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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.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]
2.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]
3.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]
4.Which of the following points lie on the locus of 3x 2 + 3y 2 – 8x – 12y + 17 = 0 (1) (0, 0) (2) (-2, 3) (3) (1, 2) (4) (0, – 1)[1]
5.Let A and B be two sets such that n (A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.[1]
6.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]
7.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]
8.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]
9.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]
10.If P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8, find P(A/B) and P(A ∪ B).[1]
11.Which one of the following is not true about the matrix \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) ?. (1) a scalar matrix (2) a diagonal matrix (3) an upper triangular matrix (4) a lower triangular matrix[1]
12.Simplify (a) (125) 2/3 (b) 16 -3/4 (c) (- 1000) -2/3 (d) (3 -6 ) 1/3 (e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)[1]
13.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]
14.If A and B are two independent events such that P(A ∪ B) = 0.6, P(A) = 0.2, find p(B).[1]
15.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]
Part II — Short Answer Questions 14 × 2 = 28
Answer briefly. (Answer all questions.)
16.In an ∆ ABC, prove the following, (i) a sin \(\left(\frac{\mathbf{A}}{2}+\mathbf{B}\right)\) = (b + c). sin \(\frac{\mathbf{A}}{2}\)[2]
17.If f(x) = x + 2, then f’ (f(x)) at x= 4 is (1) 8 (2) 1 (3) 4 (4) 5[2]
18.y = \(\sqrt[3]{1+x^{3}}\)[2]
19.One of the diagonals of parallelogram ABCD with \(\vec{a}\) and \(\vec{b}\) as adjacent sides is \(\vec{a}\) + \(\vec{b}\). The other diagonal BD is (1) \(\vec{a}\) – \(\vec{b}\) (2) \(\vec{b}\) – \(\vec{a}\) (3) \(\vec{a}\) + \(\vec{b}\) (4) \(\frac{\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}}{2}\)[2]
20.α β x α-1 e -β x α[2]
21.Write a brief description of the meaning of the notation \(\lim _{x \rightarrow 8}\) f(x) = 25[2]
22.y = sin (tan (\(\sqrt{\sin x}\)))[2]
23.If denotes the greatest integer less than or equal to the real number under consideration and – 1 ≤ x< 0, 0 ≤ y < 1, 1 ≤ z < 2, then the value of the determinant (1) (2) (3) (4)[2]
24.If the square of the matrix \(\left[ \begin{matrix} α & β \\ γ & -α \end{matrix} \right] \) is the unit matrix of order 2, then α, β, and γ should (1) 1 + α 2 + βγ = 0 (2) 1 – α 2 – βγ = 0 (3) 1 – α 2 + βγ = 0 (4) 1 + α 2 – βγ = 0[2]
25.If y = sin -1 x then find y”.[2]
26.cos -1 \(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)[2]
27.If \(\vec{a}\) and \(\vec{b}\) having same magnitude and angle between them is 60° and their scalar product \(\frac{1}{2}\) is then |\(\vec{a}\)| is (1) 2 (2) 3 (3) 7 (4) 1[2]
28.\(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)[2]
29.The value of the determinant of A = \(\left[ \begin{matrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{matrix} \right] \) is (1) -2 abc (2) abc (3) 0 (4) a 2 + b 2 + c 2[2]
Part III — Long Answer Questions 10 × 5 = 50
Answer in detail. (Answer all questions.)
30.A ∆ OPQ is formed by the pair of straight lines x 2 – 4xy + y 2 = 0 and the line PQ. The equation of PQ is x + y – 2 = 0, Find the equation of the median of the triangle ∆ OPQ drawn from the origin O[5]
31.The inverse of (1) (2) (3) (4)[5]
32.Prove that sin 75° – sin 15° = cos 105° + cos 15°[5]
33.Using the mathematical induction, show that for any natural number n,[5]
34.(i) 12 3 (ii) \(\frac{x^{24}}{x^{25}}\) (iii) e x[5]
35.Find the equation of a straight line parallel to 2x + 3y = 10 and which is such that the sum of its intercepts on the axes is 15.[5]
36.If Q is a point on the locus of x 2 + y 2 + 4x – 3y +7 = 0, then find the equation of locus of P which divides segment OQ externally in the ratio 3:4 where O is origin.[5]
37.Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is (1) 11 (2) 12 (3) 10 (4) 6[5]
38.The distance of an object falling is a function of time t and can be expressed as s ( t) = – 16t 2. Graph the function and determine if it is one – to – one.[5]
39.If A and B are symmetric matrices of order n, where (A ≠ B), then (1) A + B is skew – symmetric (2) A + B is symmetric (3) A + B is a diagonal matrix (4) A + B is a zero matrix[5]
🔑 Show Answer Key — Set 1
- 1. (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
- 2. (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.
