Students can access the CBSE Sample Papers for Class 11 Maths with Solutions and marking scheme Set 1 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Maths Set 1 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question paper contains five sections A, B, C, D and E. Each section is compulsory. However, there are internal choices in some questions.
- Section A has 18 MCQ’s and 2 Assertion-Reason based questions of 1 mark each.
- Section B has 5 Very Short Answer (VSA)-type questions of 2 marks each.
- Section C has 6 Short Answer (SA)-type questions of 3 marks each.
- Section D has 4 Long Answer (LA)-type questions of 5 marks each.
- Section E has 3 source based/case based/passage based/integrated units of assessment of 4 marks each with sub-parts.
Section-A
(Multiple Choice Questions) Each question carries 1 mark
Question 1.
If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, then what is A ∪ B?
(A) {1, 2, 3, 4, 5, 6}
(B) {1, 2, 3, 4}
(C) {3, 4}
(D) {5, 6}
Answer:
(A) {1, 2, 3, 4, 5, 6}
Explanation:
The union of two sets A and B, denoted by A ∪ B, is the set of all elements that are in either A or B (or both). In this case, A is the set {1, 2, 3, 4} and B is the set {3, 4, 5, 6}.
The elements that are in either A or B (or both) are 1, 2, 3, 4, 5 and 6, so A ∪ B = {1, 2, 3, 4, 5, 6}.
Question 2.
Three planes x + y = 0, y + z = 0 and x + z = 0
(A) meet in a line
(B) meet in a unique point
(C) meet taken two at a time in parallel lines
(D) none of these
Answer:
(B) meet in a unique point
Explanation:
Given, three planes are
x + y = 0 ……(i)
y + z = 0 …….(ii)
and x + z = 0 ……(iii)
On adding equations (i), (ii) and (iii), we get
2(x + y + z) = 0
⇒ x + y + z = 0 ……(iv)
From equation (i),
0 + z = 0
⇒ z = 0
From equation (ii),
x + 0 = 0
⇒ x = 0
From equation (iii),
y + 0 = 0
⇒ y = 0
So, (x, y, z) = (0, 0, 0)
Hence, the three planes meet in a unique point.
Question 3.
The locus of a point, whose abscissa and ordinate are always equal is
(A) x – y = 0
(B) x + y = 1
(C) x + y + 1 = 0
(D) None of these
Answer:
(A) x – y = 0
Explanation:
Let the abscissa and ordinate of a point “P” be (x, y).
Given condition: Abscissa = Ordinate (i.e.) x = y
Hence, the locus of a point is x – y = 0.
Question 4.
If A = {x | x is a multiple of 3} and B = {x | x is a multiple of 5}, then A ∩ B is
(A) {3, 5}
(B) {15, 30}
(C) {0}
(D) {3, 5, 15, 30}
Answer:
(B) {15, 30}
Explanation:
The intersection of two sets A and B, denoted by A ∩ B, is the set of elements that are common to both A and B. In this case, A is the set of multiples of 3 and B is the set of multiples of 5. The elements that are common to both sets are 15 and 30 (which are multiples of both 3 and 5), so A ∩ B = {15, 30}.
Question 5.
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
(A) 2
(B) 2.57
(C) 3
(D) 3.75
Answer:
(B) 2.57
Explanation:
Mean of given data,
\(\bar{x}=\frac{3+10+10+4+7+10+5}{7}\) = \(\frac{49}{7}\) = 7
∴ Mean deviation (MD) = \(\frac{\Sigma d_i}{n}\)
= \(\frac{18}{7}\)
= 2.57
Question 6.
Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is
(A) mn
(B) nm – 1
(C) mn – 1
(D) 2mn – 1
Answer:
(D) 2mn – 1
Explanation:
Given,
n(A) = m and n(B) = n
n(A × B) = n(A) . n(B) = mn
Now, total no. of relations from A to B = 2n(A × B) – 1 = 2mn – 1
Question 7.
