MATHEMATICS-25-02- 11th (PQRS & J) Partial Marking

MATHEMATICS FINALTEST Instruction 1. The question paper contains 15 question. All questions are compulsory. 2. Only answers are to be written in the same order in which they appear in the question paper. 3. Each subjective question should begin after the end of the previous question after drawing a line. 4. Sub part in respect of a subjective question should be done at the same place (if applicable). 5. Use of Calculator, Log table and Mobile is not permitted. 6. Legibility and clarity in answering the question will be appreciated. 7. Put a cross ( × ) on the rough work done by you. PART-A Q.1 Three straight lines l1, l2 and l3 have slopes 1/2, 1/3 and 1/4 respectively. All three lines have the same y-intercept. If the sum of the x-intercept of three lines is 36 then find the y-intercept. [5] [Ans. – 4] 1 [Sol. l1 : y = 2 x + c  x-intercept is – 2c   1 l2 : y = 3 x + c  x-intercept is – 3c 1      {3 Marks} l3 : y = 4 x + c  x-intercept is – 4c   – 2c – 3c – 4c = 36  – 9c = 36  c = – 4 Ans. ]  {2 Marks} 1+ sin x + sin2 x + sin3 x + ...... + sinn x + .... 4 Q.2 Find the general solution of the equation 1 sin x + sin2 x  sin3 x + ..(1)n sinn x + ... = 1+ tan2 x where x  k +  , k  I. [Ans. n + (–1)n  , n  I] [5] [Sol. Nr of LHS = 2 1 1 sin x ; Dr of LHS = 1 1+ sin x 6  {2 Marks} hence 1+ sin x 1 sin x 4 = sec2 x = 4 cos2x = 4(1 – sin x)(1 + sin x) hence 4(1 – sin x)2 = 1 (1 – sin x)2 = 1  (1 – sin x) = 4 1 1 2 or – 2  sin x = 1 or sin x = 3 (rejected)  {2 Marks} 2 2  sin x = sin   x = n + (–1)n  , n  I Ans. ]  {1 Mark} 6 6 B C 1  b + c sin A Q.3 In a triangle ABC if 2 cos 2 cos 2 = +   2  a  2 then find the measure of angle A. [5] B C 1  b + c sin A [Sol. Given 2 cos 2 cos 2 = +   2  a  2  B + C   B  C  1  sin B + sin C sin A or cos  2  + cos 2  = 2 +   sin A 2       2 sin B + C cos B  C sin A A  B  C  1  2   2  2 sin 2 + cos   = 2  2 +     2 sin A cos A 2 2 cos A cos B  C  A  B  C  1 2  2  sin 2 + cos   = 2  2 +   cos A 2 A cos B  C  1 cos B  C  A 1 sin 2 +   = +  2     2   sin 2 = 2   A/2 = 30°   A = 60° Ans. ] Q.432/1 Let p & q be the two roots of the equation, mx2 + x (2  m) + 3 = 0. Let m1, m2 be the two values of m satisfying p + q 2 m1 + m2 . [Ans. 99 ] [5] q p = 3 . Determine the numerical value of m2 2 [Sol. mx2 + (2 – m) x + 3 = 0 m  2 p + q = m 3 ; pq = m  {1 Mark} p + q = 2 p2 + q2 = 2 now m1 and m2 satisfies q p 3  pq 3 (p + q)2  2pq = 2  {1 Mark} pq 3  m  2 2 6 2 3 2 (m  2)2 8      m = . =  = 3 m m m2 m m2 – 4m + 4 = 8m  m2 – 12m + 4 = 0  m1 + m2 = 12 and m1 m2 = 4 m m m3 + m3 (m + m )3  3m m (m + m ) now 1 + 2 = 1 2 = 1 2 1 2 1 2  {2 Marks} 2 1 (m1m2 ) (m1m2 ) 123 12.12 = 16 122 .11 = 16 = 99 Ans. ]  {1 Mark} Q.5 If Cr denotes the combinatorial co-efficient in the expansion of (1 + x)n, n  N then using algebraic approach prove that C + C1 0 2 + C2 3 + ........ + Cn n +1 = 2n+1 1 n +1 . [5] [Sol. Let S = C0 + C1 + 2 n C C2 + + 3 Cn n +1 general term T = r r +1  {1 Mark} n! n!(n +1) Tr = r!(n  r)!(r + 1)! = (n + 1)[(n  r)!(r + 1)!] 1  (n +1)!   Tr = n +1 (n  r)!(r +1)!   n+1C T = r+1 n +1  {2 Marks}  S = 1 n +1 n  r=0 n+1 Cr+1  {1 Mark} n+1C S = 0 + n+1C +n+1 C + n+1C n+1 )n+1 C 2n+1 1 = Hence proved.]  {1 Mark} n +1 n + 1 19 + 7i 20 + 5i Q.6 Let 'A' denotes the real part of the complex number z = 9  i + 7 + 6i and 'B' denotes the sum of the imaginary parts of the roots of the equation z2 – 8(1 – i)z + 63 – 16i = 0 and 'C' denotes the sum of the series, 1 + i + i2 + i3 + ..... + i2008 where i = . and 'D' denotes the value of the product (1 + )(1 + 2)(1 + 4)(1 + 8) where  is the imaginary cube root of unity. Find the value of [Sol. A = Re (z) A  B . [Ans. 6] [5] C + D (19 + 7i)(9 + i) now z = 82 (20 + 5i)(7  6i) + 85 171+ 82i  7 = 82 140 120i + 35i + 30 + 85 164 + 82i = 82 170  85i + 85 = 2 + i + 2 – i z = 4 + 0i  Let  = x + iy  = a + ib  +  = (x + a) + i(y + b)= 8 – 8i  y + b = – 8  {1 Mark}  sum of the imaginary parts of the roots of the equation = – 8   {1 Mark} 'C' S = 1 + i + i2 + i3 + + i2008 = (1 i2009 ) 1  i 1 i = 1 i = 1   {1 Mark} obvious  {1 Mark} hence A  B C + D = 4 + 8 2 = 6 Ans. ]  {1 Mark} PART-B Q.7 (a) If  and  are the roots of the equation x2 + 5x – 49 = 0 then find the value of cot(cot–1 + cot–1). (b) 178/6 Prove that the sum to 'n' terms of the series, 1 1 1 1 1 1 1 1 n + 2 –1 tan 3 + tan 7 + tan 13 + tan 21 +...... = cot  n  [3+3] [Sol.(a)  +  = – 5;  = – 49 let y = cot(cot–1 + cot–1) = cot(cot1 )·cot(cot1 ) 1 cot(cot1 ) + cot(cot1 )  {2 Marks}  1 =  +  = – 50 – 5 = 10 Ans.  {1 Mark} [Sol.(b) S = tan 1 1 1 + 1 . 2 + tan 1 1 1 + 2 . 3 + tan 1 1 1 + 3 . 4 +....... Now , T = tan 1 1  = tan 1  (n + 1)  n  n 1 + n (n + 1)  1 + n (n + 1)      = tan 1 (n + 1)  tan 1 n T1 = tan12  tan11 T2 = tan–13 – tan–12 M Tn = tan–1(n + 1) – tan–1n ——————————  {1 Mark}  Sn = tan1 (n + 1) 4  {1 Mark} (n +1) 1  n   n + 2  Sn = tan–1 1+ (n +1) = tan–1  n + 2  = cot–1  n  Hence proved. ]  {1 Mark}     Q.8 In acute angled triangle ABC, a semicircle with radius ra is constructed with its base on BC and tangent to the other two sides. rb and rc are defined similarly. If r is the radius of the incircle of triangle ABC then prove that 2 = 1 + 1 + 1 [6] r ra rb rc [Sol. ra  c + ra  b =  (where  is the area of triangle ABC) [T/S, ph-3] 2 2 ra(b + c) = 2   |||ly rb(c + a) = 2  rc(a + b) = 2  {3 Marks} adding 2 + 2 + 2 = 2(a + b + c) ra rb rc 1 + 1 + 1 a + b + c 2s 2 = = = ra rb rc   r i.e. 2 = 1 + 1 + 1 Hence proved ]  {3 Marks} r ra rb rc Q.9 Find the equation of the circle passing through the point (–6 , 0) if the power of the point (1, 1) w.r.t. the circle is 5 and it cuts the circle x2 + y2 – 4x – 6y – 3 = 0 orthogonally. [6] [Ans. x2 + y2 + 6x – 3y = 0] [T/S, Q.2, Ex-2, circle] [Sol. Let the circle be x2 + y2 + 2gx + 2fy + c = 0 passes through the point (– 6, 0)  36 – 12g + c = 0 12g – c = 36 (1)      {1 Mark} and power of point (1, 1) w.r.t. circle is 5   1 + 1 + 2g + 2f + c = 5  2g + 2f + c = 3 ....