Sub Code: 0081 NEB MODEL QUESTION (079/023) SOLUTION GRADE: XII Subject: Mathematics Candidates are required to give their answers in their own words as far as practicable. The figures in the margin indicate full marks. Time: 3 hrs. Full Marks: 75 Attempt all questions. Group "A" 1) 2) 3) 4) 5) What is an arrangement of the n natural numbers called? A) Induction B) Permutation C) Combination D) Expectation Let 1, 𝜔, 𝜔" be the cube roots of unity. Under which operation is the set A= {1, 𝜔, 𝜔" } closed? A) Addition B) Subtraction C) Multiplication D) None of these What is the domain of sin-1 𝑥 ? A) 𝑥 ≥ 1 or 𝑥 ≤ −1 B) (−∞, ∞) C) −1 < 𝑥 < 1 D) −𝟏 ≤ 𝒙 ≤ 𝟏 ABCD is a parallelogram, which one of the following represents area of the parallelogram? A) Magnitude of vector product of two vectors along AB and BD. B) Magnitude of vector product of two vectors along AB and DC. C) Magnitude of vector product of two vectors along AC and BC. D) Magnitude of vector product of two vectors along AB and AD. a2 – b2 , what is the equation of that conic section? a x2 y2 x2 y2 x2 y2 B) 2 + 2 = 1 C) 2 + 2 = 1 D) 2 + 2 = 1 a b a a b b If a conic section has eccentricity(e) = x2 y2 – =1 a2 b2 1 If cos Ө = – , for integer (n), what is the general value of Ө? 2 2p p p p A) 2np ± B) np + (–1) n c) np – D) np + 3 3 3 3 1 3 2 Let A and B be two dependent events. If P (A) = , P (B) = and P (A Ç B) = , what is the value 2 4 5 of P(A/B)? A) 6) 7) [ 11 × 1 = 11] A) equal to P(B/A) B) equal to P(A) C) less than P(A ∩ B) D) less than P(B/A) 8) The edge of a cube increases from 10 cm to 10.025 cm. What would be the approximate increment in volume? A) 103 cm3 B) 10.0253 cm3 C) 7.5187 cm3 D) 7.5 cm3 9) What is the integrating factor of the differential equation cos2x A) tan x B) etanx C) sec2x dy + y = 1? dx D) esecx 10) What is the number of solutions of the system of linear equations x + y = 5 and x + y = 7? A) One solution B) No solution C) Infinite solutions D) More than one solution 11) Forces P and Q are acting along ceiling and floor of a rectangular room. What is the nature of the forces? A) Like If yt = yt+1 – yt , then A) yt + 2 – yt +1 B) unlike 2 C) collinear OR D) parallel yt is equal to … B) yt+1 – yt C) yt+2 – yt+1 + yt D) yt+2 – 2yt+1 + yt Group "B" [ 8 × 5 = 40] 12) For any positive integer n, (a + x)n = C0 an+ C1 an-1 x+ C2 an-2 x2+………+ Cn xn a) How many terms are there in the expansion? [1] Ans: The total number of terms in the expansion is n + 1. b) Write the binomial coefficients in the expansion. [1] Ans: The binomial coefficients in the expansion are C0, C1, C2, …, Cn. c) Write the general term of the expansion. [1] n–r r Ans: The general term of the above expansion is tr + 1 = C(n, r) a x . d) Write the relation among C(n, r–1), C(n+1, r) and C(n, r). [1] Ans: The relation among C(n, r–1), C(n+1, r) and C(n, r) is C(n, r–1) + C(n, r) = C(n+1, r). e) What is the value of C0 + C1+ C2+…+Cn ? [1] n Ans: The value of C0 + C1+ C2 +…+ Cn = 2 . 