Three identical dice are rolled. What is the probability that the same number will appear on each of them?
(A) \(\frac{1}{6}\)
(B) \(\frac{1}{36}\)
(C) \(\frac{1}{18}\)
(D) \(\frac{3}{28}\)
Answer:
(B) \(\frac{1}{36}\)
Explanation:
Total number of cases = 63 = 216
The same number can appear on each of the dice in the following ways:
(1, 1, 1), (2, 2, 2),……, (3, 3, 3)
So, favourable number of cases = 6
Hence, required probability = \(\frac{6}{216}=\frac{1}{36}\)
Question 8.
Let F1 be the set of parallelograms, F2 be the set of rectangles, F3 be the set of rhombuses, F4 be the set of squares and F5 the set of trapeziums in a plane. Then F1 may be equal to
(A) F2 ∩ F3
(B) F3 ∩ F4
(C) F2 ∪ F5
(D) F2 ∪ F3 ∪ F4 ∪ F1
Answer:
(D) F2 ∪ F3 ∪ F4 ∪ F1
Explanation:
Every rectangle, rhombus and square in a plane is a parallelogram but every trapezium is not a parallelogram. Thus, F1 = F2 ∪ F3 ∪ F4 ∪ F1
Question 9.
Range of f(x) = \(\frac{1}{1-2 \cos x}\) is
(A) [\(\frac{1}{3}\), 1]
(B) [-1, \(\frac{1}{3}\)]
(C) (-∞, -1] ∪ [\(\frac{1}{3}\), ∞)
(D) [\(-\frac{1}{3}\), 1]
Answer:
(B) [-1, \(\frac{1}{3}\)]
Explanation:
We have, f(x) = \(\frac{1}{1-2 \cos x}\)
Since, -1 ≤ cos x ≤ 1
⇒ 1 ≥ cos x ≥ -1
⇒ -2 ≤ -2cos x ≤ 2
⇒ -2 + 1 ≤ 1 – 2 cos x ≤ 2 + 1
⇒ -1 ≤ 1 – 2 cos x ≤ 3
⇒ \(-1 \leq \frac{1}{1-2 \cos x} \leq \frac{1}{3}\)
⇒ -1 ≤ f(x) ≤ \(\frac{1}{3}\)
Or, Range of f(x) = [-1, \(\frac{1}{3}\)]
Question 10.
The number of different four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is
(A) 120
(B) 96
(C) 24
(D) 100
Answer:
(C) 24
Explanation:
Required no. of 4-digit numbers than can be formed with the digits 2, 3, 4, 7 using each digit only once = 4P4
= \(\frac{4 !}{(4-4) !}\)
= 4!
= 4 × 3 × 2 × 1
= 24
Question 11.
Given that x, y and b are real numbers and x < y, b < 0, then
(A) \(\frac{x}{b}<\frac{y}{b}\)
(B) \(\frac{x}{b} \leq \frac{y}{b}\)
(C) \(\frac{x}{b}>\frac{y}{b}\)
(D) \(\frac{x}{b} \geq \frac{y}{b}\)
Answer:
(C) \(\frac{x}{b}>\frac{y}{b}\)
Explanation:
According to inequality rule, when an inequality is divided by a negative number, then sign of inequality is reversed.
Thus, if x < y, then \(\frac{x}{b}>\frac{y}{b}\) as b < 0.
Question 12.
Following are the marks obtained by 9 students in a mathematics test 50, 69, 20, 33, 53, 39, 40, 65, 59. The mean deviation from the median is
(A) 9
(B) 10.5
(C) 12.67
(D) 14.76
Answer:
(C) 12.67
Explanation:
Here, n = 9
In ascending order marks are 20, 33, 39, 40, 50, 53, 59, 65, 69
∴ Median = \(\frac{9+1}{2}\) = 5th term i.e., 50
Now,
Thus, MD = \(\frac{\Sigma d_i}{n}\)
= \(\frac{114}{9}\)
= 12.67
Question 13.
