# Dpp (10-12) 11th J-Batch Maths.pdf

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### Dpp (10-12) 11th J-Batch Maths.pdf

• 1. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-10 Q.1 Number of real x satisfying the equation | x – 1 | = | x – 2 | + | x – 3 | is (A) 1 (B*) 2 (C) 3 (D) more than 3 [Hint: x = 2 and x = 4 only solution ] Q.2 A diameter and a chord of a circle intersect at a point inside the circle. The two parts of the chord are length 3 and5 and one part of the diameter is lengthunity.The radius ofthe circle is (A*) 8 (B) 9 (C) 12 (D) 16 [Hint: x × 1 = 3 · 5  x = 15  diameter = 16 radius = 8 cm] Q.3 Smallest positive solution of the equation, 4       x sin2 16 =   x sin 6 2 , is (A) 2  (B) 3 2 (C) 6 5 (D*) none [Hint: x = 6  , 6 5 or 2   6   (D) ] Q.4log If log3(x) =p and log7(x) =q, whichof the followingyields log21(x)? (A) pq (B) q p 1  (C*) 1 1 q p 1    (D) 1 1 q p pq    [Hint: log3x = p and log7x = q now log21x = 21 log 1 x = 7 log 3 log 1 x x  = q 1 p 1 1  = 1 1 q p 1    Ans. ] *Q.5ph-1 The value of the expression 1 ) 44 cos ....... .......... 2 cos 1 (cos 2 ) 89 sin ....... 3 sin 2 sin 1 (sin 2                equals (A*) 2 (B) 2 1 (C) 2 1 (D) 1 [Sol. Nr = 2[(sin 1° + sin 89°) + (sin 2° + sin 88°) + ......... + (sin 44° + sin 46°) + sin 45°] = 2[sin 45°(2(cos 44° + cos 43° + ...... + cos1°) + 1] (using sin C + sin D)  r r D N = 2 sin 45° = 2 Ans ] *Q.6 If x satisfies log2x + logx2 = 4, then log2x can be (A*) tan(/12) (B) tan(/8) (C*) tan (5/12) (D) tan(3/8)
• 2. Q.7 Find the number of degree in the acute angle  satisfying cos  = 2 1 2 2  ? [Ans.  = 22.5°] [Sol. 4cos2 – 2 = 2  2(cos2 – 1) = 2  cos 2 = 2 1  2 = 4    = 8  = 22.5° Ans.] Ans.] Q.8 Find all values of asuch thatthe three equations ax + y = 1 x + y = 2 x – y = a aresimultaneouslysatisfied bysame orderedpair (x,y). [Ans. a = 0 or a = – 1] [Sol. from (1) and (2) adding, x = 2 2 a  subtracting y= 2 a 2  substituting in(1) a(a + 2) + (2 – a) = 2 a2 + 2a – a = 0  a2 + a = 0  a = 0 or a = 1 Ans. ] Q.9 Let Dbeanypoint on thebaseof anisosceles triangleABC.AC is extended to E so that CE = CD. ED is extended to meetAB at F. If angle CED = 10°, find the cosine of the angle BFD. [Ans. – 2 3 ] [Sol. Asshown cos(180 – ) = – cos  = – cos 30° = – 2 3 ] Q.10 Inthefigure,EisthemidpointofABandFisthemidpointofAD. Ifthe area of FAEC is 13 sq. units, find the area of the quadrilateralABCD. [Ans. 26] [Sol. Area = 2(A1 +A2) (Thing !) = 2 × 13 = 26 ] Q.11 In the figure, 'O' is the centre of the circle and A, B and C are three points on the circle. Suppose that OA =AB = 2 units and angle OAC = 10°. Find the length of the arc BC. [Ans. 9 10 units] [Sol. arc BC = l = r l = 2 · 180  · 100 = 9 10 Ans. ]