1 13) a) Using the principle of mathematical induction, show that: 1 + 2 + 3 + …. + n < (2n + 1)2. [3] 8 1 Ans: Given statement is 1+ 2 + 3+ …. + n < (2n + 1)2. 8 For n = 1, 1 1 < (2 × 1 + 1)2 8 9 1 < , which is true 8 So, the given statement is true for n = 1. Now, suppose the given statement is true for n = k, where k Î ℕ. 1 i.e., 1+ 2 + 3+ …. + k < (2k + 1)2 … (1) 8 For n = k+1 1 1+ 2 + 3+ …. + k + (k + 1) < (2k + 1)2 + (k + 1) [ ∵ using (i) ] 8 1 = (4k2 + 4k + 1) + (k + 1) 8 4k2 + 4k + 1 + 8k + 8 = 8 2 4k + 12k + 9 = 8 1 = (2k + 3)2 8 1 i.e., 1+ 2 + 3+ …. + k + (k + 1) < (2(k + 1) + 1)2 8 This shows that given statement is true for n = k + 1, whenever it is true for n = k Î ℕ. So, from the principle of mathematical induction that, given statement is true for all n Î ℕ. b) Find the quadratic equation with rational coefficients whose one of the roots is 2 + 3 . [2] Ans: The irrational roots of quadratic equation with rational coefficients always appears in conjugate pair. So, the roots of quadratic equation are 2 + 3 and 2 – 3 . We have, the required quadratic equation is given by x2 – (sum of roots) x + (product of roots) = 0 or, x2 – (2 + 3 + 2 – 3 ) x + (2 + 3 ) (2 – 3 ) = 0 or, x2 – 4x + (4 – 1) = 0 \ x2 – 4x + 1 = 0 is the required quadratic equation. 14) a) Given y = sin-1x and y > 0, express cos y and tan y in terms of x. Ans: Given, y = sin–1x ⇒ sin y = x So, cos y = 1 – sin2 y = 1 – x2 sin y x And, tan y = = cos y 1 – x2 ® ® ® ® ® ® ® ® b) If a , b and c are any three vectors such that a × b = a × c for a ¹ (0,0), ® ® ® ® ® show b = c or a is parallel to b – c . ® ® ® ® Ans: Here, a , b and c are three vectors and a ¹ 0 and ® ® ® ® a × b = a × c ® ® ® ® or, a × b – a × c =0 ® ® ® or, a ×(b – c )=0 Ù ® ® ® or, | a | | b – c | sin q n = 0 ® ® Ù ® or, | b – c | sin q = 0 [∵ as n , a ¹ 0] ® ® ® ® Either, | b – c | ⇒ b = c Or, sin q = 0 ⇒ q = 0 or p. ® ® ® ® ® So, this this proves that either b = c or a is parallel to b – c . [3] [2] 15) The price in Rupees (X) and demand in unit (Y) of 6 days of a week is given as: [5] X 10 12 13 12 16 15 Y 40 38 43 45 37 43 Calculate the Pearson’s coefficient of correlation and the regression coefficient of X on Y. Ans: Calculation of Karl Pearson's correlation coefficient and regression coefficient of X on Y X Y X2 Y2 XY 10 12 13 12 16 15 40 38 43 45 37 43 100 1600 400 144 169 144 256 225 1444 1849 2025 1369 1849 456 559 540 592 645 ∑𝑋𝑌 = 3192 ∑𝑋 = 78 ∑𝑌 = 246 ∑𝑋 " = 1038 ∑𝑌 " = 10136 Here, no. of pair observations (n) = 6 We have, Karl Pearson's coefficient of correlation is r n åXY – åX åY = n åX2 – (åX) 2 n åY2 – (åY) = 6 × 3192 – 78 × 246 6 × 1038 – 782 6 × 10136 – 2462 = –36 144 300 2 = – 0.17 And, the regression coefficient of X on Y is bXY = n åXY – åX åY n åY2 – (åY) 2 = 6 × 3192 – 78 × 246 6 × 10136 – 2462 = –36 300 = – 0.12 \ The Karl Pearson's coefficient of correlation is – 0.17 and the regression coefficient of X on Y is – 0.12. 