(2)  it cuts the circle x2 + y2 – 4x – 6y – 3 = 0 orthogonally  2(–2g – 3f) = c – 3  – 4g – 6f = c – 3  – 4g – 6f = 12g – 39 (using (1))  + 16g + 6f = 39  and 2g = 2f + 12g – 36 = 3  14g + 2f = 39  42g + 6f = 117  – 16g – 6f = – 39  ——————— 26g = 78  g = 3,  {1 Mark}  {1 Mark}  c = 0  {1 Mark} 6 + 2f = 3  2f = – 3  f = – 3/2  {1 Mark}  required equation of circle is x2 + y2 + 6x – 3y = 0 Ans. ]  {1 Mark} Q.10 Ten dogs encounter 8 biscuits. Dogs do not share biscuits. In how many different ways can the biscuits be consumed (a) if we assume that the dogs are distinguishable, but the biscuits are not. (b) if we assume that both dogs and biscuits are different and any dog can receive any number of biscuits. (c) if dogs and biscuits are different and every dog can get atmost one biscuit. [2+2+2] [Ans. (a) 17C8; (b) 108; (c) 10C8 · 8! ] [Sol. (a) dogs different D1, D2 , , D10 biscuits alike B1 4B.2....4 3.B 8 8 alike object to be distributed in 10 different dogs Total ways = 17C8 Ans. when any number of biscuits can be had by any dog. 01 40.2....4..3..0 8 Ø1 4Ø.2....4 3.Ø 9 (b) dogs are different D1, D2 , , D10 biscuits are different B1, B2 , , B8 1st biscuits can be given in 10 ways 2nd biscuits can be given in 10 ways hence total ways = 108 Ans. (c) Select 8 dogs in 10C8 and give them 1 biscuit each. Distribution of biscuits in 8! ways. Total ways = 10C8 · 8! Ans. ] Q.11 Given 35 sin 5k = tan m  , where angles are measured in degrees, and m and n are relatively prime n k=1   positive integers that satisfy m < 90, find the value of (m + n). [6] n [Ans. 177] [Sol. LHS: S = sin 5 + sin 10 + sin 15 + + sin 170 + sin 175  2 sin 5  5 S  2  = 2 sin 2 [sin 5 + sin 10 + + sin 175] 5 T1 = cos 2 15 T2 = cos 2 15 – cos 2 25 – cos 2 345 355 T = cos – cos  {2 Marks} 35 2 2 ————————  2 sin 5  5 355 180 175 175  2  · S = cos 2 – cos 2 = 2 sin 2 · sin 2 = 2 sin 2 sin 175 S = 2 sin 175 = 2 sin 175 = 2 = tan175  = tan m   {1 Mark} 5  5  175  2   n  sin 2 cos90     cos 2      m = 175 and n = 2  {1 Mark}  m + n = 177 Ans. ]  {1 Mark} PART-C Q.12 A point moving around circle (x + 4)2 + (y + 2)2 = 25 with centre C broke away from it either at the point A or point B on the circle and moved along a tangent to the circle passing through the point D (3, – 3). Find the following. (i) Equation of the tangents at A and B. (ii) Coordinates of the points A and B. (iii) Equation of the circle circumscribing the DAB and also the intercepts made by this circle on the coordinate axes. [T/S, Q.12, Ex-1, circle] [2.5 + 2.5 + 2.5] [Ans. (i) 4x + 3y = 3; 3x – 4y = 21 (ii) A(0, 1) and B (–1, – 6); (iii) x2 + y2 + x + 5y – 6, x intercept 5; y intercept 7 ] [Sol. (i) Equation of tangent from point (3, –3) to the given circle is y + 3 = m(x – 3) mx – 3m – y – 3 = 0 and also (1 + 7m)2 = 25(1 + m2)   (4m – 3)(3m + 4) = 0 = 