Practice Paper
TERM II (2021 – 2022)
Class – XI
Mathematics (041)
Time: 2 hours Maximum Marks: 40
General Instructions:
1. This question paper contains three sections – A, B and C. Each part is compulsory.
2. Section – A has 6 short answer type (SA1) questions of 2 marks each.
3. Section – B has 4 short answer type (SA2) questions of 3 marks each.
4. Section – C has 4 long answer type questions (LA) of 4 marks each.
5. There is an internal choice in some of the questions.
6. Q14 is a case-based problem.
SECTION – A
1. If the angles of a triangle are in the ratio 3: 4: 5, then find the smallest angle in degree and the greatest angle in radians.
OR
If the arcs of the same lengths in two circles subtend angles 65° and 110° at the centre, find the ratio of their radii.
2. Prove that cos 20° cos 40° cos 80° = 1/8.
3. Solve the linear inequality:
4. If the letters of the word ‘EXAMINATION’ are arranged in all possible ways as listed in dictionary. How many words are there in the list in which the first word start with ‘A’?
Total number of letters in the word ‘EXAMINATION’ are 11 with 2 A’s, 2I ’s and 2N ’s. Now when word starts with A i.e., A is fixed at the beginning of the word. Then we have to arrange the remaining 10 letters with 2I’s, 2N’s. Then words starting with A = 10/ (2! 2!) = 907200 ∴ Required number of words = 907200
5. Differentiate tan2x using first principle.
6. From a group of 2 boys and 3 girls, two children are selected at random. Consider the following events:
(i) A: Event that both the selected children are girls
(ii) B: Event that the selected group consists of one boy and one girl
(iii) C: Event that at least one boy is selected
Which pairs of events are mutually exclusive?
Let us name the boys as B1 and B2, and the girls as G1, G2 and G3.
Then, S = {B1B2, B1G1, B1G2, B1G3, B2G1, B2G2, B2G3, G1G2, G1G3, G2G3}.
We have, (i) A = {G1G2, G1G3, G2G3}
(ii) B = {B1G1, B1G2, B1G3, B2G1, B2G2, B2G3}
(iii) C = {B1G1, B1G2, B1G3, B2G1, B2G2, B2G3, B1B2}
Clearly, A ∩ B = φ and A ∩ C = φ.
Hence, (A, B) and (A, C) are mutually exclusive events.
SECTION – B
7. Solve the following system of inequalities graphically: x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, x – y ≥ 1
Converting the given inequations into equations, we get
x – 2y = 3 …(i)
3x + 4y = 12 …(ii)
x – y = 1 …(iii)
x = 0 …(iv)
Now, we draw the graphs of (i), (ii), (iii) and (iv) as shown below.
So, the shaded region including all the points on the lines represents the solution set of the given system of linear inequalities.
8. Find the equation of the parabola whose focus is at (–1, –2) and the directrix is the line x – 2y + 3 = 0.
Let P(x, y) be any point on the parabola whose focus is S(–1, –2) and the directrix x – 2y + 3 = 0. Draw PM perpendicular from P(x, y) on the directrix x – 2y + 3 = 0.
OR
Find the equation of ellipse with centre at origin, major axis along x-axis, foci (± 2, 0) and passing through the point (2, 3).
9. Find the equation of the circle which passes through the points (2, –2) and (3, 4) and whose centre lies on x + y = 1.
Let circle be (x – h)2 + (y – k)2 = r2 …(i)
Circle (i) passes through the points (2, –2) and (3, 4).
∴ (2 – h) 2 + (–2 – k) 2 = r 2 …(ii)
and (3 – h) 2 + (4 – k) 2 = r 2 …(iii)
Also centre (h, k) lies on x + y = 1
∴ h + k = 1 …(iv)
From (ii) and (iii), we get
(2 – h) 2 + (–2 – k) 2 = (3 – h) 2 + (4 – k) 2
⇒ 4 – 4h + h 2 + 4 + 4k + k 2 = 9 – 6h + h 2 + 16 – 8k + k 2
⇒ 2h + 12k = 17 …(v)
Solving (iv) and (v), we get, h = −1/2, k = 3/2
Substituting for h, k in (iii), we get
10. Find the ratio in which YZ-plane divides the line segment joining points (–2, 4, 7) and (3, –5, 8). Also, find the coordinates of the point of intersection.
Let point R(x, y, z) divides the join of (–2, 4, 7) and (3, –5, 8) in the ratio k : 1, then point of division is
OR
If the origin is the centroid of the triangle PQR with vertices P(2a, 2, 6), Q(–4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.
Vertices of ∆PQR are P(2a, 2, 6), Q(–4, 3b, –10) and R(8, 14, 2c).
SECTION – C
11. 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 girl?
A team consisting no girl can be selected 7C5 ways i.e., (7 × 6)/(2 x 1) = 21 ways
(ii) Atleast one boy and one girl?
Number of ways of selecting at least one boy and one girl
= 7C1 × 4C4 + 7C2 × 4C3 + 7C3 × 4C2 + 7C4 × 4C1
= 7 + 84 + 210 + 140 = 441
(iii) Atleast 3 girls?
A team having alteast 3 girls will consist of:
3 girls, 2 boys; 4 girls, 1 boy.
∴ Required no. of ways = 4C3 × 7C2 + 4C4 × 7C1
= 4 × 21 + 1 × 7 = 91 ways.
OR
If nCr : nCr + 1 = 1 : 2 and nCr + 1 : nCr + 2 = 2 : 3, then find the values of n and r.
12. If A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that
(i) cos A + cos B + cos C + cos D = 0
: (i) Since A, B, C, D are the angles of a cyclic quadrilateral, we have
A + C = 180° and B + D = 180°
⇒ cos A = cos(180° – C) and cos B = cos(180°– D)
⇒ cos A = –cos C …(i)
and cos B = –cos D …(ii) [∵ cos(180° – θ) = – cos θ]
Addition. (i) & (ii), we get
cos A + cos B = –(cos C + cos D)
⇒ cos A + cos B + cos C + cos D = 0
(ii) cos(180° + A) + cos(180° + B) + cos(180° + C) – sin(90° + D) = 0.
L.H.S. = –cos A – cos B – cos C – cos D [∵ cos(180° + θ) = – cos θ]
= –cos A – cos B – cos(180° – A) – cos(180° – B)
= –cos A – cos B – (–cos A) – (–cos B)
= –cos A – cos B + cos A + cos B = 0 = R.H.S
13. Find the derivative of
(i) (6x3 + 9x)(5x + 10)
(ii) (5x + 4)/(x – 3)
CASE BASED
14. Nishtha and Naira are sisters. They purchased two dice of different colour. Nishtha selected the red coloured dice and Naira selected the black coloured dice. They and were playing with two dice. They decided to threw both the dice simultaneously and note down the sum of numbers which come up on two dice.
(i) Find the probability that the sum is even.
Let A be event of the sum is even
⇒ A = {(1,1), (2,2), (1,3), (1,5), (2,4) (2,6)(3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)}
Thus P(A) = 18/36 = 1/2
(ii) Find the probability that the sum is multiple of 3.
Let B be event of the sum to be multiple of 3 ⇒ B = {(1,2), (2,1), (1,5), (5,1), (2,4), (4,2), (3,3), (3,6), (6,3), (4,5), (5,4), (6,6)}
Thus P(B) = 12/36 =1/3