lines represented by 2x² – 3xy – 9y² = 0.
∴ m₁ + m₂ = \(-\frac{2 h}{b}\) = \(-\frac{(-3)}{-9}\) = \(-\frac{1}{3}\)
and m₁m₂ = \(\frac{a}{b}\) = \(\frac{2}{-9}\)
Since, the required lines are perpendicular to these lines.
∴ Slope of the required lines are \(-\frac{1}{m_1}\) and \(-\frac{1}{m_2}\)
Required lines also pass through the origin, therefore their equations are
y = \(-\frac{1}{m_1}\)x and y = \(-\frac{1}{m_2}\)x
∴ x + m₁y = 0 and x + m₂y = 0
∴ The joint equation of these lines is
(x + m₁y) (x + m₂y) = 0
∴ x² + (m₁ + m₂) xy + m₁m₂y² = 0
∴ x² + \(\left(-\frac{1}{3}\right)\)xy + \(\left(-\frac{2}{9}\right)\)y² = 0
∴ 9x² – 3xy – 2y² = 0

Question 20.
If G(a, 2, -1) is the centroid of the triangle with vertices P(1, 3, 2), Q(3, b, -4) and R(5, 1, c), then find the values of a, b and c. (3)
Answer:
Let \(\overline{\mathrm{p}}\), \(\overline{\mathrm{q}}\) and \(\overline{\mathrm{r}}\) be the position vectors of points P, Q and R respectively.
G(\(\overline{\mathrm{g}}\)) is the centroid of APQR.
\(\overline{\mathrm{g}}\) = \(\frac{\bar{p}+\bar{q}+\bar{r}}{3}\)
∴ a\(\hat{i}\) + 2\(\hat{j}\) – \(\hat{k}\) = \(\frac{(\hat{i}+3 \hat{j}+2 \hat{k})+(3 \hat{i}+b \hat{j}-4 \hat{k})+(5 \hat{i}+\hat{j}+c \hat{k})}{3}\)
= \(\frac{(1+3+5) \hat{i}+(3+b+1) \hat{j}+(2-4+c) \hat{k}}{3}\)
By equality of vectors,
a = \(\frac{1+3+5}{3}\) = \(\frac{9}{3}\) = 3,
2 = \(\frac{3+b+1}{3}\)
∴ 6 = 4 + b
∴ b = 2
and -1 = \(\frac{2-4+c}{3}\)
∴ -3 = -2 + c
∴ c = -1

Question 21.
If x = f (t) and y = g(t) are two differentiable functions of parameter t, then prove that y is a differentiable function of x and \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = \(\left(\begin{array}{l}
\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{~d} x}{\mathrm{dt}}}
\end{array}\right)\), \(\frac{\mathrm{dx}}{\mathrm{dt}}\) ≠ 0.
Answer:
x and y are differentiable functions of t.
Let there be a small increment δt in the value of t.
Correspondingly, there should be a small increments δx, δy in the values of x and y respectively.
As δt → 0, δx → 0, δy → 0
Consider, \(\frac{\delta y}{\delta x}\) = \(\frac{\frac{\delta y}{\delta t}}{\frac{\delta x}{\delta t}}\)
Taking \(\lim _{\delta t \rightarrow 0}\) on both sides, we get
\(\lim _{\delta t \rightarrow 0} \frac{\delta y}{\delta x}\) = \(\frac{\lim _{\delta \dagger \rightarrow 0} \frac{\delta y}{\delta t}}{\lim _{\delta \dagger \rightarrow 0} \frac{\delta x}{\delta t}}\)
Since x and y are differentiable functions of t, \(\lim _{\delta t \rightarrow 0} \frac{\delta y}{\delta t}\) = \(\frac{\mathrm{d} y}{\mathrm{~d} t}\) exists and is finite.

∴ Limits on right hand side exists and are finite.
∴ Limits on the left hand side should also exists and be finite.
∴ \(\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}\) = \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) exists and is finite.