- 3. (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)
- 4. (3) (1, 2) Explaination: The point that satisfies the given equations (0, 0) ⇒ 17 ≠ 0 (-2, 3) ⇒ 3 (4) + 3 (9) + 16 – 36 + 17 ≠ 0 (1, 2) ⇒ 3 + 3 (4) – 8 (1) – 12 (2) + 17 32 – 32 = 0, 0 = 0
- 5. n(A) = 3 ⇒ set A contains 3 elements n(B) = 2 ⇒ set B contains 2 elements – we are given (x, 1), (y, 2), (z, 1) are elements in A × B ⇒ A = {x, y, z} and B = {1, 2}
- 6. (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).
- 7. 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 possible (i.e.) for any event A, (0 ≤ P(A) ≤ 1) ∴ The assignment of probability is not permissible.
- 8. (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.
- 9. 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?
- 10. 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
- 11. (1) a scalar matrix Explaination: Let A = \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) (1) a scalar matrix – not true (2) a diagonal matrix – true (3) an upper triangular matrix – true (4) a lower triangular matrix – true [(1) A square matrix A = [a ij ] m × n is called a diagonal matrix if a ij = 0 whenever i ≠ j (2) A diagonal matrix whose entries along the principle diagonal are equal is called a scalar matrix. (3) A square matrix is said to be an upper triangular matrix if all the elements below the main diagonal are zero. (4) A square matrix is said to be a lower triangular matrix if all the elements above the main diagonal are zero.]
- 12. (a) (125) 2/3 (b) 16 -3/4 (c) (- 1000) -2/3 (d) (3 -6 ) 1/3 (e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)
- 13. 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) Next we find the length of the string when a weight of 6 kg is added. Put x = 6 in equation (1) 6 – 2y + 4 = 0 ⇒ 2y = 10 ⇒ y = 5cm ∴ Required length is 5 cm.
- 14. 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
- 15. 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
- 16. (ii) a (cos B + cos C) = 2(b + c) sin 2 \(\frac{\mathbf{A}}{2}\) (iii) (iv) (v)
- 17. (2) 1 Explaination: f(x) x + 2 f’ (f(x)) = \(\frac{\mathrm{d}}{\mathrm{d} x}\) (f(x)) = \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x + 2) = 1
- 18. y = \(\sqrt[3]{1+x^{3}}\) y = (1 + x 3 ) 1/3 [ y = f(g(x) \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
- 19. (2) \(\vec{b}\) – \(\vec{a}\) Explaination:
- 20. ∫α β x α-1 e -β x α Put β x α = u α β x α-1 dx = du
- 21. Given \(\lim _{x \rightarrow 8}\) f(x) = 25 By the definition of limit ∴ f(8 – ) = f(8 + ) = 25
- 22. y = sin (tan (\(\sqrt{\sin x}\))) y = f(g(x)) \(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = f'(g(x)). g'(x)]
- 23. (1) Explaination:
- 24. (1) 1 + α 2 + βγ = 0 Explaination: – α 2 – βγ = 1 α 2 + βγ + 1 = 0
- 25. y = sin -1 x
- 26. Let y = cos -1 \(\left(\frac{1-x^{2}}{1+x^{2}}\right)\) Put x = tan θ y = cos -1 (cos 2θ) y = 2θ y = 2 tan -1 x
- 27. (4) 1 Explaination:
- 28. Let y = \(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\) [1 – cos 2θ = 2 sin 2 θ and 1 + cos 2θ = 2 sin 2 θ]
- 29. (3) 0 Explaination: = 0 – a(0 – bc) – b (ac – 0) = abc – abc = 0
- 30. The equation of the given pair of lines is x 2 – 4xy + y 2 = 0 ……….. (1) The equation of the line PQ is x + y – 2 = 0 y = 2 – x ……….. (2) To find the coordinates of P and Q,. Solve equations (1) and (2) (1) ⇒ x 2 – 4x ( 2 – x) + ( 2 – x) 2 = 0 x 2 – 8x + 4x 2 + 4 – 4x + x 2 = 0 6x 2 – 12x + 4 = 0 3x 2 – 6x + 2 = 0 The midpoint of PQ is The equation of the median drawn from 0 is the equation of the line joining 0 (0, 0) and D (1, 1) ∴ The required equation is x = y