5 1 + 49m2 + 14m = 25 + 25m2  12m2 + 7m – 12 = 0  m = 3/4 or m = – 4/3  {0.5 Mark}  equation of tangent at point Aand B are 4 3 y + 3 = – 3 (x – 3) and y + 3 = 4 (x – 3) 3y + 9 = – 4x + 12 4y + 12 = 3x – 9 4x + 3y = 3  {1 Mark} 3x – 4y = 21  {1 Mark} (ii) Equation of normals to these 2 tangents are 3 4 y + 2 = 4 (x + 4) and y + 2 = – 3 (x + 4) 4y + 8 = 3x + 12 3y + 6 = – 4x – 16 3(3x – 4y + 4 = 0) 4(4x + 3y= – 22) 9x – 12y = – 12 16x + 12y = – 88 16x + 12y = 12  {1 Mark} 9x – 12y = 63  {1 Mark} —————— —————— x = 0;  y = 1 25x = – 25 x = – 1;  y = – 6  points A and B are (0, 1) and (–1, – 6) Ans.  {0.5 Mark} (iii) Circle circumscribing  DAB will have points Aand B as its diametrical extremities x2 + y2 – x(–1) – y(–5) – 6 = 0 x2 + y2 + x + 5y – 6 = 0 Ans.  {0.5 Mark} x-intercept = 2 = 2 = 5 Ans.  {1 Mark} y-intercept = 2 = 2 = 7 Ans. ]  {1 Mark} n n k  Q.13 Let f (n) =  r  . Find the total number of divisors of f (9). [7.5] r=0 k=r   n [Sol.  k C = rC + r + 1Cr + r + 2Cr + + nCr  {1 Mark} k=r = 1 + r + 1C1 + r + 2C2 + r + 3C3 + + nCn – r  {1 Mark} now, n+1Cn – r = n+1Cr+1 n  {2.5 Marks}  f (n) =  n+1C + r=0 = n+1C1 + n+1C2 + n+1C3 + ..... + n+1Cn+1   {1 Mark} f (n) = (2n+1) – 1 = n+1C0 + n+1C1 + n+1C2 + + n+1Cn+1 – 1  f (9) = 210 – 1 = 1023 = 3 · 11 · 31  {1 Mark} hence number of divisors are (1 + 1)(1 + 1)(1 + 1) = 8 Ans. ]  {1 Mark} Q.14 Given below is a partial graph of an even periodic function f whose period is 8. If [*] denotes greatest integer function then find the value of the expression.  f  7  f (–3) + 2 | f (–1) | +   8  + f (0) + arc cos ( f (2)) + f (–7) + f (20) [7.5]    [Sol. f (–3) = f (3) = 2  {1 Mark} [ f (x) is an even function,  f (– x) = f (x) ] again f (–1) = f (1) = – 3  2 | f (–1) | = 2 | f (1) | = 2 | – 3 | = 6 f  7   {1 Mark} from the graph, – 3 <  8  < – 2    f  7     8  = – 3  {1 Mark}    f (0) = 0 (obviously from the graph)  {1 Mark} cos–1 (f (2)) = cos–1 (f (2)) = cos–1(1) =0  {1 Mark} f (–7) = f (– 7 + 8) = f (1) = – 3 [f (x) has period 8]  {1 Mark} f (20) = f (4 + 16) = f (4) = 3 [ f (nT + x) = f (x) ] sum = 2 + 6 – 3 + 0 + 0 – 3 + 3  {1 Mark}  sum = 5 Ans. ]  {0.5 Mark} Q.15 Asquare ABCD lying in I-quadrant has area 36 sq. units and is such that its side AB is parallel to x-axis. Vertices A, B and C are on the graph of y = logax, y = 2 logax and y = 3 logax respectively then find the value of 'a'. [7.5] [Sol. AB : y = c (c > 0) Length of the side of square = 6 A has y-coordinate = c and it lies on y = logax  x-coordinate = ac  point A is (ac, c)  {1 Mark} |||ly B is (ac/2, c) and BC  AB  {1 Mark}  C has x-coordinate = ac/2 and it lies on y = 3 logax 3c y = 3 logaac/2 = 2 ac 2 , 3c   point C is   | AB | = 6   {2 Marks}   ac – ac/2 = 6 a > 0, c > 0 let ac/2 = t t2 – t – 6 = 0  (t – 3)(t + 2) = 0  t = 3 or t = – 2 (rejected)  t = 3  ac/2 = 3  ac = 9 also | BC | = 6 3c  {2 Marks} 2 – c = 6;  c = 12  {0.5 Mark}  a12 = 9  a6 = 3;  a = Ans. ]  {1 Mark}