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Maths — Practice Paper · Set 1
Part I — Multiple Choice Questions 15 × 1 = 15
Choose the correct answer. (Answer all questions.)
1.An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C, and D. Check whether the following assignments of probability are permissible. (i) P(a) = 0.15, P(b) = 0.30, P(c) = 0.43, P(d) = 0.12[1]
2.If A and B are two independent events such that P(A ∪ B) = 0.6, P(A) = 0.2, find p(B).[1]
3.Find the condition that one of the roots of ax 2 + bx + c may be (a) negative of the other (b) thrice the other (c) reciprocal of the other.[1]
4.A student appears in an objective test which contain 5 multiple choice, questions. Each question has 4 choices, out of which one correct answer. (i) What is the maximum number of different answers can the students give? (ii) How,will the answer change if each question may have more than one correct answer ?[1]
5.A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time shown in the following table.(a) Draw a graph showing the results.(b) Find the equation relating the length of the spring to the weight on it.(c) What is the actual length of the spring?(d) If the spring stretches to 9 cm long, how much weight should be added?(e) How long will the spring be when 6 kilograms of weight on it?[1]
6.A test consists of 10 multiple choice questions. In how many ways can the test be answered if (i) Each question has four choices ? (ii) The first four questions have three choices and the remaining have five choices? (iii) Question number n has n + 1 choices ?[1]
7.Let A and B be two symmetrh matrices of same order. T hen which one of the following statement is not true? (1) A + B is a symmetric matrix (2) AB is a symmetric matrix (3) AB = (BA) T (4) A T B = MI T[1]
8.If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are the position vectors of three collinear points, then which of the following is true? (1) \(\overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\) (2) \(2 \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\) (3) \(\overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}}\) (4) \(4 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=0\)[1]
9.Which of the following equation is the locus of (at 2, 2at) (1) \(\frac{x^{2}}{\mathbf{a}^{2}}-\frac{\mathbf{y}^{2}}{\mathbf{b}^{2}}\) = 1 (2) \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 (3) x 2 + y 2 = a 2 (4) y 2 = 4ax[1]
10.If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct? (1) P(A/B) = \(\frac{\mathbf{P}(\mathbf{A})}{\mathbf{P}(\mathbf{B})}\) (2) P(A/B) < P(A) (3) P(A/B) ≥ P(A) (4) P(A/B) >P(A)[1]
11.Which of the following is not true? (1) sin θ = – \(\frac{3}{4}\) (2) cos θ = – 1 (3) tan θ = 25 (4) sec θ = \(\frac{1}{4}\)[1]
12.If A and B are any two events, then the probability that exactly one of them occur is (1 P(A ∪ B̅) + P(A̅ ∪ B) (2) P(A ∩ B̅) + P(A̅ ∩ B) (3) P(A) + P(B) – P(A ∩ B) (4) P(A) + P(B) + 2P(A ∩ B)[1]
13.Let R be the set of all real numbers. Consider the following subsets of the plane R × R: S = { (x, y): y = x + 1 and 0 < x < 27 and T = {(x, y): x – y is an integer} Then which of the following is true? (1) T is an equivalence relation but S Is not an equivalence relation (2) Neither S nor T is an equivalence relation (3) Both S and T are equivalence relation (4) S is an equivalence relation but T is not an equivalence relation.[1]
14.If P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8, find P(A/B) and P(A ∪ B).[1]
15.The relation R defined on a set A = {0, -1, 1, 2} by x R y if |x 2 + y 2 | ≤ 2, then which one of the following is true. (1) R = {(0,0), (0,-1), (0,1), (-1,0), (-1,1), (1,2), (1,0)} (2) R -1 = {(0, 0), (0, -1), (0, 1), (-1, 0), (1, 0)} (3) Domain of R is {0,- 1, 1, 2} (4) Range of R is {0, -1, 1}[1]
Part II — Short Answer Questions 14 × 2 = 28
Answer briefly. (Answer all questions.)