16) a) Define L-Hospital’s rule. Ans: Let f(x) and g(x) be two differentiable functions of x at x = a and Hospital's rule states that [1] f(a) 0 ¥ = or . Then Lg(a) 0 ¥ lim f(x) lim f'(x) f'(a) = = , provided that g'(a) ¹ 0. x ® a g(x) x ® a g'(x) g'(a) b) Write the slope of the tangent and normal to the curve y = f(x) at (x1, y1). [1+1] Ans: The slope of tangent to the curve y = f(x) at (x1, y1) is given by f'(x1) and the slope of normal is 1 given by – . f'(x1) c) 1 Write the integral ofó 2 dx . õx + a2 [1] 1 1 x Ans: ó 2 dx = tan–1 + C a a õx + a2 d) What is the integral of ó õ sinh x dx . [1] ó sinh x dx = cosh x + C Ans: õ 17) a) Solve: dy y = dx x [2] Ans: Given, dy y = dx x dy dx = y x or, Integrating, or, ln y = ln x + ln C ln y = ln Cx \ y = Cx is the required solution of given differential equation. b) Verify the Rolle’s theorem for f(x) = x2 + 3x – 4 in [–4, 1]. 2 [3] Ans: Given f(x) = x + 3x – 4 in [–4, 1] i. f(x) = x2 + 3x – 4 is a polynomial function so it is continuous in [–4, 1]. ii. f'(x) = 2x + 3 exists for all x Î (–4, 1). So, f(x) is differentiable in (–4, 1). iii. f(–4) = (–4)2 + 3 (–4) – 4 = 16 – 12 – 4 = 0 f(1) = 12 + 3 – 4 = 0 i.e. f(–4) = f(1) So, f(x) satisfies all three conditions of Rolle's mean value theorem. So, there must exists a point c Î (–4, 1) such that f'(c) = 0 or, 2c + 3 = 0 or, c = –3 Î (–4, 1) 2 This verifies the Rolle's theorem. 18) Using simplex method, maximize P(x, y) = 15x + 10y subject to 2x + y £ 10, x + 3y £ 10, x, y Ans: Given, Max. P = 15x + 10 y s.t. 2x + y £ 10 x + 3y £ 10 x, y By introducing non-negative slack variables ‘r’ and ‘s’ above LLP in standard form can be written as 2x + y + r = 10 x + 3y + s = 10 –15x – 10y + P = 0, x, y, r, s ³ 0 The initial simplex tableau corresponding to the above system is B.V. r s x 2 1 –15 y 1 3 –10 r 1 0 0 s 0 1 0 RHS 10 10 0 The most negative entry on last row is –15, so x column is the pivot column. Also, 10 10 < so r row is the pivot row and 2 is the pivot element. 2 1 Performing R1 ® R1 , we get 2 B.V. x x 1 s 1 –15 y 1 2 3 –10 r 1 2 0 0 s RHS 0 5 1 0 10 0 Performing R2 ® R2 – R1 and R3 ® R3 + 15 R1, we get B.V. x x 1 s 0 0 y 1 2 5 2 5 –2 r 1 2 1 –2 15 2 s RHS 0 5 1 5 0 75 . [5] 5 The most negative entry on last row is – , so y column is the pivot column. 2 Also, 5 5 5 =2< = 10 so, s row is the pivot row and is the pivot element. 5/2 1/2 2 2 Performing R2 ® R2 , we get 5 B.V. x y 1 x 1 2 y 0 1 0 5 –2 Performing R1 ® R1 – r 1 2 1 –5 15 2 s RHS 0 5 2 5 2 0 75 1 5 R and R3 ® R3 + R2, we get 2 2 2 B.V. x y x 1 0 y 0 1 0 0 r 3 5 1 –5 7 s 1 – 5 2 5 1 RHS 4 2 80 Now, all the entries of last row become non-negative. Thus, we obtained the maximum solution. \ Max. P = 80 at (x, y) = (4, 2). 