If a + ib = c + id, then
(A) a2 + c2 = 0
(B) b2 + c2 = 0
(C) b2 + d2 = 0
(D) a2 + b2 = c2 + d2
Answer:
(D) a2 + b2 = c2 + d2
Explanation:
Given,
a + ib = c + id
⇒ |a + ib| = |c + id|
⇒ \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)
On squaring both sides, we get
a2 + b2 = c2 + d2
Question 14.
If sin θ + cos θ = 1, then the value of sin 2θ is equal to
(A) 1
(B) \(\frac{1}{2}\)
(C) 0
(D) -1
Answer:
(C) 0
Explanation:
We have, sin θ + cos θ = 1
⇒ sin2θ + cos2θ + 2 sin θ . cos θ = 1 (On squaring both sides)
⇒ 1 + sin 2θ = 1
⇒ sin 2θ = 0
Question 15.
If f(x) = x100 + x99 + ……. + x + 1, then f'(1) is equal to
(A) 5050
(B) 5049
(C) 5051
(D) 50051
Answer:
(A) 5050
Explanation:
f(x) = x100 + x99 + …… + x + 1
f'(x) = 100x99 + 99x98 + ….. + 1 + 0
f'(1) = 100(1)99 + 99(1)98 + ….. + 1 = 100 + 99 + …… + 1
This is an AP with common difference -1,
a = 100, n = 100 and l = 1
So, the sum of this AP = (\(\frac{100}{2}\)) (100 + 1)
= 50(101)
= 5050
Question 16.
The number of tangents that can be drawn from (1, 2) to x2 + y2 = 5 is
(A) 0
(B) 1
(C) 2
(D) More than 2
Answer:
(B) 1
Explanation:
Given point (1, 2) and equation of circle is x2 + y2 = 5
Now, x2 + y2 – 5 = 0
Put (1, 2) in this equation, we get
12 + 22 – 5 = 1 + 4 – 5 = 5 – 5 = 0
So, the point (1, 2) lies on the circle.
Hence, only one tangent can be drawn.
Question 17.
The equation of the line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0 is
(A) y – x + 1 = 0
(B) y – x – 1 = 0
(C) y – x + 2 = 0
(D) y – x – 2 = 0
Answer:
(B) y – x – 1 = 0
Explanation:
Any line perpendicular to line
x + y + 1 = 0 is x – y + k = 0 …… (i)
Also, line (i) passes through the point (1, 2) then
1 – 2 + k = 0
⇒ k = 1
From equation (i), we get
x – y + 1 = 0 or y – x – 1 = 0
Question 18.
Given the integers r > 1, n > 2, and coefficients of (3r)th and (r + 2)th terms in the binomial expansion of (1 + x)2n are equal, then
(A) n = 2r
(B) n = 3r
(C) n = 2r + 1
(D) none of these
Answer:
(A) n = 2r
Explanation:
Assertion-Reason Based Questions
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Question 19.
Assertion (A): The region containing all the solutions of an inequality is called the solution region.
Reason (R): The values of x, which make an inequality a true statement, are called solutions of the inequality.
Answer:
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
Question 20.
Assertion (A): Consider z1 and z2 as two complex numbers such that |z1| = |z2| + |z1 – z2|, then Im\(\left(\frac{z_1}{z_2}\right)\) = 0
Reason (R): arg(z) = 0 ⇒ z is purely real.
Answer:
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
Given, |z1| = |z2| + |z1 – z2|
⇒ |z1 – z2|2 = (|z1| – |z2|)2
⇒ |z1|2 + |z2|2 – 2|z1||z2|cos(θ1 – θ2)
⇒ |z1|2 + |z2|2 – 2|z1||z2|
⇒ cos (θ1 – θ2) = 1
⇒ θ1 – θ2 = 0
⇒ arg(z1) – arg(z2) = 0
⇒ \(\arg \left(\frac{z_1}{z_2}\right)\) = 0
Section-B
[This section comprises very short answer-type questions (VSA) of 2 marks each]
Question 21.