16.If A + 1 = \(\left[ \begin{matrix} 3 & -2 \\ 4 & 1 \end{matrix} \right]\), then (A + I) (A – I) is equal to (1) (2) (3) (4)[2]
17.If y = cos (sin x 2 ), then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) at x = \(\sqrt{\frac{\pi}{2}}\) is (1) – 2 (2) 2 (3) – 2 \(\sqrt{\frac{\pi}{2}}\) (4) 0[2]
18.The coefficient of x 8 y 12 in the expansion of (2x + 3y) 20 is (1) 0 (2) 2 8 3 12 (3) 2 8 3 12 + 2 12 3 8 (4) 20C 8 2 8 3 12[2]
19.y = cos x – 2 tan x[2]
20.If tan α and tan β are the roots of x 2 + ax + b = 0 then \(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}\) is equal to (1) \(\frac{\mathbf{b}}{\mathbf{a}}\) (2) \(\frac{\mathbf{a}}{\mathbf{b}}\) (3) –\(\frac{\mathbf{a}}{\mathbf{b}}\) (4) –\(\frac{\mathbf{b}}{\mathbf{a}}\)[2]
21.In a ∆ ABC, ∠A = 60°. Prove that b + c = 2a cos \(\left(\frac{\mathbf{B}-\mathbf{C}}{2}\right)\)[2]
22.(a) Find the left and right limits of[2]
23.\(\frac{d}{d x}\) (e x + 5 log x ) is (1) e x. x 4 (x + 5) (2) e x. x (x + 5) (3) e x + \(\frac{5}{x}\) (4) e x – \(\frac{5}{x}\)[2]
24.Show that the relation xy = – 2 is a function for a suitable domain. Find the domain and the range of the function.[2]
25.Show that tan 75° + cot 75° = 4[2]
26.x = a cos t; y = a sin 3 t[2]
27.(i) \(\frac{1}{4-x^{2}}\)[2]
28.If \(\vec{a}\) and \(\vec{b}\) are two vectors of magnitude 2 and inclined at an angle 60°, then the angle between \(\vec{a}\) and \(\vec{a}\) + \(\vec{b}\) is (1) 30° (2) 60° (3) 45° (4) 90°[2]
29.(i) \(\frac{1}{\sqrt{1-(4 x)^{2}}}\) (ii) \(\frac{1}{\sqrt{1-81 x^{2}}}\) (iii) \(\frac{1}{1+36 x^{2}}\)[2]
Part III — Long Answer Questions 10 × 5 = 50
Answer in detail. (Answer all questions.)
30.y = sin 2 (cos kx)[5]
31.State how continuity is destroyed at x = x 0 for each of the following graphs.[5]
32.If A and B are square matrices of order 3 such that |A| = -1 and |B| = 3, find the value of |3 AB|.[5]
33.Solve x 2 + 3x – 2 ≥ 0[5]
34.For what values of k does the equation 12x 2 + 2kxy + 2y 2 +11x – 5y + 2 = 0 represent two straight lines.[5]
35.A single card is drawn from a pack of 52 cards. What is the probability that (i) the card is an ace or a king (ii) the card w11 be 6 or smaller (iii) the card is either a queen or 9?[5]
36.In a ∆ ABC, if cos c = \(\frac{\sin \mathbf{A}}{2 \sin B}\) show that the triangle is isosceles.[5]
37.How many three-digit numbers, which are divisible by 5, can be formed using the digits 0, 1, 2, 3, 4, 5 if (i) Repetition of digits is not allowed? (ii) Repetition of digits is allowed?[5]
38.How many numbers are there between 1 and 1000 (both inclusive) which are divisible neither by 2 nor by 5 ?[5]
39.How many strings can be formed using the letters of the word LOTUS if the word (i) either start with L or end with S? (ii) neither starts with L nor ends with S?[5]
🔑 Show Answer Key — Set 1
- 1. P(A) = 0.15, P(B) = 0.30, P(C) = 0.43, P(D) = 0.12 Now P(A) + P(B) + P(C) + P(D) = 0.15 + 0.30 + 0.43 + 0.12 = 1 0.15 + 0.30 + 0.43 + 0.12 = 1 ∴ The assignment of probability is permissible. (ii) P (A) = 0.22, P (B) = 0.38, P (C) = 0.16, P (D) = 0.34 Given that P (A) = 0.22 ≥ 0, P (B) = 0.38 ≥ 0, P(C) = 0.16 ≥ 0, P (D) = 0.34 ≥ 0 P(S) = P (A) + P(B) + P(C) + P(D) = 0.22 + 0.38 + 0.16 + 0.34 = 1.1 > 1 Therefore the assignment of probability isn’t permissible (iii) P(A) = \(\frac{2}{5}\), P(B) = \(\frac{3}{5}\), P(C) = – \(\frac{1}{5}\), P(D) = \(\frac{1}{5}\) P(A) = \(\frac{2}{5}\), P(B) = \(\frac{3}{5}\), P(C) = \(-\frac{1}{5}\), P(D) = \(\frac{1}{5}\) P(C) = \(-\frac{1}{5}\) which is not…