19) A particle is projected with a velocity ‘ ’ and greatest height is ‘ ’, prove the horizontal range R is: NO R = 4 KH M"P − HQ Ans: Given, Initial velocity = 𝑣 Let angle of projection = 𝛼 We know, Greatest height (H) = Horizontal range(R) = N O TUVOW "X N O TUV "W X NO We need to show that, R = 4 KH M"P − HQ Taking R.H.S. NO 4 KH M"P − HQ = 4 K N O TUVOW N O "X M "P − N O TUVO W "X Q =4K =4K =4 N O TUVOW "X N O TUVOW "X × × NO "X NO "X N O × TUVW × `aTW "X (1 − 𝑠𝑖𝑛" 𝛼 ) × 𝑐𝑜𝑠 " 𝛼 = 4 KM =2 N O × TUVW × `aTW X N O × TUV W × `abW " = Q "X N O × " × TUVW × `aTW X = N O TUV"W X =R NO Hence, R = 4 KH M"P − HQ. OR, The cost function C (x) in thousands of rupees for producing x units (in thousands) of maths textbooks is given by C(x) = 30 + 20x – 0.5 x2, 0 £ x £ 15 a) Find the marginal cost function. Ans: Given that, Cost function C(x) = 30 + 20x – 0.5 x2, 0 £ x £ 15 Differentiating w.r.t x, we get d C(x) = 20 – x dx dC \ Marginal cost function MC = = 20 – x. dx b) Find the marginal cost for producing 12,000 maths textbooks. Ans: Number of maths textbooks is 12,000. So, x = 12. Then [3] [2] MC = 20 – x = 20 – 12 =8 \ The marginal cost of producing 12,000 maths textbooks is Rs. 8,000. Group "C" [ 3 × 8 = 24] 20) a) Using matrix methods, solve the following system of linear equations: x + y + z = 4, 2x + y – 3z = –9, 2x – y + z = –1 [Matrix methods means either Inverse Matrix method or Row equivalent method] Ans: Given system of linear equations is x+y+z=4 2x + y – 3z = –9 2x – y + z = –1 The augmented matrix corresponding to the above system is é1 1 1 : 4ù ê2 1 –3 : –9ú ê ú ë2 –1 1 : –1û [4] Applying R2 ® R2 – 2R1 and R3 ® R3 – 2R1, we get é 1 1 1 : 4 ù ~ êê0 –1 –5 : –17úú ë 0 –3 –1 : –9 û Applying R2 ® –1 × R2, we get é1 1 1 : 4ù ~ êê 0 1 5 : 17 úú ë0 –3 –1 : –9û Applying R1 ® R1 – R2 and R3 ® R3 + 3R2, we get é1 0 –4 : –13ù ~ êê 0 1 5 : 17 úú ë 0 0 14 : 42 û 1 Applying R3 ® 14 × R3, we get é1 0 –4 : –13ù ~ êê 0 1 5 : 17 úú ë0 0 1 : 3û Applying R1 ® R1 + 4R3 and R2 ® R2 – 5R3, we get é1 0 0 : –1ù ~ êê 0 1 0 : 2 úú ë 0 0 1 : 3û \ The requires solution to the system is x = –1, y = 2 and z = 3. b) Apply De-moivre’s theorem to find the value of [2 (cos 15o + i sin 15o)]6 Ans: Given complex is, [2 (cos 15 + i sin 15 )] Applying De-moivre's theorem, we get, 26 [ cos (6 × 15o) + i sin (6 × 15o)] = 64 [ cos 90o + i sin 90o] = 64 (0 + i) o o = 64i. 1 1 1 1 1 1 öæ ö 1! + 2! + 3! + …÷ø çè1 – 1! + 2! – 3! + …÷ø = 1 1 1 1 æ ö d) Ans: We have, e = ç1 + + + + …÷ 1! 2! 3! è ø 1 1 1 and e–1 = æç1 – 1! + 2! – 3! + …ö÷ è ø 1 1 1 1 1 1 Then, æç1 + 1! + 2! + 3! + …ö÷ æç1 – 1! + 2! – 3! + …ö÷ = e × e–1 = 1. Proved. è øè ø æ è c) Prove that ç1 + [2] 6 [2] 21) a) Find the direction cosines of the line joining the points (4, 4, –10) and (–2, 2, 4). [2] Ans: Given points are (4, 4, –10) = (x1, y1, z1) and (–2, 2, 4) = (x2, y2, z2). We have, (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2 r= = (–2 – 4)2 + (2 – 4)2 + (4 + 10)2 = 36 + 4 +196 = 236 = 2 59 Now, the direction cosines are, l= x2 – x1 –2 – 4 – 3 = = r 2 59 59 m= n= y2 – y1 –2 –1 = = r 2 59 59 z2 – z1 14 7 = = r 2 59 59 \The required direction cosines are –3 –1 7 , and . 