Students can access the CBSE Sample Papers for Class 11 Maths with Solutions and marking scheme Set 2 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Maths Set 2 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 MCQs 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 Ehas3 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 f(x) = cos2x + sec2x, then
(A) f(x) < 1
(B) f(x) = 1
(C) 2 < f(x) < 1
(D) f(x) ≥ 2
Answer:
(D) f(x) ≥ 2
Explanation:
Question 2.
The real value of a for which the expression \(\frac{1-i \sin \alpha}{1+2 i \sin \alpha}\) is purely real is
(A) (n + 1)\(\frac{\pi}{2}\)
(B) (2n + 1)\(\frac{\pi}{2}\)
(C) nπ
(D) None of these
Answer:
(C) nπ
Explanation:
As, z is purely real, then
\(\frac{3 \sin \alpha}{1+4 \sin ^2 \alpha}\) = 0
⇒ 3 sin α = 0
⇒ sin α = 0
⇒ sin α = sin nπ, n ∈ N
⇒ α = nπ, n ∈ N
Question 3.
If 9 times the 9th term of an A.P. is equal to 13 times the 13th term, then the 22nd term of the A.P. is
(A) 0
(B) 22
(C) 220
(D) 198
Answer:
(A) 0
Explanation:
Given that 9(a + 8 d) = 13(a + 12d)
⇒ 4a = -84d
⇒ a = -21d
⇒ a + 21d = 0
Hence, the 22nd term is zero.
Question 4.
The value of cos 12° + cos 84° + cos 156° + cos 132° is
(A) \(\frac{1}{2}\)
(B) 1
(C) \(-\frac{1}{2}\)
(D) 18
Answer:
(C) \(-\frac{1}{2}\)
Explanation:
cos 12° + cos 84° + cos 156° + cos 132°
= (cos 12° + cos 132°) + (cos 84° + cos 156°)
= 2 cos 72° cos 60° + 2 cos 120° cos 36°
[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{D-C}{2}\)]
= 2 cos 72° × \(\frac{1}{2}\) + 2(\(-\frac{1}{2}\)) cos 36°
= cos 72° – cos 36°
= cos(90° – 18°) – cos 36°
= sin 18° – cos 36°
= \(\left(\frac{\sqrt{5}-1}{4}\right)-\left(\frac{\sqrt{5}+1}{4}\right)\)
= \(\frac{\sqrt{5}-1-\sqrt{5}-1}{4}\)
= \(-\frac{1}{2}\)
Question 5.
What is the distance between the points (2, 1, 3) and (-2, -1, 3)?
(A) 2√15 units
(B) 2√5 units
(C) 25 units
(D) √5 units
Answer:
(B) 2√5 units
Explanation:
Let the points be P(2, 1, 3) and Q(-2, -1, 3).
By using the distance formula,
Therefore, the required distance is 2√5 units.
Question 6.
The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is
(A) 16C11
(B) 16C5
(C) 16C9
(D) 20C9
Answer:
(C) 16C9
Explanation:
Given, the total no. of players = 22
Also, 2 players are always included and 4 players are always excluded or never included, then players remaining for selection = 22 – 2 – 4 = 16
Required no. of ways of selection = 16C9
Question 7.
If |x + 2| ≤ 9, then
(A) x ∈ (-7, 11)
(B) x ∈ [-11, 7]
(C) x ∈ (-∞, -7) ∪ (11, ∞)
(D) x ∈ (-∞, -7) ∪ [11, ∞)
Answer:
(B) x ∈ [-11, 7]
Explanation:
We have, |x + 2| ≤ 9
∴ -9 ≤ x + 2 ≤ 9 [∵ |x| ≤ a ⇒ -a ≤ x ≤ a]
⇒ -9 – 2 ≤ x + 2 – 2 ≤ 9 – 2
⇒ -11 ≤ x ≤ 7
Thus, x ∈ [-11, 7]
Question 8.