- 2. Given A and B are independent. ⇒ P(A ∪ B) = P(A).P(B) Here P(A ∪ B) = 0.6 and P(A) = 0.2 To find P(B): Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) (i.e.,) P(A ∪ B) = P(A) + P(B) – P(A). P(B) (i.e.,) 0.6 = 0.2 + P(B) (1 – 0.2) P(B) (0.8) = 0.4 ⇒ P(B) = \(\frac{0.4}{0.8}=\frac{4}{8}=\frac{1}{2}\) = 0.5
- 3. The given quadratic equation is ax 2 + bx + c = 0 ——- (1) Let α and β be the roots of the equation (1) then Sum of the roots α + β = ——- (2) Product of the roots αβ = ——- (3) (a) Given one root is the negative of the other β = – α (2) ⇒ α + (-α) = – \(\frac{b}{a}\) 0 = – \(\frac{b}{a}\) ⇒ b = 0 (3) ⇒ α(-α) = \(\frac{c}{a}\) – α 2 = \(\frac{c}{a}\) Hence the required condition is b = 0 (b) Given that one root is thrice the other β = 3α When is the required condition? (c) One root is reciprocal of the other When is the required condition?
- 4. (i) What is the maximum number of different answers can the students give? Selecting a correct answer from the 4 answers can be done in 4 ways. Total number of questions = 5 So they can be answered in 45 ways (ii) How will the answer change if each question may have more than one correct answers? Since each question may have more than one correct answer, each question can have the possibilities 1, 2, 3 or 4 correct answers. ∴ Number of ways of answering each question = 4C 1 + 4C 2 + 4C 3 + 4C 4 = 4 + 6 + 4 + 1 = 15 Thus, the answer will change as 15 5 (i.e, Total number of ways of answering five questions).
- 5. Choose the weight along the x-axis and Length along the y-axis. (a) (b) The points are(2, 3), (4, 4), (5, 4.5), (8, 6) The relation connecting weight and Length is the equation of the straight line joining the points (2, 3) and (4, 4) x – 2 = 2(y – 3) x – 2 = 2y – 6 x – 2y + 6 – 2 = 0 x – 2y + 4 = 0 —– (1) which the required relation connecting weight and length. (c) To find the actual length of the spring, put weight x = 0 in equation (1) 0 – 2y + 4 = 0 ⇒ 2y = 4 ⇒ y = 2 ∴ The actual length of the spring is 2 cm. (d) If the spring stretch to 9 cm long, To find the required weight, put y = 9, in equation (1) (1) ⇒ x – 2 (9) + 4 = 0 x – 18 +4 = 0 ⇒ x = 14 Weight to be added is 14 kg. (e) Ne…
- 6. Each question has 4 choices. So each question can be answered in 4 ways. Number of Questions = 10 So they can be answered in 410 ways (ii) The first four questions have 3 choices. So they can be answered in 3 4 ways. The remaining 6 questions have 5 choices. So they can be answered in 5 6 ways. So all 10 questions can be answered in 3 4 × 5 6 ways. (iii) Given question n has n + 1 choices
- 7. (2) AB is a symmetric matrix Explaination: Given A and B are two symmetric matrices of the same order. A = A T, B = B T (1)(A+B) T = A T + B T = A + B A + B is symmetric. (2) (AB) T = B T A T BA Thus (AB) T ≠ AB Hence, AB is not symmetric. (3) AB = (BA) T = A T B T = AB Statement is true. (4) A T B = AB T Since A T = A B = B T Statement is true.
- 8. (2) \(2 \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\) Explaination:
- 9. (4) y 2 = 4ax Explaination: y 2 = 4ax ⇒ Equation that satisfies the given point (at 2, 2at)
- 10. (3) P(A/B) ≥ P(A) Explaination: Given A and B are two events such that A ⊆ B and P(B) ≠ 0 then P(A/B) ≥ P(A)
- 11. (4) sec θ = \(\frac{1}{4}\) Explaination: We know |cos θ| < 1 sec θ = \(\frac{1}{4}\) ⇒ \(\frac{1}{\cos \theta}\) = \(\frac{1}{4}\) ⇒ cos θ = 4 which is not possible.