59 59 59 b) Find the angle between the two diagonals of a cube. Ans: Consider a cube having one of the corner O as origin. Let OA = OB = OC = a, be three adjacent edges of a cube as shown in figure. Then the coordinates of the points A, B, C, D, E, F & G be (a, 0, 0), (0, a, 0), (0, 0, a), (a, a, 0), (0, a, a), (a, 0, a) and (a, a, a) respectively. Let OG and AE be any two diagonals of cube. Then from figure, OG = (a – 0)2 + (a – 0)2 + (a – 0)2 = a 3 In a cube OG = AE, so OG = AE = a 3 Now the d.c.s of OG are, similarly; the d.c.s of AE are, a–0 a–0 a–0 1 1 1 , , i.e., , , a 3 a 3 a 3 3 3 3 0–a a–0 a–0 –1 1 1 , , i.e., , , a 3 a 3 a 3 3 3 3 Let q be the angle between the two diagonals OG & AE then we have, cos q = l1l2 + m1m2 + n1n2 = = 1 –1 1 1 1 1 . + . + . 3 3 3 3 3 3 1 3 1 \ q = cos–1 æ ö is the required angle between the diagonals of cube. è3ø [4] c) Find the vertices of the conic section: 16(y – 1)2 – 9(x – 5)2 = 144. [2] Ans: Given, 16(y – 1)2 – 9(x – 5)2 = 144 or, 9 (x – 5)2 – 16 (y – 1)2 = – 144 or, 9 (x –5)2 16 (y –1)2 – =–1 144 144 or, (x – 5)2 (y – 1)2 – = –1 16 9 Comparing with (x – h)2 (y – k)2 – = –1 , we get a2 b2 h = 5, k = 1, a2 = 16 ⇒ a = 4, b2 = 9 ⇒ b = 3 Now, vertices = (h, k ± b) = (5, 1 ± 3) \ The vertices are (5, 4) and (5, – 2). 22) a) If the limiting value of ( f(x) – 5) x–3 at x = 3 is 2 by using L-Hospital’s rule, find the appropriate value of f(x). [2] Ans: According to the question lim f(x) – 5 = 2 using L-Hospital's rule x®3 x–3 or, lim f'(x) =2 x®3 1 or, f'(x) = 2 Integrating, we get at x = 3 ó õ f'(x) dx = ó õ 2 dx or, f(x) = 2x + C … (i) Also, at x = 3 given expression f(x) – 5 0 must take an indeterminate form . x–3 0 i.e., f(x) – 5 = 0 at x = 3 or, f(3) – 5 = 0 or, 2 × 3 + C – 5 = 0 [∵ from (i)] or, 1 + C = 0 or, C = – 1 \ f(x) = 2x – 1 is one appropriate value of f(x). b) Write any one homogeneous differential equation in (x, y) and solve it. dy y Ans: An example of homogeneous differential equation is dx = x . [3] Now, dy y = dx x or, dy dx = y x Integrating, or, ln y = ln x + ln C ln y = ln Cx \ y = Cx is the required solution of above homogeneous differential equation. c) The concept of anti-derivative is necessary for solving a differential equation. Justify this statement with example. [3] Ans: Differential equation is an equation which involves the differential coefficient (i.e. the derivative of dependent variable w.r.t an independent variable). In order to find the solution of differential equation we have to remove the derivative term, in that process we must have to use antiderivative. Thus, the concept of antiderivative is necessary for solving a differential equation. Which is also illustrated by the following example. Let us consider a differential equation dy = a ⇒ dy = a dx dx To remove the differential, we now use antiderivative ó dy = õ óa dx õ \ y = ax + C -0- Prepared by: Dept. of Mathematics Little Angels Secondary School(+2)