Question 22.
A stone is dropped into a quiet lake and waves in the form of circles are generated, radius of the circular wave increases at the rate of 5 cm/ sec. At the instant when the radius of the circular wave is 8 cm, how fast the area enclosed is increasing? (3)
Answer:
Let R be the radius and A be the area of the circular wave.
Then, \(\frac{d R}{d t}\) = 5 cm/sec, R = 8 cm ………(i)
and A = πR²
Differentiating w.r.t. t, we get
\(\frac{d A}{d t}\) = π\(\frac{d}{d t}\)(R²)
\(\frac{d A}{d t}\) = 2πR\(\frac{d R}{d t}\)
= 2π(8)(5)
= 80π cm²/ sec
Hence, when the radius of the circular wave is 8 cm, the area of the circular wave is increasing at the rate of 80π cm²/ sec.

Question 23.
Find the approximate value of loge (9.01), given that log 3 = 1.0986. (3)
Answer:
Let f(x) = log_e x
∴ f'(x) = \(\frac{1}{x}\)
x = 9.01 = 9 + 0.01 = a + h
Here, a = 9 and h = 0.01
f(a) = f(9) = log_e9 = log_e3² = 2 log_e 3
= 2 × 1.0986
= 2.1972
f ‘(a) = f'(9) = \(\frac{1}{9}\) = 0.1111
f(a + h) ≈ f (a) + h f ‘(a)
∴ log_e(9.01) ≈ 2.1972 + (0.01)(0.1111)
≈ 2.1972 + 0.001111
∴ log_e(9.01) ≈ 2.198311

Question 24.
Show that : \(\int_a^b f(x) \mathrm{d} x[latex] = [latex]\int_a^b f(a+b-x) d x[latex] (3)
Answer:
Consider R.H.S.: [latex]\int_a^b\)f(a + b – x)dx
Let I = \(\int_a^b\)f(a + b – x)dx
Put a + b – x = t
∴ -dx = dt
∴ dx = -dt
When x = a, t = a + b – a = b and when x = b,t = a + b – b = a

Question 25.
A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of at least 5 successes. (3)
Answer:
Let X denote the number of odd numbers.
P(getting an odd number) = p = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ a = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Given, n = 6
∴ X ~ B(6, \(\frac{1}{2}\))
The p.m.f. of X is given by
P(X = x) = ⁶C_x\(\left(\frac{1}{2}\right)^x\left(\frac{1}{2}\right)^{6-x}\), x = 0, 1, 2, ….., 6
P(at least 5 successes) = P(X ≥ 5)
= P(X = 5 or X = 6)
= P(X = 5) + P(X = 6)
= \({ }^6 C_5\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)\) + = \({ }^6 C_6\left(\frac{1}{2}\right)^6\)
= \(\frac{6}{2^6}\) + \(\frac{1}{2^6}\)
= \(\frac{7}{64}\)

Question 26.
Evaluate: \(\int_0^1 \log \left(\frac{1}{x}-1\right) d x\) (3)
Answer:

Section – D [20 Marks]

Attempt any FIVE of the following questions:

Question 27.
Solve the following equations by the method of reduction:
x + 3y + 2z = 6, 3x – 2y + 5z = 5 and 2x – 3y + 6z = 7 (4)
Answer:
Matrix form of the given system of equations is

∴ By equality of matrices, we get
x + 3y + 2z = 6 ……… (i)
-11y – z = -13 ………(ii)
31z = 62 ………(iii)
i.e., z = 2
Substituting z = 2 in equation (ii), we get
-11y – 2 = -13
∴ y = 1
Substituting y = 1 and z = 2 in equation (i), we get
x + 3(1) + 2(2) = 6
∴ x = -1
∴ x = -1, y = 1 and z = 2 is the required solution.

