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.
- 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 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 xey + yex = 1. (2)
Answer:
xey + yex = 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) = loge 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) = loge9 = loge32 = 2 loge 3
= 2 × 1.0986
= 2.1972
f ‘(a) = f'(9) = \(\frac{1}{9}\) = 0.1111
f(a + h) ≈ f (a) + h f ‘(a)
∴ loge(9.01) ≈ 2.1972 + (0.01)(0.1111)
≈ 2.1972 + 0.001111
∴ loge(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) = 6Cx\(\left(\frac{1}{2}\right)^x\left(\frac{1}{2}\right)^{6-x}\), 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 = -2ex
∴ y = 4 – x – 2ex, 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