Evaluate (101)4 using the binomial theorem.
OR
For the pool, the average water acidity is considered normal when in the range of 7.2 and 7.8 pH when measured for three days continuously. What will be the range of pH for the third day when acidity for the first two days is 7.52 and 7.1. Consider the pH of the pool in normal.
Answer:
Given: (101)4
Here, 101 can be written as the sum or the difference of two numbers, such that the binomial theorem can be applied.
Therefore, 101 = 100 + 1
Hence, (101)4 = (100 + 1)4
Now, by applying the binomial theorem, we get:
(101)4 = (100 + 1)4
⇒ (101)4 = 4C0 (100)4 + 4C1 (100)3 (1) + 4C2 (100)2 (1)2 + 4C3 (100) (1)3 + 4C4 (1)4
⇒ (101)4 = (100)4 + 4(100)3 + 6(100)2 + 4(100) + (1)4
⇒ (101)4 = 100000000 + 4000000 + 60000 + 400 + 1
⇒ (101)4 = 104060401
Hence, the value of (101)4 is 104060401.
OR
Let, reading for third day be x3.
Given, reading for first two days: 7.52 and 7.1
Therefore, 7.2 < \(\frac{7.52+7.1+x_3}{3}\) < 7.8
21.6 < 14.62 + x3 < 23.4
(21.6 – 14.62) < x3 < (23.4 – 14.62)
6.98 < x3 < 8.78
The range of pH value for the third day lies between 6.98 to 8.78.
Question 22.
Find the symmetric difference of sets A = {1, 3, 5, 6, 7} and B = {3, 7, 8, 9}.
Answer:
Given, sets are A = {1, 3, 5, 6, 7} and B = {3, 7, 8, 9}
Now, A – B = {1, 3, 5, 6, 7} – {3, 7, 8, 9} = {1, 5, 6}
and B – A = {3, 7, 8, 9} – {1, 3, 5, 6, 7} = {8, 9}
∴ Required symmetric difference, A ∆ B = (A – B) ∪ (B – A)
= {1, 5, 6} ∪ {8, 9}
= {1, 5, 6, 8, 9}
Question 23.
Show that if A ⊂ B, then C – B ⊂ C – A.
OR
If A = {x : x ∈ W, x < 2}, B = {x : x ∈ N, 1 < x < 5} and C = {3, 5}, then find
(i) A × (B ∩ C)
(ii) A × (B ∪ C)
Answer:
Let x ∈ C – B
x ∈ C and x ∉ B
x ∈ C and x ∉ A (∴ A ⊂ B)
x ∈ (C – A)
(C – B) ⊂ (C – A)
OR
We have, A = {x : x ∈ W, x < 2} = {0, 1}
B = {x : x ∈ N, 1 < x < 2} = {2, 3, 4}
and C = {3, 5}
(i) Now, B ∩ C = {2, 3, 4} ∩ {3, 5} = {3}
∴ A × (B ∩ C) = {0, 1} × {3} = {(0, 3), (1, 3)}
(ii) Now, B ∪ C = {2, 3, 4} ∪ {3, 5} = {2, 3, 4, 5}
∴ A × (B ∪ C) = {0, 1} × {2, 3, 4, 5} = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}
Question 24.
Solve the equation √3x2 – √2x + 3√3 = 0.
Answer:
Given equation: √3x2 – √2x + 3√3 = 0
Comparing it with ax2 + bx + c = 0, we get
a = √3, b = -√2, c = 3√3
Discriminant = b2 – 4ac
= (-√2)2 – 4 × √3 × 3√3
= 2 – 36
= -34
∴ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2 \sqrt{3}}\)
= \(\frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}}\)
Hence, the roots of the given equation are \(\frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}}\).
Question 25.
