# MATHS- 11th (PQRS & J) Partial Marking.pdf

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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

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### MATHS- 11th (PQRS & J) Partial Marking.pdf

• 1. 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] [Sol. l1 : y = 2 1 x + c  x-intercept is – 2c l2 : y = 3 1 x + c  x-intercept is – 3c l3 : y = 4 1 x + c  x-intercept is – 4c  – 2c – 3c – 4c = 36  – 9c = 36  c = – 4 Ans. ] Q.2 Findthegeneralsolutionoftheequation               ... x sin ) 1 ..( x sin x sin x sin 1 .... x sin ...... x sin x sin x sin 1 n n 3 2 n 3 2 = x tan 1 4 2  where x  k + 2  , k  I. [Ans. n + (–1)n 6  , n  I] [5] [Sol. Nr of LHS = x sin 1 1  ; Dr of LHS = x sin 1 1  hence x sin 1 x sin 1   = x sec 4 2 = 4 cos2x = 4(1 – sin x)(1 + sin x) hence 4(1 – sin x)2 = 1 (1 – sin x)2 = 4 1  (1 – sin x) = 2 1 or – 2 1  sin x = 2 1 or sin x = 2 3 (rejected)  sinx= sin 6   x = n + (–1)n 6  , n  I Ans. ] Class: XI(PQRS & J) Time: 90 Minutes Max. Marks:90 M A T H E M A T I C S F I N A L T E S T Instruction 1. The question paper contains 15 question. All questions are compulsory. 2. Onlyanswers are to be written in the same order in which theyappear in the question paper. 3. Each subjectivequestionshouldbegin aftertheend oftheprevious question afterdrawingaline. 4. Sub part in respect of a subjective question should be doneat the same place(if applicable). 5. Use of Calculator,Logtable and Mobileis not permitted. 6. Legibilityandclarityinansweringthequestionwillbeappreciated. 7. Put a cross ( × ) on the rough work done by you.           {3 Marks}  {2 Marks}  {2 Marks}  {2 Marks}  {1 Mark} (ONLYFOR FACULTY) DO NOT DISPLAYFOR STUDENTS
• 2. Q.3 In a triangleABC if 2 cos 2 B cos 2 C = 2 1 + 2 A sin a c b        then find the measure of angleA. A. [5] [Sol. Given 2 cos 2 B cos 2 C = 2 1 + 2 A sin a c b        or cos        2 C B + cos        2 C B = 2 1 + 2 A sin A sin C sin B sin        sin 2 A +        2 C B cos = 2 1 + 2 A cos 2 A sin 2 2 A sin 2 C B cos 2 C B sin 2               sin 2 A +        2 C B cos = 2 1 + 2 A cos 2 C B cos 2 A cos        sin 2 A +        2 C B cos = 2 1 +        2 C B cos  sin 2 A = 2 1  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 q p  = 2 3 .Determinethenumericalvalueof m m m m 1 2 2 2 1 2  . [Ans. 99 ] [5] [Sol. mx2 + (2 – m) x + 3 = 0 p + q = m 2 m  ; pq = m 3 now m1 and m2 satisfies 3 2 p q q p    3 2 pq q p 2 2   3 2 pq pq 2 ) q p ( 2    m 2 m 3 . 3 2 m 6 m 2 m 2            m 8 m ) 2 m ( 2 2   m2 – 4m + 4 = 8m  m2 – 12m + 4 = 0  m1 + m2 = 12 and m1 m2 = 4 now 2 1 2 2 2 1 m m m m  = 2 2 1 3 2 3 1 ) m m ( m m  = 2 2 1 2 1 2 1 3 2 1 ) m m ( ) m m ( m m 3 ) m m (    = 16 12 . 12 123  = 16 11 . 122 = 99 Ans. ] Decide yourself  {1 Mark}  {1 Mark}  {2 Marks}  {1 Mark}
• 3. Q.5 If Cr denotes the combinatorial co-efficient in the expansion of (1 + x)n, n  N then using algebraic approach prove that C0 + 2 C1 + 3 C2 + ........ + 1 n Cn  = 1 n 1 2 1 n    . [5] [Sol. Let S = C0 + 2 C1 + 3 C2 + ........ + 1 n Cn  generalterm Tr = 1 r Cr n  Tr = )! 1 r ( )! r n ( ! r ! n   = ] )! 1 r ( )! r n )[( 1 n ( ) 1 n ( ! n      Tr = 1 n 1           )! 1 r ( )! r n ( )! 