Question 28.
Prove that three vectors \(\overline{\mathrm{a}}\), \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) are coplanar, if and only if there exists a non-zero linear combination x\(\overline{\mathrm{a}}\) + y\(\overline{\mathrm{b}}\) + z\(\overline{\mathrm{c}}\) = 0 with (x, y, z) ≠ (0, 0, 0). (4)
Answer:
Assume that \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) are coplanar.
Case – 1:
Suppose that any two of \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) are collinear vectors, say \(\bar{a}\) and \(\bar{b}\).
∴ There exist scalars x, y at least one of which is non-zero such that x\(\bar{a}\) + y\(\bar{b}\) = 0
i.e., x\(\bar{a}\) + y\(\bar{b}\) + 0\(\bar{c}\) = 0 and (x, y, 0) is the required solution for x\(\bar{a}\) + y\(\bar{b}\) + z\(\bar{c}\) = \(\bar{0}\).
Case – 2:
No two vectors \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) are collinear.
As \(\bar{c}\) is coplanar with \(\bar{a}\) and \(\bar{b}\),
∴ we have scalars x, y such that \(\bar{c}\) = x\(\bar{a}\) + y\(\bar{b}\).
∴ x\(\bar{a}\) + y\(\bar{b}\) – \(\bar{c}\) = \(\bar{0}\) and (x, y, -1) is the required solution for x\(\bar{a}\) + y\(\bar{b}\) – z\(\bar{c}\) = \(\bar{0}\).
Conversely, suppose x\(\bar{a}\) + y\(\bar{b}\) = \(\bar{0}\) where one of x, y, z is non-zero, say z ≠ 0.
∴ \(\bar{c}\) = \(\frac{-x}{z} \bar{a}\) – \(\frac{y}{z} \bar{b}\)
∴ \(\bar{c}\) is coplanar with \(\bar{a}\) and \(\bar{b}\).
∴ \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) are coplanar vectors.

Question 29.
Show that lines \(\overline{\mathbf{r}}\) = (\(\hat{\mathrm{i}}\) + \(\hat{\mathrm{j}}\) – \(\hat{\mathrm{k}}\)) + λ(2\(\hat{\mathrm{i}}\) – 2\(\hat{\mathrm{j}}\) + \(\hat{\mathrm{k}}\)) and \(\overline{\mathbf{r}}\) = (4\(\hat{\mathrm{i}}\) – 3\(\hat{\mathrm{j}}\) + 2\(\hat{\mathrm{k}}\)) + μ(\(\hat{\mathrm{i}}\) – 2\(\hat{\mathrm{j}}\) + 2\(\hat{\mathrm{k}}\)) are coplanar. Find the equation of the plane determined by them. (4)
Answer:

The equation of the plane determined by them is
\(\left(\overline{\mathrm{r}}-\overline{\mathrm{a}}_1\right)\) ⋅ \(\left(\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2\right)\) = 0
∴ \(\bar{r}\) = \(\left(\bar{b}_1 \times \bar{b}_2\right)\) = \(\overline{a_1} \cdot\left(\bar{b}_1 \times \bar{b}_2\right)\)
∴ \(\bar{r}\) = (-2\(\hat{\mathbf{i}}\) – 3\(\hat{\mathbf{j}}\) – 2\(\hat{\mathbf{k}}\)) = (\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)) ⋅(-2\(\hat{\mathbf{i}}\) –
3\(\hat{\mathbf{j}}\) – 2\(\hat{\mathbf{k}}\))
∴ \(\bar{r}\)⋅(-2\(\hat{\mathbf{i}}\) – 3\(\hat{\mathbf{j}}\) – 2\(\hat{\mathbf{k}}\)) = -3
∴ \(\bar{r}\)⋅(2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) + 2\(\hat{\mathbf{k}}\) = 3

Question 30.
A firm manufacturing two types of electrical items A and B, can make a profit of ₹ 20/- per unit of A and ₹ 30/- per unit of B. Both A and B make use of two essential components a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each units of B requires 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should be manufactured per month to maximize profit? How much is the maximum profit? (4)
Answer:
Let the firm manufacture x units of A and y units of B.
The profit is ₹ 20 per unit of A and ₹ 30 per unit of B.
∴ Total profit = ₹ (20 x + 30 y).
We construct a table with the constraints of number of motors and transformers needed.

From the table, the total motors required is (3x + 2y) and total motor required is (2x + 4y).
But total supply of components per month is restricted to 210 motors and 300 transformers.
∴ The constraints are 3x + 2y ≤ 210 and 2x + 4y ≤ 300.
As x, y cannot be negative, we have x ≥ 0 and y ≥ 0.
Hence the given LPP can be formulated as follows:
Maximize Z = 20x + 30y
Subject to
3x+ 2y ≤ 210,
2x + 4y ≤ 300,
x ≥ 0, y ≥ 0.
For graphical solutions of the inequalities, consider lines L₁ : 3x + 2y = 210 and 2x + 4y = 300