- 12. (2) P(A ∩ B̅) + P(A̅ ∩ B) Explaination: Let A and B be an two events The probability that exactly one of them occur is = P(A ∩ B̅) + P(A̅ ∩ B)
- 13. (1) T is an equivalence relation but S Is not an equivalence relation Explanation: (0, 1), (1, 2) it is not an equivalence relation T is an equivalence relation
- 14. Given P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8 P(A ∩ B) = P(B/A) P(A) Substituting in equation (1) we get P(A/B) = 0.5 P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ………. (2) P(A ∩ B) = P(A/B). P(B) = 0.5 × 0.8 P(A ∩ B) = 0.40 (2) ⇒ P(A ∪ B) = 0.5 + 0.8 – 0.40 = 1.3 – 0.40 P(A ∪ B) = 0.90
- 15. (4) Range of R is {0, -1, 1} Explaination: A= {0, -1, 1, 2} |x 2 + y 2 | ≤ 2 The values of x and y can be 0, -1 or 1 So range = {0, -1, 1}
- 16. (1) Explaination:
- 17. (4) 0 Explaination:
- 18. (4) 20C 8 2 8 3 12 Explaination:
- 19. y = cos x – 2 tan x \(\frac{d y}{d x}\) = – sin x – 2 sec 2 x
- 20. (3) –\(\frac{\mathbf{a}}{\mathbf{b}}\) Explaination: x 2 + ax + b = 0 Given tan α and tan β are the roots of the above equation. Then
- 21. Given ∠A = 60° A + B + C = 180° 60° + B + C = 180° B + C = 180° – 60° = 120°
- 22. (b) f(x) = tan x at x = \(\frac{\pi}{2}\) To find the left limit of f(x) at x = \(\frac{\pi}{2}\) Evaluate the following limits.
- 23. (1) e x. x 4 (x + 5) Explaination:
- 24. xy = – 2 ⇒ y = -2/x which is a function The domain is (-∞, 0) ∪ (0, ∞) and range is R – {0}
- 25. tan 75° = tan (45° + 30°)
- 26. x = a cos t, y = a sin 3 t
- 27. (ii) \(\frac{1}{25-4 x^{2}}\) (iii) \(\frac{1}{9 x^{2}-4}\)
- 28. (1) 30° Explaination:
- 29. (i) \(\frac{1}{\sqrt{1-(4 x)^{2}}}\) (ii) \(\frac{1}{\sqrt{1-81 x^{2}}}\) (iii) \(\frac{1}{1+36 x^{2}}\)
- 30. y = sin 2 (cos kx) y = f(g(x) \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)] \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 2 sin (cos kx) × cos (cos kx) × -sin kx × k × 1 \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sin (2 cos kx) × -k sin kx \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = -k sin kx. sin (2 cos kx)
- 31. (a) The left – hand limit and right hand limit does not coincide at x = x 0 (b) The function f(x) is not defined at x = x 0 and hence the continuity is destroyed at x = x 0 (c) The limit of f(x) does not exist at x = x 0 (d) The left hand limit and right – hand limit does not coincide at x = x 0
- 32. Given |A| = -1: |B| = 3 Given A and B are square matrices of order 3. ∴ |kAB| = k 3 |AB| Here k = 3 ∴ |3AB| = 3 3 |AB| = 27 |AB| = 27 (-1) (3) = -81
- 33. The given inequality is – x 2 + 3x – 2 ≥ 0 x 2 – 3x + 2 < 0 ——– (1) x 2 – 3x + 2 = x 2 – 2x – x + 2 = x(x – 2) – 1(x – 2) x 2 – 3x + 2 = (x – 1) (x – 2) ——— (2) The critical numbers are x – 1 = 0 or x – 2 = 0 The critical numbers are x = 1 or x = 2 Divide the number line into three intervals (- ∞, 1), (1, 2) and (2, ∞). (i) (- ∞, 1) When x < 1 say x = 0 The factor x – 1 = 0 – 1 = – 1 < 0 and x – 2 = 0 – 2 = – 2 < 0 x – 1 < 0 and x – 2 < 0 ⇒ (x – 1)(x – 2) > 0 Using equation (2) x 2 – 3x + 2 > 0 ∴ The inequality x 2 – 3x + 2 ≤ 0 is not true in the interval (- ∞, 1 ) (ii) (1, 2) When x lies between 1 and 2 say x = \(\frac{3}{2}\) The factor x – 1 = \(\frac{3}{2}\) – 1 = \(\frac{1}{2}\) > 0…