\(\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}\) is equal to
(A) \(\frac{4}{9}\)
(B) \(\frac{1}{2}\)
(C) \(-\frac{1}{2}\)
(D) -1
Answer:
(A) \(\frac{4}{9}\)
Explanation:
Question 9.
If the sum of n terms of an A.P. is given by Sn = 3n + 2n2, then the common difference of the A.P. is
(A) 3
(B) 2
(C) 6
(D) 4
Answer:
(D) 4
Explanation:
Given that Sn = 3n + 2n2
Common difference = Twice of coefficient of n2 in standard form
So, Common difference = 4
Question 10.
The domain and range of the real function f defined by f(x) = \(\frac{4-x}{x-4}\) is given by
(A) Domain = R, Range = {-1, 1}
(B) Domain = R – {1}, Range = R
(C) Domain = R – {4}, Range = R – {-1}
(D) Domain = R – {-4}, Range = {-1, 1}
Answer:
(C) Domain = R – {4}, Range = R – {-1}
Explanation:
We have, f(x) = \(\frac{4-x}{x-4}\)
f(x) is defined if x – 4 ≠ 0 or x ≠ 4
So, Domain of f(x) = R – {4}
For range,
Let f(x) = y = \(\frac{4-x}{x-4}\)
⇒ yx – 4y = 4 – x
⇒ yx + x = 4y + 4
⇒ x(y + 1) = 4(y + 1)
⇒ x = \(\frac{4(y+1)}{y+1}\)
If x is a real number, then 1 + y ≠ 0 ⇒ y ≠ -1
∴ Range of f(x) = R – {-1}
Question 11.
The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 are
(A) (-6, 5)
(B) (5, 6)
(C) (-5, 6)
(D) (6, 5)
Answer:
(B) (5, 6)
Explanation:
Let (h, k) be the coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0.
Then, the slope of the perpendicular line is k – 3h – 2.
Again, the slope of the given line x + y – 11 = 0 is -1.
Using the condition of perpendicularity of lines, we have
\(\frac{k-3}{h-2}\) × (-1) = -1
Or, k – h = 1 …..(i)
Since, (h, k) lies on the given line, we have,
h + k – 11 = 0
or h + k = 11 ……(ii)
Solving (i) and (ii), we get h = 5 and k = 6.
Thus (5, 6) are the required coordinates of the foot of the perpendicular.
Question 12.
Domain of \(\sqrt{a^2-x^2}\) (a > 0) is
(A) (-a, a)
(B) [-a, a]
(C) [0, a]
(D) (-a, 0]
Answer:
(B) [-a, a]
Explanation:
Let f(x) = \(\sqrt{a^2-x^2}\)
From domain of f(x), f(x) ≥ 0
\(\sqrt{a^2-x^2}\) ≥ 0
⇒ x2 – a2 ≤ 0
⇒ x2 ≤ a2
⇒ x ≤ ±a
⇒ -a ≤ x ≤ a
Therefore, Domain of \(\sqrt{a^2-x^2}\) = [-a, a]
Question 13.
If the mean of 100 observations is 50 and their standard deviation is 5, then the sum of all squares of all the observations is
(A) 50000
(B) 250000
(C) 252500
(D) 255000
Answer:
(C) 252500
Explanation:
Given, \(\bar{x}\) = 50, S.D. = 5
We know that,
Question 14.
If the line \(\frac{x}{a}+\frac{y}{b}=1\) passes through the points (2, -3) and (4, -5), then (a, b) is
(A) (1, 1)
(B) (-1, 1)
(C) (1,-1)
(D) (-1, -1)
Answer:
(D) (-1, -1)
Explanation:
Line passing through points (2, -3) and (4, -5) is
y + 3 = \(\frac{-5 + 3}{4 – 2}\) (x – 2)
⇒ y + 3 = \(\frac{-2}{2}\) (x – 2)
⇒ x + y = -1
⇒ \(\frac{x}{-1}\) + \(\frac{y}{-1}\) = 1
On comparing with \(\frac{x}{a}+\frac{y}{b}\) = 1, we get
a = -1 and b = -1
Question 15.