L₁ passes through A (0, 105) and B (70, 0)
L₂ passes through P (0, 75) and Q (150, 0)
Solving both lines, we get x = 30, y = 60
The coordinates of origin O(0, 0) satisfies both the inequalities.
∴ The required region is on origin side of both the lines L₁ and L₂.
As x ≥ 0, y ≥ 0 ; the feasible region lies in the first quadrant.
OBRPO is the required feasible region.
At O(0, 0), Z = 0 + 0 = 0
At B (70, 0), Z = 20 (70) + 0 = 1400
At R (30, 60), Z = 20 (30) + 30 (60) = 2400
At P (0, 75), Z = 0 + 30 (75) = 2250
The maximum value of Z is 2400 and it occurs at R (30, 60)
Thus 30 units of A and 60 units of B must be manufactured to get maximum profit of ₹ 2400.

Question 31.
If u and v are two functions of x, then prove that ∫u • vdx = u∫vdx – ∫(\(\frac{\mathrm{d}}{\mathrm{~d} x}\)(u) • ∫vdx)dx.
Hence, evaluate ∫x log x dx (4)
Answer:

Question 32.
Find the area of the region lying between the parabolas 4y² = 9x and 3x² = 16y.
Answer:
Given equations of the parabolas are
4y² = 9x ……..(i)
and 3x² = 16y
∴ y = \(\frac{3 x^2}{16}\) …. (ii)
From (i), we get
y² = \(\frac{9}{4}\)x
∴ y = \(\frac{3}{2} \sqrt{x}\) ….(iii) [∵ In first quadrant, y> 0]
Find the points of intersection of 4y² = 9x and 3x² = 16y.
Substituting (ii) in (i), we get
4\(\left(\frac{3 x^2}{16}\right)^2\) = 9x
∴ x⁴ = 64x
∴ x(x³ – 64) = 0
∴ x = 0 or x³ = 64 = 4³
∴ x = 0 or x = 4
When x = 0, y = 0 and when x = 4, y = 3
∴ The points of intersection are 0(0,0) and B(4, 3).
Draw BD ⊥ OX.
Required area = area of the region OABCO
= area of the region ODBCO – area of the region ODBAO
= area under the parabola 4y²
= 9x – area under the parabola 3x² = 16y

Question 33.
A curve passes through the point (0, 2). The sum of the co-ordinates of any point on the curve exceeds the slope of the tangent to the curve at that point by 5. Find the equation of the curve.
Answer:
Let (x, y) be any point on the curve y = f (x).
The slope of the tangent at point (x, y) is \(\frac{\mathrm{d} y}{\mathrm{~d} x}\).
∴ \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) + 5 = x + y
∴ \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) – y = (x – 5)
This equation is of the form \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) + Py = Q,
where P = -1 and Q = (x – 5)
∴ I.F. = \(e^{\int P d x}\) = \(e^{\int-1 \cdot d x}\) = e^-x
∴ Solution of the given equation is
∴ y (I.F.) = ∫Q(I.F.) dx + c
∴ y e^-x = ∫(x – 5)e^-x dx+ c

But the required curve passes through the point (0, 2).
∴ 0 + 2 – 4 = ce°
∴ c = -2
Substituting c = -2 in (i), we get
x + y – 4 = -2e^x
∴ y = 4 – x – 2e^x, which is the required equation of the curve.

Question 34.
A player tosses two coins. He wins ₹ 10 if 2 heads appears, ₹ 5 if 1 head appears and ₹ 2 if no head appears. Find the expected winning amount and variance of winning amount.
Answer:
Let X denote the winning amount.
∴ Possible values of X are 2, 5, 10
Let P(getting head) = p = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ P(X = 2) = P(no head) = qq = q² = \(\frac{1}{4}\)
P(X = 5) = P(one head) = pq + qp
= 2pq = 2 × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{2}{4}\)
P(X = 10) = P(two heads) = pp = p² = \(\frac{1}{4}\)
∴ The probability distribution of X is as follows

Variance of winning amount = Var(X)
= E(X²) – [E(X)]²
= 38.5 – (5.5)²
= 38.5 – 30.25
= ₹ 8.25

Maharashtra Board Class 12 Maths Previous Year Question Papers