- 31. (1) Explaination: Let f(x) = x if x < 1 —— (1) Put y = x then (1) ⇒ f(x) = y ⇒ x = f -1 (y) if y < 1 ⇒ y = f -1 (y) if y < 1 ⇒ f -1 (x) = x if x < 1 Let f(x) = x 2 if 1 ≤ x ≤ 4 —– (2) Put y = x 2 ⇒ x = √y, if 1 ≤ y ≤ 16 then (2) ⇒ f(x) = y ⇒ x = f -1 (y) if 1 ≤ y ≤ 16 ⇒ √(y) = f -1 (y) if 1 ≤ y ≤ 16 ⇒ √x = f -1 (y) if 1 ≤ x ≤ 16 Let f(x) = 8√x if x > 4 ———– (3) Put y = 8√x ⇒ y 2 = 64 x ⇒ x = \(\frac{y^{2}}{64}\) if y>16 then (3) ⇒ f(x) = y ⇒ x = f -1 (y) if y > 16 ⇒ √y = f -1 (y) if y > 16 ⇒ \(\frac{y^{2}}{64}\) = f -1 (x) if x > 16
- 32. RHS = cos 105° + cos 15° = cos (90° + 15°) + cos (90° – 75°) = -sin 15° + sin 75° = sin 75° – sin 15° = LHS
- 33. Step 1: First, let us verify the result for n = 1 ∴ The given result is true for n = 1 Step 2: Let us assume the result for n = k Step 3: Let us prove the result for n = k + 1 Factorizing k 3 + 6k 2 + 9k + 4 f (k) = k 3 + 6k 2 + 9k + 4 f(- 1) = (- 1) 3 + 6(- 1) 2 + 9(- 1) + 4 f(- 1) = – 1 + 6 – 9 + 4 = 0 ∴ (k + 1) is a factor of f(k) k 3 + 6k 2 + 9k + 4 = (k + 1) (k 2 + 5k + 4). = (k + 1) (k 2 + 4k + k + 4) = (k + 1) [k(k + 4) + 1(k + 4)] = (k + 1) (k + 4) (k + 1) This implies P(k + 1) is true. ∴ Thus, we have proved the result for n = k + 1. Hence by the principle of mathematical induction, the result is true for all natural numbers n. is true for all natural numbers n
- 34. (i) 12 3 ∫12 3 dx = 12 3 ∫dx = 12 3 x + c (ii) \(\frac{x^{24}}{x^{25}}\) (iii) e x ∫e x dx = e x + c
- 35. The equation of the given line is 2x + 3y = 10 ………….. (1) The equation of any line parallel to (1) is 2x + 3y = k …………. (2) Given that the sum of the intercepts of the line (2) on the axes is 15 ∴ The equation of the required line is 2x + 3y = 18
- 36. Let Q be (a, b) lying on the locus x 2 + y 2 + 4x – 3y + 7 = 0 ∴ a 2 + b 2 + 4a – 3b + 7 = 0 Let the movable point P be (h, k) Given P divides OQ externally in the ratio 3: 4 Substituting in equation (1) we have h 2 + k 2 – 12h + 9k + 63 = 0 The locus of P(h, k) is obtained by replacing h by x and k by y. ∴ The required locus is x 2 + y 2 – 12x + 9y + 63 = 0
- 37. (2) 12 Explaination: Number of shake hands = 66 Let the number of persons = n First person shakes hands with the remaining n – 1 persons. Second person shakes hands with the remaining n – 2 persons. ∴ Total number of shake hands = (n – 1) + (n – 2) + …………. + 2 + 1 = \(\frac{(\mathrm{n}-1)(\mathrm{n}-1+1)}{2}\) Given 66 = \(\frac{(\mathrm{n}-1) \mathrm{n}}{2}\) n 2 – n = 132 n 2 – n – 132 = 0 (n – 12) (n + 11) = 0 n = 12 or n = – 11 n = – 11 is not possible. ∴ n = 12
- 38. Given s (t) = – 16t 2 s (t 1 ) = s (t 2 ) ⇒ – 16t 1 2 = – 16t 2 2 ⇒ t 1 2 = t 2 2 ⇒ ± t 1 = ± t 2 Since s (t 1 ) = s (t 1 ) 14 t 1 = t 2 ∴ The function s(t) is not one-one Graph of s(t) = – 16t 2 Take the time along x – axis and distance along y – axis.
- 39. (2) A + B is symmetric Explaination: Given A and B are symmetric matrices of order n. ∴ A T = A and B T = B A matrix A is skew symmetric if A T = – A (1)(A + B) T = A T + B T = A + B A + B is not skew symmetric. (2)(A + B) T = A T +B T = A + B ∴ A + B is symmetric. (3) A + B is a diagonal matrix is incorrect. (4) A + B is a zero matrix is incorrect.