The toted number of terms in the expansion of (x + a)100 + (x – a)100 after simplification is
(A) 50
(B) 202
(C) 51
(D) None of these
Answer:
(C) 51
Explanation:
We have,
Therefore, no. of terms = 51
Question 16.
A bag contains 5 brown and 4 white socks. Ram pulls out two socks. What is the probability that both socks are of the same colour?
(A) \(\frac{9}{20}\)
(B) \(\frac{2}{9}\)
(C) \(\frac{3}{20}\)
(D) \(\frac{4}{9}\)
Answer:
(D) \(\frac{4}{9}\)
Explanation:
Total number of socks = 5 + 4 = 9
Two socks are pulled.
Now, P(Both are same colour) = \(\frac{5 C_2+4 C_2}{9 C_2}\)
Question 17.
If x1, x2, x3 and x4 be the observations with mean m and standard deviation s then, the standard deviation of the observations kx1, kx2, kx3, kx4 and kx5 is
(A) k + s
(B) sk
(C) ks
(D) s
Answer:
(C) ks
Explanation:
Here,
Question 18.
The locus of the point from which the tangent to the circles x2 + y2 – 4 = 0 and x2 + y2 – 8x + 15 = 0 are equal is given by the equation
(A) 8x + 19 = 0
(B) 8x – 19 = 0
(C) 4x – 19 = 0
(D) 4x + 19 = 0
Answer:
(B) 8x – 19 = 0
Explanation:
Given equation of circles are x2 + y2 – 4 = 0 and x2 + y2 – 8x + 15 = 0.
Now, the required line is the radical axis of the two circles is (x2 + y2 – 4) – (x2 + y2 – 8x + 15) = 0.
⇒ x2 + y2 – 4 – x2 – y2 + 8x – 15 = 0
⇒ 8x – 19 = 0
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): Let A = {x: x is a natural number greater than 1} is an empty set.
Reason (R): A set which does not contain any element is called an empty set.
Answer:
(D) (A) is false but (R) is true.
Explanation:
Natural numbers start from 1. So, a set containing natural numbers greater than 1 is not an empty set.
Question 20.
Assertion (A): If tan(\(\frac{\pi}{2}\) sin θ) = co(\(\frac{\pi}{2}\) cos θ), then sin θ + cos θ = ±√2
Reason (R): -√2 ≤ sin θ + cos θ ≤ √2
Answer:
(D) (A) is false but (R) is true.
Explanation:
tan(\(\frac{\pi}{2}\) sin θ) = co(\(\frac{\pi}{2}\) cos θ)
tan(\(\frac{\pi}{2}\) sin θ) = \(\tan \left(\frac{\pi}{2}-\frac{\pi}{2} \cos \theta\right)\)
Therefore, \(\frac{\pi}{2}\) sin θ = nπ + \(\frac{\pi}{2}-\frac{\pi}{2} \cos \theta\)
⇒ sin θ + cos θ = 2n + 1, n ∈ Z
Therefore, -√2 ≤ sin θ + cos θ ≤ √2
n = 0, -1
sin θ + cos θ = 1 or sin θ + cos θ = -1
Section-B
[This section comprises very short answer-type questions (VSA) of 2 marks each]
Question 21.
Find the coefficient of x5 in the expansion of the product (1 + 2x)6 (1 – x)7.
OR
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Answer:
= [1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + ……] × [1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + ……]
∴ Coefficient of x5 is the product = [1 × (-21) + (12 × 35) + 60 × (-35) + 160 × 21 + 240 × (-7) + 192 × 1]
= [-21 + 420 – 2100 + 3360 – 1680 + 192]
= 171
OR
A three-digit even number is to be formed from given 6 digits 1, 2, 3, 4, 5, 6.