- 34. The given equation of the pair of straight line is 12x 2 + 2kxy + 2y 2 + 11x – 5y + 2 = 0 ……… (1) Compare this equation with the equation ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 ………. (2) a = 12, 2h = 2k, b = 2, 2g = 11, 2f = – 5, c = 2, a =12, h = k, b = 2, g = \(\frac{11}{2}\), f = –\(\frac{5}{2}\), c = 2, The condition for a second degree equation in x and y to represent a pair of straight lines is abc + 2fgh – af 2 – bg 2 – ch 2 = 0 96 – 55k – 150 – 121 – 4k 2 = o – 4k 2 – 55k – 175 = 0 4k 2 + 55k + 175 = 0 ∴ The given equation represents a pair of straight lines when k = – 5 and k = \(\frac{-35}{4}\)
- 35. S be the sample space one card is drawn from a pack of 52 cards. ∴ n(S) = 52C 1 n(S) = 52 (i) The card is an ace or a king Let A be the event of getting an ace. n(A) = 4C 1 = 4 Let B be the event of getting a king. n(B) = 4C 1 = 4 P (getting an ace or a king) = P(A or B) = P(A ∪ B) = P(A) + P(B) (since A and B are mutually exclusive events, A ∩ B = Φ) (ii) The card will be 6 or smaller: Let A be the event of getting a number 6. ∴ n(A) = 4C 1 = 4 Let B be the event of getting numbers less than 6. n(B) = 16C 1 = 16 P (the card will be 6 or less than 6) = P(A or B) = P(A ∪ B) = P(A) + P(B) (since A and B are mutually exclusive events A ∩ B = Φ) (iii) The card is either a queen or 9? Let A be…
- 36. a 2 + b 2 – c 2 = a 2 b 2 – c 2 = 0 b 2 = c 2 ⇒ b = c Two sides of is ∆ ABC are equal. ∴ ∆ ABC is an isosceles triangle.
- 37. The given digits are 0, 1, 2, 3, 4, 5 To find the 3 – digit numbers formed by using the digits 0, 1, 2, 3, 4, 5 which are divisible by 5. (i)The repetition of digits are not allowed: Since the 3 – digit number is divisible by 5, the unit place can be filled in 2 ways using the digits 0 or 5 Case (i) When the unit place is filled with the digit 0. The hundreds place can be filled in 5 ways using the digits 1, 2, 3, 4, 5 and the ten’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 and the digit which is placed in hundred’s place. Therefore, by fundamental principle of multiplication, the number of 3 – digit numbers divisible by 5 is = 5 × 4 × 1 = 20 Case (ii) Wh…
- 38. From 1 to 1000, the numbers ÷ by 2 = 500 the number ÷ by 5 = 200 and the numbers ÷ by 10 = 100(5 × 2 = 10) So number ÷ by 2 or 5 = 500 + 200 – 100 = 600 Total numbers from 1 to 1000 = 1000 So the number of numbers which are ÷ neither by 2 nor by 5 = 1000 – 600 = 400 Three-digit numbers: The unit place can be filed in 4 ways using the digits 1, 3, 7, 9. Hundred’s place can be filled in 9 ways excluding 0. Ten’s place can be filled in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Therefore, the required number of 3 digit numbers neither divisible by 2 nor by 5 is = 9 × 10 × 4 = 360. There is only one 4 – digit number, but it is divisible by 2 and 5. Therefore, required numbers usin…
- 39. (i) Either starts with L or ends with S The first box is filled with the letter L. The second box can be filled with the remaining letters O, T, U, S in 4 ways. The third box can be filled with the remaining letters excluding L and the letter placed in box 2 in 3 ways. The fourth box can be filled with the remaining letters excluding L and the letters placed in a box – 2 and box – 3 in 2 ways. The fifth box can be filled with the remaining one letter excluding L and the letters placed in a box – 2 and box – 3, box – 4 in 1 way. Therefore, by fundamental principle of multiplication, the number of words start with L is = 1 × 4 × 3 × 2 × 1 = 24 Since the word ends with S, the fifth box can b…