• 3. Q.1 If logab+ logbc + logca vanishes where a, b andc are positivereals different than unitythen thevalue of (logab)3 + (logbc)3 + (logca)3 is (A*) an odd prime (B) aneven prime (C) an odd composite (D)anirrationalnumber [Hint: x + y + z = 0  x3 + y3 + z3 = 3xyz  3  (A) ] *Q.2 Each ofthe four statements givenbelow areeitherTrueor False. I. sin765° = – 2 1 II. cosec(–1410°) = 2 III. tan 3 13 = 3 1 IV. cot         4 15 = – 1 Indicate the correct order of sequence, where 'T' stands for true and 'F' stands for false. (A) F T F T (B) F F T T (C) T F F F (D*) F T F F Q.3 The value of p which satisfies the equation 122p–1 = 5(3p ·7p)is (A) 12 n 21 n 12 n 5 n l l l l   (B) 21 n 12 n 5 n 12 n l l l l   (C*) 21 n 144 n 12 n 5 n l l l l   (D) 21 n 5 12 n 12 n l l l  *Q.482/ph-1 The expression      20 sin · 20 tan 20 sin 20 tan 2 2 2 2 simplifies to (A)arationalwhichis notintegral (B) a surd (C)anaturalwhichis prime (D*) anaturalwhich is not composite [Hint: tan220° – sin220° = tan220° (1 – cos220°) = tan220° sin220° Hence Nr = Dr  (D) ] Q.5 If tan 2  = m, then the value of     sin 1 ) 2 / ( sin 2 1 2 is (A) m 1 m 2  (B*) m 1 m 1   (C) m 1 m 1   (D) m 2 m 1 [Sol. y =    sin 1 cos = ) 2 / sin( ) 2 / cos( ) 2 / sin( ) 2 / cos(       = ) 2 / tan( 1 ) 2 / tan( 1     = m 1 m 1   Ans. ] *Q.668/ph-1 The value of 3 76 16 76 16       cot cot cot cot is : (A*) tan46º (B) tan44º (C) cot 46º (D) cot2º [Sol. Using 0 0 0 0 0 0 0 0 16 cos 76 sin 16 sin 76 cos 16 cos 76 cos 16 sin . 76 sin 3   = 0 0 0 0 0 0 0 92 sin ] 16 cos 76 cos 16 sin 76 [sin 16 sin 76 sin 2   =       92 sin 60 cos 92 cos 60 cos CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-11
• 4. = 0 0 92 sin 92 cos 1 = 0 0 0 2 46 cos 46 sin 2 46 sin 2 = tan460 = cot440 Ans ] *Q.7 An unknown polynomial yields a remainder of 2 upon division by x – 1, and a remainder of 1 upon division byx – 2. Ifthis polynomial is divided by(x – 1)(x – 2), then theremainderis (A) 2 (B) 3 (C*) – x + 3 (D) x + 1 [Sol. We have P (x) = Q1(x – 1) + 2 ....(1) P (x) = Q2(x – 2) + 1 ....(2) P (x) = Q3(x – 1)(x – 2) + ax + b ....(3) From (1), P(1) = 2 = a + b ....(4) from (2), P (2) = 1 = 2a + b .....(5) hence a = – 1 and b = 3 Hence the remainder is 3 – x  (D) ] Q.8 The difference (sin8 75° – cos8 75°) is equal to (A) 1 (B) 8 3 3 (C) 16 3 3 (D*) 16 3 7 [Hint: (sin4  – cos4 )(sin4  + cos8 ) where  = 75° (– cos150°)(1 – 2 sin2 cos2) + 2 3         150 sin 2 1 1 2 = 2 3        8 1 1 = 2 3 · 8 7 = 16 3 7 Ans. ] Ans. ] *Q.9circle There is an equilateral triangle with side 4 and a circle with the centre on one of the vertex of that triangle.Thearcofthat circledivides thetriangleinto twoparts of equal area. Howlongis the radius of thecircle? (A*)  3 12 (B)  3 24 (C)  3 30 (D)  3 6 [Sol. 4 3 · 16 = 2 · 2 1 r2 · 3  [11th J-Batch (15-01-2006)] r2 = 12 3  r =  3 12 Ans. ] Q.10 In a triangleABC, BC = 8, CA= 6 andAB = 10.Aline dividing the triangleABC into two regions of equal area is perpendicular toAB at the point X. Find the length BX. [Sol. 2 · 2 y · x = 2 6 · 8 = 24 x · x tan B = 24 x2 · 4 3 = 24 x2 = 32  x = 2 4 Ans. ]
• 5. Q.11 If sec x + tan x = 7 22 , find the value of tan 2 x . Use it to deduce the value of cosec x + cot x. [3] [Sol. x cos x sin 1 = 7 22 [Ans. 29 15 , 15 29 ] 7 22 t 1 t 1 t 1 t 2 1 2 2 2      where t = tan 2 x 2 2 t 1 ) t 1 (   = 7 22  t 1 t 1   = 7 22 ) t 1 ( ) t 1 ( ) t 1 ( ) t 1 (       = 7 22 7 22   = 15 29  t 2 2 = 15 29  t = 29 15 Ans. now cosec x + cot x = x sin x cos 1 =       2 x cos 2 x sin 2 2 x cos 2 2 =       2 x cot =   2 x tan 1 = 15 29  cosec x + cot x = 15 29 Ans. ]