Since, for the number is to be even, so one’s place can be filled by 2, 4 or 6. So, there are 3 ways to fill ones place.
Since, repetition is allowed, so tens place can also be filled by 6 ways.
Similarly, hundreds place can also be filled by 6 ways.
So, number of ways in which three-digit even numbers can be formed from the given digits is 6 × 6 × 3 = 108.
Question 22.
What is represented by the shaded regions in each of the following Venn-diagrams.
Answer:
(i)
∴ (A – B) ∪ (B – A)
(ii)
∴ (A ∩ B) ∪ (A ∩ C) Or, A ∩ (B ∪ C)
Question 23.
Find x and y, if (x + iy)(2 – 3i) = 4 + i.
OR
Solve the following inequalities for real x: \(\frac{2 x+3}{4}-3<\frac{x-4}{3}-2\), x ∈ R.
Answer:
Given, (x + iy)(2 – 3i) = 4 + i
⇒ (2x + 3y) + i(-3x + 2y) = 4 + i
⇒ 2x + 3y = 4 and -3x + 2y = 1
On solving, we get
x = \(\frac{5}{13}\) and y = \(\frac{14}{13}\)
OR
Question 24.
Find the range of the following relations: R = {(a, b) : a, b ∈ N and 2a + b = 10}
Answer:
Given, R = {(a, b) : a, b ∈ N and 2a + b = 10}
∵ 2a + b = 10
⇒ b = 10 – 2a
Thus, R = {(1, 8), (2, 6), (3, 4), (4, 2)}
∴ Range of R = (8, 6, 4, 2} or (2, 4, 6, 8}
Question 25.
If tan A = \(\frac{a}{a+1}\) and tan B = \(\frac{1}{2 a+1}\), then find the value of A + B.
Answer:
We know that,
Section-C
[This section comprises short answer type questions (SA) of 3 marks each]
Question 26.
Evaluate: \(\lim _{x \rightarrow 0} \frac{\sec 4 x-\sec 2 x}{\sec 3 x-\sec x}\)
Answer:
Question 27.
Given, P = {x: 5 < 2x – 1 ≤ 11, x ∈ R} and Q = {x : -1 ≤ 3 + 4x < 23, x ∈ I}. Represent P and Q on the number line. Write down the elements of P ∩ Q.
Answer:
Given, P = (x : 5 < 2x – 1 ≤ 11, x ∈ R} and Q = (x : -1 ≤ 3 + 4x < 23, x ∈ I}
Solving for P,
5 < 2x – 1 ≤ 11
⇒ 5 + 1 < 2x ≤ 11 + 1
⇒ 6 < 2x ≤ 12
⇒ 3 < x ≤ 6
Hence, P = {x : 3 < x ≤ 6, x ∈ R}
The solution P is represented on number line (a).
Next, solving for Q,
-1 ≤ 3 + 4x < 23
⇒ -1 – 3 ≤ 4x < 23 – 3
⇒ -4 ≤ 4x < 20
⇒ -1 ≤ x < 5
As Q ∈ I
Hence, solution Q = {-1, 0, 1, 2, 3, 4}
The solution Q is represented on number line (b).
Therefore, P ∩ Q = {4}
Question 28.
Using the digits 0, 1, 2, 2, 3; how many numbers greater than 20000 can be made?
OR
Show that 32n+2 – 8n – 9 is divisible by 8.
Answer:
Total number of digits = 0, 1, 2, 2, 3
Total number formed by these digit = \(\frac{5 !}{2 !}\) = \(\frac{120}{2}\) = 60
Total number formed by starting 0 = \(\frac{4 !}{2 !}=\frac{24}{2}\) = 12
Total number formed by starting 1 = \(\frac{4 !}{2 !}=\frac{24}{2}\) = 12
Total number formed greater than 20000 = 60 – 12 – 12 = 36
OR
We have 32n+2 which can be written as \(9^{n+1}=(8+1)^{n+1}\)
According to binomial theorem this can be expanded as
∴ 9n+1 – 8n – 9 is divisible by 64.