Prove that: sin2 6x – sin2 4x = sin 2x sin 10x
Answer:
L.H.S. = sin2 6x – sin2 4x
= sin (6x + 4x) sin (6x – 4x) [∵ sin2 A – sin2 B = sin (A + B) sin (A – B)]
= sin 10x sin 2x
= sin 2x sin 10x
= R.H.S.
Hence proved.
Section-C
[This section comprises short answer type questions (SA) of 3 marks each]
Question 26.
Expand using binomial theorem \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^4\), x ≠ 0.
Answer:
Question 27.
Prove that: \(\frac{\sec 8 \theta-1}{\sec 4 \theta-1}=\frac{\tan 8 \theta}{\tan 2 \theta}\)
Answer:
We have,
Hence proved.
Question 28.
The ratio of the sum of n terms of two A.P.’s is (7n – 1) : (3n + 11), Find the ratio of their 10th terms.
OR
A student has to answer 10 questions, choosing at least 4 from each of part A and B. If there are 6 questions in part A and 7 in part B. In how many ways can the student choose 10 questions?
Answer:
Let a1, a2 be the first terms and d1, d2 the common differences of the two given A.P.’s. Then, the sums of their n terms are given by
It is given that
To find the ratio of the 10th terms of the two A.P.’s, we replace n by (2 × 10 – 1) i.e., ’19’ in (i)
Replacing n by ’19’ in (i), we get,
∴ Ratio of the 10th terms of the two A.P.’s = \(\frac{33}{17}\)
OR
Combination from A and from B: \(\frac{4}{6} ; \frac{5}{5} ; \frac{6}{4}\)
Number of ways to get \(\frac{4}{6}\) pattern = 6C4 × 7C6
= \(\frac{6 !}{2 ! 4 !} \times \frac{7 !}{1 ! 6 !}\)
= \(\frac{6 \cdot 5}{2 \cdot 1} \times \frac{7}{1}\)
= 15 × 7
= 105
Number of ways to get \(\frac{5}{5}\) pattern = 6C5 × 7C5
= \(\frac{6 !}{1 ! 5 !} \times \frac{7 !}{2 ! 5 !}\)
= \(\frac{6}{1} \times \frac{7 \times 6}{2 \times 1}\)
= 6 × 21
= 126
Number of ways to get the \(\frac{6}{4}\) pattern = 6C6 × 7C4
= \(\frac{6 !}{0 ! 6 !} \times \frac{7 !}{3 ! 4 !}\)
= 1 × \(\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}\)
= 35
Hence, total number of ways = 105 + 126 + 35 = 266
Question 29.
If f(x) = \(\frac{x-1}{x+1}\), then show that
(i) f(\(\frac{1}{x}\)) = -f(x)
(ii) \(f\left(-\frac{1}{x}\right)=-\frac{1}{f(x)}\)
OR
If sec x = √2 and \(\frac{3 \pi}{2}\) < x < 2π, find the value of \(\frac{1+\tan x+{cosec} x}{1+\cot x-{cosec} x}\)
Answer:
Question 30.
A circle has radius 3 unit and its centre lies on the line y = x – 1. If it passes through the point (7, 3), then find the equation of the circle.
OR
Find the equation for the ellipse that satisfies the given conditions: vertices (±6, 0), foci (±4, 0).
Answer:
Let the equation of the circle be
(x – h)2 +(y – k)2 = 32 ………(i)
Since, (h, k) lies on y = x – 1
∴ k = h – 1
⇒ h – k – 1 = 0 …..(ii)
Also, circle (i) passes through (7, 3),
(7 – h)2 + (3 – k)2 = 9
⇒ 49 – 14h + h2 + 16 – 8h + h2 = 9
⇒ 2h2 – 22h + 65 = 9
⇒ 2h2 – 22h + 56 = 0
⇒ h2 – 11h + 28 = 0
⇒ (h – 4)(h – 7) = 0
∴ h = 4 or h = 7
⇒ k = 3 or k = 6
∴ Equation of the circle is
(x – 4)2 + (y – 3)2 = 32 or (x – 7)2 + (y – 6)2 = 32
⇒ x2 + y2 – 8x – 6y + 16 = 0 or x2 + y2 – 14x – 12y + 76 = 0
OR
Given, vertices (±6, 0), foci (±4, 0).