1 n ( Tr = 1 n C 1 r 1 n     S = 1 n 1      n 0 r 1 r 1 n C S =   1 n C C ..... C C C 0 1 n 1 n 1 n 2 1 n 1 1 n 0 1 n            = 1 n 1 2 1 n    Hence proved.] Q.6 Let 'A' denotes the real part of the complex number z = i 9 i 7 19   + i 6 7 i 5 20   and 'B' denotes thesum of the imaginaryparts of the roots ofthe equation z2 – 8(1 – i)z + 63 – 16i = 0 and 'C' denotes the sum of the series, 1 + i + i2 + i3 + ..... + i2008 where i = 1  . 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 valueof D C B A   . [Ans. 6] [5] [Sol. A = Re (z) now z = 82 ) i 9 )( i 7 19 (   + 85 ) i 6 7 )( i 5 20 (   = 82 7 i 82 171   + 85 30 i 35 i 120 140    = 82 i 82 164  + 85 i 85 170  = 2 + i + 2 – i z = 4 + 0i  4 A  Let  = x + iy  = a + ib  +  = (x + a) + i(y + b)= 8 – 8i  y + b = – 8  sum of the imaginaryparts of the roots of the equation = – 8  8 B    {1 Mark}  {2 Marks}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}
• 4. 'C' S = 1 + i + i2 + i3 + ........ + i2008 = i 1 ) i 1 ( 2009   = i 1 i 1   = 1  1 C  1 D  obvious hence D C B A   = 2 8 4  = 6 Ans. ] 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, tan1 1 3 + tan1 1 7 + tan1 1 13 + tan1 1 21 +...... = cot–1        n 2 n [3+3] [Sol.(a)  +  = – 5;  = – 49 let y = cot(cot–1 + cot–1) = ) cot(cot ) cot(cot 1 ) )·cot(cot cot(cot 1 1 1 1           =      1 = 5 50   = 10 Ans. [Sol.(b) S = tan 1 1 1 1 2  . + tan 1 1 1 2 3  . + tan 1 1 1 3 4  . +....... Now , Tn = tan 1 1 1 1         n n ( ) = tan 1 ( ) ( ) n n n n           1 1 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 —————————— Sn = tan1 (n + 1)  4  Sn = tan–1 ) 1 n ( 1 1 ) 1 n (     = tan–1        2 n n = cot–1        n 2 n Hence proved. ] Q.8 In acute angledtriangleABC, a semicircle with radius ra is constructedwith its base onBC andtangent totheothertwosides. rb andrc aredefinedsimilarly. Ifris theradius oftheincircleoftriangleABCthen prove that r 2 = c b a r 1 r 1 r 1   [6] [Sol. 2 b r 2 c r a a    =  (where  is the area of triangleABC) [T/S, ph-3] ra(b + c) = 2 |||ly rb(c + a) = 2 rc(a + b) = 2  {1 Mark}  {1 Mark}  {1 Mark}  {2 Marks}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}       {3 Marks}
• 5. adding c b a r 2 r 2 r 2      = 2(a + b + c) c b a r 1 r 1 r 1   =    c b a =  s 2 = r 2 i.e. r 2 = c b a r 1 r 1 r 1   Hence proved ] Q.9 Findthe equationof the circlepassingthrough thepoint (–6, 0) ifthepowerof 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) 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,  c = 0 6 + 2f = 3  2f = – 3  f = – 3/2  required equation of circle is x2 + y2 + 6x – 3y = 0 Ans. ] Q.10 Tendogs encounter8 biscuits. Dogs donot sharebiscuits. In howmanydifferent ways can the biscuits be consumed (a) if weassume that thedogsare distinguishable,but thebiscuits arenot. (b) if we assume that both dogs andbiscuits are different and anydog can receive anynumber of biscuits. (c) if dogs andbiscuits are different andeverydog can get atmost one biscuit. [2+2+2] [Ans. (a) 17C8; (b) 108; (c) 10C8 · 8! ] [Sol. (a) dogs different D1, D2 , ......., D10 biscuitsalike      8 B ......... B B 8 alike object to be distributed in 10 different dogs     8 0 ......... 0 0      9 Ø ......... Ø Ø Total ways = 17C8 Ans. when anynumber of biscuits can be had byanydog.  {3 Marks}           {1 Mark}             {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark} Give marks only if both reasoning and answer are correct (1 Mark forreasoning and 1 Mark forAns.)