Hence, 32n+2 – 8n – 9 is divisible by 8.
Question 29.
Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
OR
Let A = {9, 10, 11, 12, 13} and let f: A → N defined by f(n) = the highest prime factor of n. Find the range of f.
Answer:
Let x ∈ B …..(i)
⇒ x ∈ A ∪ B
[Since, B ⊂ A ∪ B, all elements of B are in A ∪ B]
⇒ x ∈ A ∪ C (Given, A ∪ B = A ∪ C)
⇒ x ∈ A or x ∈ C ……(ii)
Taking x ∈ A
Also, x ∈ B [From (i)]
∴ x ∈ A ∩ B
So, x ∈ A ∩ C (Given, A ∩ B = A ∩ C)
i.e., x ∈ A and x ∈ C
i.e., x ∈ C
∴ If x ∈ B, then x ∈ C
i.e., If an element belongs to set B, then it must belong to set C also,
∴ B ⊂ C
Similarly, we can prove C ⊂ B
Since, B ⊂ C and C ⊂ B
⇒ B = C
Hence Proved.
OR
Given, A= {9, 10, 11, 12, 13}
and f: A → N and f(n) = the highest prime factor of ‘n’
For n = 9, 9 = 1 × 3 × 3
⇒ Highest prime factor of 9 = 3
For n = 10, 10 = 1 × 2 × 5
⇒ Highest prime factor of 10 = 5
For n = 11, 11 = 1 × 11
⇒ Highest prime factor of 11 = 11
For n = 12, 12 = 1 × 2 × 2 × 3
⇒ Highest prime factor of 12 = 3
For n = 13, 13 = 1 × 13
⇒ Highest prime factor of 13 = 13
∴ f(n) = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13)}
⇒ Range = {3, 5, 11, 13}
Question 30.
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
OR
The mid-points of the sides of a triangle are (5, 7, 11), (0, 8, 5) and (2, 3, -1). Finds its vertices.
Answer:
Let the image of point P(3, 8) be Q(a, b) with respect to line AB which is given by
x + 3y – 7 = 0 ………(i)
The mid-point of P and Q is \(\left(\frac{3+a}{2}, \frac{8+b}{2}\right)\), which lies on line (i).
(i) ⇒ \(\frac{3+a}{2}+3 \cdot \frac{8+b}{2}-7\) = 0
⇒ 3 + a + 24 + 3b – 14 = 0
⇒ a + 3b + 13 = 0 ……..(ii)
Now, slope of AB = \(-\frac{1}{3}\)
and slope of PQ = \(\frac{b-8}{a-3}\)
∴ \(-\frac{1}{3} \times \frac{b-8}{a-3}\) = -1
⇒ 3a – b – 1 = 0 ……(iii)
Solving (ii) and (iii), we arrive at a = -1, b = -4
∴ Image of (3, 8) is (-1, -4).
OR
Let vertices of the ΔABC are
A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3), D, E & F are the midpoints of BC, CA & AB respectively.
Equations (i) and (iv),
x1 + 2x2 + x3 = 14 …….(x)
Equations (ii) and (v),
y1 + 2y2 + y3 = 20
Equations (iii) and (v),
z1 + 2z2 + z3 = 20
Equations (vii) and (x),
2x2 = 14 ⇒ x2 = 7
x2 = 7, then x3 = 10 – 7 = 3
x3 = 3, then x1 = -3
x1 = -3, x2 = 7, x3 = 3
Equations (viii) and (xi),
2y2 = 4 ⇒ y2 = 2
y2 = 2, then y1 = 4
y1 = 4, then y3 = 12
y1 = 4, y2 = 2, y3 = 12
Equations (ix) and (xii),
2z2 = 10 ⇒ z2 = 5
z2 = 5, then, z1 = -7
z1 = -7, then, z3 = 17
z3 = -7, z2 = 6, z3 = 17
The vertices are A(-3, 4, -7), B(7, 2, 5) and C(3, 12, 17).