Here, the vertices are on the x-axis.
Therefore, the equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where a is the semi-major axis.
Accordingly, a = 6, c = 4.
It is known as a2 = b2 + c2
⇒ 62 = b2 + 42
⇒ 36 = b2 + 16
⇒ b2 = 36 – 16
⇒ b = √20
Thus, the equation of the ellipse is \(\frac{x^2}{6^2}+\frac{y^2}{(\sqrt{20})^2}=1\)
∴ \(\frac{x^2}{36}+\frac{y^2}{20}=1\)
Question 31.
Find the value of ‘k’ if \(\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}\)
Answer:
Given,
Section-D
[This section comprises long answer type questions (LA) of 5 marks each]
Question 32.
Calculate the mean deviation about the mean of the set of first n natural numbers when ‘n’ is an odd number.
Answer:
= \(\frac{1}{n} \frac{n^2-1}{4}\)
= \(\frac{n^2-1}{4 n}\)
Question 33.
A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected, if the team has
(i) no girls.
(ii) at least one boy and one girl.
(iii) at least three girls.
OR
If xn occurs in the expansion of \(\left(x^2+\frac{1}{x}\right)^{2 n}\), then prove that its coefficient is \(\frac{2 n !}{\frac{(4 n-p) !(2 n+p) !}{3 !} \frac{(2 n !}{3 !}}\).
Answer:
Number of girls = 4 and Number of boys = 7
We have to select a team of 5 members provided that
(i) team having no girls
∴ Required selection = 7C5
= \(\frac{7 !}{5 ! 2 !}\)
= \(\frac{7 \times 6}{2}\)
= 21
(ii) at least one boy and one girl
∴ Required selection = 7C1 × 4C4 + 7C2 × 4C3 + 7C3 × 4C2 + 7C4 × 4C1
= 7 × 1 + 21 × 4 + 35 × 6 + 35 × 4
= 7 + 84 + 210 + 140
= 441
(iii) when at least three girls are included
∴ Required selection = 4C3 × 7C2 + 4C4 × 7C1
= 4 × 21 + 7
= 84 + 7
= 91
OR
Given expansion is \(\left(x^2+\frac{1}{x}\right)^{2 n}\)
Let xp occur in the expansion of \(\left(x^2+\frac{1}{x}\right)^{2 n}\)
Question 34.
Three vertices of a parallelogram ABCD are A (3, -1, 2), B (1, 2, -4) and C (-1, 1, 2). Find the coordinates of the fourth vertex.
Answer:
The three vertices of a parallelogram ABCD are given as A(3, -1, 2), B(1, 2, -4), and C(-1, 1, 2).
Let the coordinates of the fourth vertex be D (x, y, z).
We know that the diagonals of a parallelogram bisect each other.
Therefore, in parallelogram ABCD, AC and BD bisect each other.
∴ Mid-point of AC = Mid-point of BD
⇒ \(\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)=\left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)\)
⇒ (10, 2) = \(\left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)\)
⇒ \(\frac{x+1}{2}\) = 1, \(\frac{y+2}{2}\) = 0 and \(\frac{z-4}{2}\) = 2
⇒ x = 1, y = -2 and z = 8
Thus, the coordinates of the fourth vertex are (1, -2, 8).
Question 35.
Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g(x) = αx + β, then what values should be assigned to α and β?
OR
If cos (θ + φ) = m cos (θ – φ), then prove that tan θ = \(\frac{1-m}{1+m}\) cot φ.
Answer:
Given, g = {(1, 1), (2, 3), (3, 5), (4, 7)}
Since, every element of domain has unique image under g. Hence, g is a function.