• 6. (b) dogs aredifferent D1, D2 , ......., D10 biscuitsaredifferent B1, B2 , ......., B8 1st biscuits can be given in10 ways 2nd biscuits can be given in10 ways hence total ways = 108 Ans. (c) Select 8dogs in 10C8 andgive them 1 biscuit each. Distribution of biscuits in 8! ways. Total ways = 10C8 · 8! Ans. ] Q.11 Given   35 1 k k 5 sin = tan       n m , where angles are measured indegrees, and m and n arerelativelyprime positive integers that satisfy n m <90, find thevalue of (m + n). [6] [Ans. 177] [Sol. LHS: S = sin 5 + sin 10 + sin 15 + .......... + sin 170 + sin 175 S       2 5 sin 2 = 2 sin 2 5 [sin 5 + sin 10 + ......... + sin 175] T1 = cos 2 5 – cos 2 15 T2 = cos 2 15 – cos 2 25    T35 = cos 2 345 – cos 2 355 ————————       2 5 sin 2 · S = cos 2 5 – cos 2 355 = 2 sin 2 180 · sin 2 175 = 2 sin 2 175 S = 2 5 sin 2 175 sin =        2 5 90 cos 2 175 sin = 2 175 cos 2 175 sin = tan       2 175 = tan       n m  m = 175 and n = 2  m + n = 177 Ans. ] PART-C Q.12 Apointmovingaroundcircle(x +4)2 +(y+2)2 =25withcentreCbrokeawayfromiteitheratthe point A or point B on the circleand moved along atangent to the circle passing through the point D(3, – 3). Findthefollowing. (i) Equation ofthe tangents atAand B. (ii) Coordinates of the pointsAand B. (iii) Equation of the circle circumscribing the DAB and also the intercepts made by this circle on the coordinateaxes. [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 ]  {2 Marks}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}
• 7. [Sol. (i) Equationoftangent from point (3, –3) to thegiven circle is y + 3 = m(x – 3) mx – 3m – y – 3 = 0 and also 2 m 1 3 2 m 3 m 4      = 5 (1 + 7m)2 = 25(1 + m2)  1 + 49m2 + 14m = 25 + 25m2  12m2 + 7m – 12 = 0  (4m – 3)(3m + 4) = 0  m = 3/4 or m = – 4/3  equation of tangent at pointAand B are y + 3 = – 3 4 (x – 3) and y + 3 = 4 3 (x – 3) 3y + 9 = – 4x + 12 4y + 12 = 3x – 9 4x + 3y = 3 3x – 4y = 21 (ii) Equation ofnormals to these 2tangents are y + 2 = 4 3 (x + 4) and y + 2 = – 3 4 (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 9x – 12y = 63 —————— —————— x = 0;  y = 1 25x = – 25 x = – 1;  y = – 6  points Aand B are (0, 1) and (–1, – 6) Ans. (iii) CirclecircumscribingDABwill havepointsAandBas its diametricalextremities x2 + y2 – x(–1) – y(–5) – 6 = 0 x2 + y2 + x + 5y – 6 = 0 Ans. x-intercept = c g 2 2  = 6 ) 4 1 ( 2  = 5 Ans. y-intercept = c f 2 2  = 6 ) 4 25 ( 2  = 7 Ans. ]  {0.5 Mark}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}  {0.5 Mark}  {0.5 Mark}  {1 Mark}  {1 Mark}
• 8. Q.13 Let f (n) =           n 0 r n r k r k . Find the total number of divisors of f (9). [7.5] [Sol.   n r k r k C = rCr + r + 1Cr + r + 2Cr + ....... + nCr = 1 + r + 1C1 + r + 2C2 + r + 3C3 + ....... + nCn – r now, n+1Cn – r = n+1Cr+1  f (n) =     n 0 r 1 r 1 n C = n+1C1 + n+1C2 + n+1C3 + ..... + n+1Cn+1 = n+1C0 + n+1C1 + n+1C2 + ..... + n+1Cn+1 – 1 f (n) = (2n+1) – 1 f (9) = 210 – 1 = 1023 = 3 · 11 · 31 hence number of divisors are (1 + 1)(1 + 1)(1 + 1) = 8Ans. ] Q.14 Givenbelow is apartial graph of an even periodic function f whose period is 8. If [*] denotes greatest integerfunctionthenfindthevalueoftheexpression. f (–3) + 2 | f (–1) | +             8 7 f + f (0) + arc cos  ) 2 ( f + f (–7) + f (20) [7.5] [Sol. f (–3) = f (3) = 2 [ 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 fromthegraph, – 3 <       8 7 f < – 2              8 7 f = – 3 f (0) = 0 (obviouslyfromthegraph) cos–1   ) 2 ( f  = cos–1   ) 2 ( f = cos–1(1) = 0 f (–7) = f (– 7 + 8) = f (1) = – 3 [f (x) has period 8]  {1 Mark}  {1 Mark}  {2.5 Marks}       {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}  {1 Mark}
• 9. f (20) = f (4 + 16) = f (4) = 3 [ f (nT + x) = f (x) ] sum = 2 + 6 – 3 + 0 + 0 – 3 + 3  sum = 5 Ans. ] Q.15 AsquareABCDlyinginI-quadrant hasarea36sq. units andis such thatits sideABis parallelto x-axis. VerticesA, B and C are on the graph of y = logax, y= 2 logax and y = 3 logax respectivelythen 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  pointAis (ac, c) |||ly B is (ac/2, c) and BC  AB  C has x-coordinate = ac/2 and it lies on y = 3 logax y = 3 logaac/2 = 2 c 3  point C is       2 c 3 , a 2 c | AB | = 6  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 2 c 3 – c = 6;  c = 12  a12 = 9  a6 = 3;  a = 6 3 Ans. ]  {1 Mark}  {0.5 Mark}  {1 Mark}  {1 Mark}  {2 Marks}  {2 Marks}  {0.5 Mark}  {1 Mark}
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