Question 31.
Prove that: \(\frac{\tan 5 \theta+\tan 3 \theta}{\tan 5 \theta-\tan 3 \theta}\) = 4 cos 2θ cos 4θ
Answer:
Section-D
[This section comprises long answer type questions (LA) of 5 marks each]
Question 32.
Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
Answer:
L.H.S. = sin x + sin 3x + sin 5x + sin 7x
= (sin 7x + sin x) + (sin 5x + sin 3x)
= 2 sin(\(\frac{7 x+x}{2}\)) cos(\(\frac{7 x-x}{2}\)) + 2 sin(\(\frac{5 x+3 x}{2}\)) cos(\(\frac{5 x-3 x}{2}\))
[∵ sin C + sin D = 2 sin(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]
= 2 sin 4x cos 3x + 2 sin 4x cos x
= 2 sin 4x (cos 3x + cos x)
= 2 sin 4x 2 cos(\(\frac{3 x+x}{2}\)) cos(\(\frac{3 x-x}{2}\))
[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]
= 4 sin 4x cos 2x cos x
= R.H.S.
Hence proved.
Question 33.
The sum of interior angles of a triangle is 180°. Show that the sum of the interior angles of polygons with 3, 4, 5, 6,… sides from an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
OR
A committee of 6 is to be chosen from 10 men and 7 women, so as to contain at least 3 men and 2 women. In how many different ways can this be done, if two particular women refuse to serve on the same committee?
Answer:
We know that, sum of interior angles of a polygon of side n = (2n – 4) × 90° = (n – 2) × 180°
Sum of interior angles of a polygon with sides 3 is 180°.
Sum of interior angles of polygon with side 4 = (4 – 2) × 180° = 360°
Similarly, sum of interior angles of polygon with side 5, 6, 7,….. are 540°, 720°, 900°,……
The series will be 180°, 360°, 540°, 720°, 900°,……
Here, n = 180° and d = 360° – 180° = 180°
Since, common difference is same between two consecutive terms of the series. So, it form an A.P.
We have to find the sum of interior angles of a 21 sides polygon.
It means, we have to find the 19th term of the above series.
∴ a19 = a + (19 – 1)d
= 180 + 18 × 180
= 3420
OR
Total number of men = 10
and total number of women = 7
We have to form a committee containing at least 3 men and 2 women.
Total number of ways = 10C3 × 7C3 + 10C4 × 7C2
= 120 × 350 + 210 × 21
= 8610
If two particular women to be always there.
∴ Number of ways two particular women to be always there = 10C4 × 5C0 + 10C3 × 5C1
Total number of committee when two particular women are never together = Total number of ways – Number of ways two particular women to be always there
= (10C3 × 7C3 + 10C4 × 7C2) – (10C4 × 5C0 + 10C3 × 5C1)
= (120 × 35 + 210 × 21) – (210 + 120 × 5)
= 4200 + 4410 – (210 + 600)
= 8610 – 810
= 7800
Question 34.
Calculate the mean deviation about the mean of the set first n natural numbers when n is an even number.
Answer:
Let n = 2k ⇒ k = \(\frac{n}{2}\)
∴ \(\bar{x}=\frac{1+2+3+\ldots+n}{n}\)
= \(\frac{n(n+1)}{2 n}\)
= \(\frac{n+1}{2}\)
= \(\frac{2 k+1}{2}\)
Question 35.
An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 in wide at the base. How wide is it 2 m from the vertex of the parabola?
OR
A ray of light passing through the point (1, 2) reflects on the X-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
Answer:
The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the positive-axis. This can be diagrammatically represented as
The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).