Also given, g(x) = αx + β
When x = 1, then g(1) = α(1) + β
⇒ 1 = α + β …….(i)
When x = 2, then g(2) = α(2) + β
⇒ 3 = 2α + β
On solving equations (i) and (ii), we get
α = 2 and β = -1
OR
Given, cos(θ + φ) = m cos(θ – φ)
⇒ \(\frac{\cos (\theta+\phi)}{\cos (\theta-\phi)}=\frac{m}{1}\)
Using componendo and dividendo rule, we get
Section-E
[This section comprises of 3 case- study/passage-based questions of 4 marks each with subparts.]
The first two case study questions have three subparts (i), (ii), (iii) of marks 1, 1, 2 respectively. The third case study question has two subparts of 2 marks each.
Question 36.
Read the following passage and answer the questions given below:
Indeterminate forms of limits: On direct evaluation, if a limit takes the form \(\frac{0}{0}, \frac{\infty}{\infty}\), 0 × ∞, ….., we use standard results for evaluating the limits.
The below figure shows a few indeterminate forms.
\(\frac{0}{0}, \infty^0, 0^0, 1^{\infty}, 0 . \infty, \frac{\infty}{\infty}, \infty-\infty\)
(i) Evaluate: \(\lim _{x \rightarrow 2} \frac{x^6-64}{x-2}\)
(ii) Evaluate: \(\lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}\)
(iii) Evaluate: \(\lim _{x \rightarrow 0} \frac{\sqrt{1+3 x}-\sqrt{1-3 x}}{x}\)
OR
Evaluate: \(\lim _{x \rightarrow 0} \frac{8^x-2^x}{x}\)
Answer:
= log\(\frac{8}{2}\)
= log 4
Question 37.
Read the following article and answer the following question:
In drilling world’s deepest hole, the Kola Super deep Borehole, the deepest man made hole on Earth and deepest artificial point on Earth, as a result of a scientific drilling project, it was found that the temperature T in degree Celsius, x km below the surface of Earth, was given by
(i) If the required temperature lies between 200°C and 300°C, then find the range for depth x.
(ii) If |x| < 5, find the intervals in which value of x lies.
(iii) Solve for x, -13x + 2 > 18 or 13x + 15 ≤ -4
OR
Represent the inequality on number line: x > -32
Answer:
(i) 200 < 30 + 25(x – 3) < 300
⇒ 170 < 25(x – 3) < 270
⇒ \(\frac{170}{25}\) < (x – 3) < \(\frac{270}{25}\)
⇒ \(\frac{34}{5}\) < x – 3 < \(\frac{54}{5}\)
⇒ 6.8 < x – 3 < 10.8
⇒ 9.8 < x < 13.8
(ii) Since |x| < 5
∴ -5 < x < 5
Thus, solution set is (-5, 5).
(iii) -13x + 2 > 18
⇒ -13x > 16
⇒ x < \(\frac{-16}{13}\)
∴ 13x + 15 ≤ -4
⇒ 13x ≤ -4 – 15
⇒ 13x ≤ -19
⇒ x ≤ \(\frac{-19}{13}\)
OR
Question 38.
Read the following article and answer the following question:
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:
Particulars | Firm A | Firm B |
No. of wage earners | 586 | 648 |
Mean of monthly wages | ₹ 5253 | ₹ 5253 |
Variance of the distribution of wages | 100 | 121 |
(i) Find the amount paid by firm A.
(ii) Find the coefficient of variation of the distribution of wages for firm A.
Answer:
(i) No. of wage earners = 586
Mean of monthly wages, \(\bar{x}\) = ₹ 5253
Amount paid by firm A = ₹(586 × 5253) = ₹ 3078258
(ii) Variance of distribution of wages, σ2 = 100
Standard deviation, σ = √100 = 10
Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100\)
= \(\frac{10}{5253}\) × 100
= 0.19