• 6. Q.1        x sin 3 4 x sin dx d 3 when x = 12° is (A) 0 (B) 4 2 6  (C*) 4 1 5  (D) 4 1 5  [Hint: y = 3 x sin 4 x sin 3 3  = 3 x 3 sin ; y' = cos 3x = cos 36° = 4 1 5  ] Q.2 A rectanglehas its sides oflength sin x and cos x for some x. Largest possible area which it can have, is (A) 4 1 (B) 1 (C*) 2 1 (D) can not be determined [Hint: A= sin x cos x = 2 1 sin 2x, henceA Amax = 2 1 ] Q.3 If logAB + logBA2 = 4 and B <A then the value of logAB equals (A) 1 2  (B) 2 2 2  (C) 3 2  (D*) 2 2  [Hint: Let logA(B) =x x + x 2 = 4  x2 – 4x + 2 = 0  x = 2 8 16 4   = 2 ± 2  x = 2 – 2 ;  B <A hence logB < logA ; logAB < 1 ] Q.4 The sum of3 real numbers is zero. If the sum of theircubes is C then their product is (A) a rational greater than 1 (B) a rational less than 1 (C*) an irrational greater than 1 (D) anirrational less than 1 [Hint: a + b + c = 0  a3 + b3 + c3 = 3abc  abc = 3  which is irrational > 1  (B) ] Q.5 The sides of a triangleABC are as shown in the given figure. Let D be any internal point of this triangle and let e, f, and g denote the distance between the point D and the sides of the triangle. The sum (5e + 12f + 13g) is equal to (A) 120 (B) 90 (C*) 60 (D) 30 [Hint: 2 f 12 2 g 13 2 e 5   = 30 ( = 2 12 · 5 = 30) hence 5e + 12f + 13g = 60 Ans. ] Q.6 The value of tan27° + tan18° + tan27° tan18°, is (A)anirrationalnumber (B)rationalwhichisnotinteger (C)integerwhichisprime (D*) integer whichis not aprime. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-12
• 7. [Sol. Consider tan(27° + 18°) =       18 tan 27 tan 1 18 tan 27 tan 1 =       18 tan 27 tan 1 18 tan 27 tan  1 – tan27° tan18° = tan27° + tan18°  tan27° + tan18° + tan27° tan18° = 1 ] Q.7 In a triangleABC, the value of C sin B sin A cos + A sin C sin B cos + B sin A sin C cos (A*)is prime (B)is composite (C)isrational whichis notaninteger (D*)an integer [Hint: – [(cos B cos C – 1) + (cos C cosA – 1) + (cos A cos B – 1)] = 2 Q.869/6 If cos( + ) + sin( – ) = 0 and tan  = 2006 1 . Find tan . [3] [Ans. – 1] [Sol. cos  · cos  – sin  · sin  + sin  · cos  – cos  · sin  = 0 cos (cos  – sin ) + sin (cos  – sin ) = 0 (cos  – sin ) × (cos  + sin ) = 0 if cos  – sin  = 0  tan  = 1 which is not possible  sin  + cos  = 0  tan  = – 1 Ans. ] Q.9 ABC is aright angledtriangle.Show that sinA·sinB·sin(A–B)+sinB·sinC·sin(B–C)+sinC·sinA·sin(C–A)+sin(A–B)·sin(B–C)·sin(C–A)=0. [3] [Sol. Let B = 90° T1 = sinA· sin B · sin(A– B) = (sinA) · (1) · sin(A– 90°) T1 = – sinAcosA T2 = sin B · sin C · sin(B – C) = (1) (cos A) · sin(90° – C) = cos Asin A T3 = sin C · sinA· sin(C –A)                A sin C cos A cos C sin 0 B cos / 90 B sin = sin(90° –A) · sinA· sin(90° – 2A) T3 = sinA· cosA· cos2A T4 = sin (A– B) · sin (B – C) · sin(C –A) = – cos A · sin A · cos 2A ] Q.10(a) Solve (x + 2)(x – 2)(x – 13) = (x + 2)(x – 7)(x – 11) for x. (b) Solve (x – 3)(x – 2)(x – 13) = (x – 3)(x – 4)(x – 11) for x. [Sol. (a) (x + 2) [(x2 – 15x + 26) – (x2 – 18x + 77)] = 0 (x + 2)(3x – 51) = 0  x = – 2 or x = 17 [Ans. (a) x = – 2 or 17, (b) x = 3 ] Q.11 If m, n > 1 and for all x > 0 and x  1 lognx = 3 logmx. Writeanequationexpressingmexplicitlyin termsof n. [Sol. lognx = 3 m log x log n n if x  1  lognm = 3 m = n3 ]