It can be clearly seen that the parabola passes through point (\(\frac{5}{2}\), 10)
\(\left(\frac{5}{2}\right)^2\) = 4a(10)
⇒ a = \(\frac{25}{4 \times 4 \times 10}=\frac{5}{32}\)
Therefore, the arch is in the form of a parabola whose equation is x2 = \(\frac{5}{8}\)y.
When y = 2m,
x2 = \(\frac{5}{8}\) × 2
⇒ x2 = \(\frac{5}{4}\)
⇒ x = \(\sqrt{\frac{5}{4}}\) m
∴ AB = 2 × \(\sqrt{\frac{5}{4}}\) m
= 2 × 1.118 m approx.)
= 2.23 m (approx.)
Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23 m.
OR
Let the coordinates of point A be (a, 0).
Draw a line (AL) perpendicular to the x-axis.
We know that angle of incidence is equal to angle of reflection.
Hence, let ∠BAL = ∠CAL = φ
Let ∠CAX = θ
∴ ∠OAB = 180° – (θ + 2φ)
= 180° – [θ + 2(90° – θ)]
= 180° – θ – 180° + 2θ
= θ
∴ ∠BAX = 180° – θ
Thus, the coordinates of point A are (\(\frac{13}{5}\), 0)
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 sub parts of 2 marks each.
Question 36.
Read the following passage and answer the questions given below:
Two friends Sumit and Pawan have just came to know about complex numbers. They themselves created a game, for which they have prepared a board having an argand plane drawn on it. (as shown in the figure).
They have taken two dice of different colours i.e., red and blue. And they have decided to take the number on red die as real part and that on blue die as imaginary part. Both have thrown that pair of dice and marked the number by each represented as A and B. Now they are asking each other some questions.
(i) Which complex numbers are represented by the points A and B?
(ii) Write the conjugates of the complex numbers at the points A and B respectively.
(iii) Write the multiplicative inverse of the complex number at A.
OR
Write the sum of A and B.
Answer:
(i) Position of A is 5 + i4 and position of B is 7 + i2.
(ii) z = a + ib
⇒ \(\bar{z}\) = a – ib
So, \(\overline{5+i 4}\) = 5 – i4 And \(\overline{7+i 2}\) = 7 – i2
OR
A + B = (5 + 7) + i(4 + 2) = 12 + i6
Question 37.
Read the following passage and answer the questions given below:
The derivative of y with respect to x is the change in y with respect to a change in x. The derivative of f(x) at x0 is given by
\(f\left(x_0\right)=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}\)
(i) Find the derivative of sin x with respect to x.
(ii) Find the derivative of cos x with respect to x.
(iii) Find the derivative of tan x with respect to x.
OR
If y = \(\frac{x}{\tan x}\), then \(\frac{d y}{d x}\) = ________
Answer:
(i) The derivative of sin x with respect to x is cos x.
(ii) \(\frac{d}{d x}\)(cos x) = -sin x
Question 38.
Read the following passage and answer the questions given below:
A child’s game has 8 triangles of which 3 are blue and rest are red and 10 squares of which 6 are blue and rest are red. One piece is lost at random.
(i) How many triangles are of red colour and how many squares are of red colour?
(ii) Find the probability that lost piece is triangle.
Answer:
(i) Since, total no. of triangles = 8
Triangles with blue colour = 3
Triangles with red colour = 8 – 3 = 5
and Total no. of squares = 10
Squares with blue colour = 6
Squares with red colour = 10 – 6 = 4
(ii) Number of favourable outcomes for the event that lost figure is triangle,
i.e., F(E) = 8
Total figures (square and triangle) = 8 + 10 = 18
i.e., T(E) = 18
Probability (getting a triangle), P(E) = \(\frac{F(E)}{T(E)}\)
= \(\frac{8}{18}\)
= \(\frac{4}{9}\)