# General Quiz (1- ) 12th ABCD.pdf

Q.1 A circular arch having width 24m and height 9m is to be constructed. What is the radius of the circle of which the arch is an arc? (A) 10m (B*) 12.5 m (C) 13.5m (D) 14m Q.2 The sum of the solutions on the interval (0, 2] to the equation: 0 = – (2 sin2 x + 2 cos2 x)  , is (cos x  0 )  cos x  (A*) 3 (B) 4 (C) 5 (D) 6 Q.3 Let XOY be a right triangle with XOY = 90°. Let M and N be the midpoints of legs OX and OY respectively. If XN = 19 and YM = 22, then XY equals (A*) 26 (B) 13 (C) 32.5 (D) 41 [Sol. (XY)2 = 4(a2 + b2) now, 4a2 + b2 = (19)2 a2 + 4b2 = (22)2  a2 + b2 = 169  4(a2 + b2) = (XY)2 = 4 × 169  XY = 2 × 13 = 26 Ans.] Q.4 Given parallelogram ABCD, with AB = 10, AD = 12 and BD = 2 . The length of (AC) is (A) 10 (B) 2 (C) 12 (D*) 2 [Hint: 2(a2 + b2) = d2 + d2 ] Q.5 ABC is a right triangle with hypotenuse AB and AC = 15 cm. Altitude CH divides AB into segments AH and HB, with HB = 16cm. The area of ABC, is (A) 120 cm2 (B) 144cm2 (C*) 150cm2 (D) 216cm2 [Sol. Equating the two values of p 15 sin  = 16 cot   15 sin2 = 16 cos  solving cos  = 3/5 (cos  = – 5/3 rejected)  sin  = 4/5 2512 x = 15 cos  = 9; p = 15 sin  = 12; Area = 2 = 150cm2] Q.6 Consider the eight digit number N = 11115556. Which of the following statements are true? I. N is divisible by 11 II. N – 9 is a prime III N is a perfect square (A) I (B) II (C*) III (D) I, II and III [Hint: I SE = 12; So = 13  not divisible by 11 II N – 9 = 11115556 – 9 = 11115547 = 3331 × 3337 Note : N – 9 = (3334)2 – 32 = (3337) × (3331)  N – 9 can not be prime III N = 11115556 is a perfect square of 3334 ] Q.7 The value of the series 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100, is (A) –100 (B) 0 (C) 1 (D*) 100 [Sol. 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100 = (1 – 3) + (2 – 4) + (5 – 7) + (6 – 8) + ... = (–2) × 50 = – 100 ] One or more than one is/are correct. Q.110/mod If x = t3 + t + 5 & y = sin t then (3t 2 + 1) sin t + 6 t cos t d2y dx2 = (3t 2 + 1) sin t + 6 t cos t (A*)  (3t 2 + 1)3 (B) (3t 2 + 1)2 (C)  (3t 2 + 1) sin t + 6 t cos t 2 (D) cost 2 Q.212/modIf y = (3t 2 + 1) 1 then 2x2 + 3x + 1 d2y dx2 at x =  2 is : 3t + 1 (A*) 38 27 1 (B)  38 27 2 (C) 27 38 L(2x + 2)  (2x + 1) (D) none [Hint : y = (2x + 1)(x + 1) = (2x + 1) (2x + 2) = 2 MN(2x + 1) (2x + 2) ] Q.3 If y2 = P(x), is a polynomial of degree 3, then 2  d  dx  y . d2y 2  equals :  dx  (A) P  (x) + P  (x) (B) P  (x) . P  (x) (C*) P (x) . P  (x) (D) a constant d [Sol: 2 (y3 y ) = 2 (y3.y + 3 y2 y y ). Now differentiate y2 = P(x) thrice)] dx 2 3 1 2 Q.425/modIf f (x) = x  2 & g (x) = f ( f (x)) then for x > 20, g  (x) = (A*) 1 (B)  1 (C) 0 (D) none [Hint: for x > 20, f (x) = x – 2 ; g () = f (x) – 2 = x – 4 ] Q.537/modPeople living at Mars, instead of the usual definition of derivative D f(x), define a new kind of derivative, D*f(x) by the form

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### General Quiz (1- ) 12th ABCD.pdf

• 1. Q.1 Acirculararchhaving width24m andheight 9m isto be constructed. Whatis the radius ofthecircleof which the archis an arc? (A)10m (B*) 12.5 m (C) 13.5m (D)14m Q.2 The sum ofthe solutions on theinterval (0, 2] to the equation: 0 = –             x cos x sin x sin x cos 2 x sin 2 3 2 2 , is (cos x  0 ) (A*) 3 (B) 4 (C) 5 (D) 6 Q.3 Let XOY be a right triangle with XOY = 90°. Let M and N be the midpoints of legs OX and OY respectively. If XN = 19 andYM = 22, then XYequals (A*) 26 (B) 5 13 (C) 32.5 (D) 41 [Sol. (XY)2 = 4(a2 + b2) now, 4a2 + b2 = (19)2 a2 + 4b2 = (22)2  a2 + b2 = 169  4(a2 + b2) = (XY)2 = 4 × 169  XY = 2 × 13 = 26 Ans.] Q.4 Given parallelogramABCD, withAB = 10,AD = 12 and BD = 13 2 .The length of (AC) is (A) 10 (B) 13 2 (C) 3 12 (D*) 109 2 [Hint: 2(a2 + b2) = 2 1 d + 2 2 d ] Q.5 ABC is arighttrianglewithhypotenuseABandAC =15cm.Altitude CH dividesABinto segmentsAH and HB, with HB =16cm.The area of ABC, is (A) 120 cm2 (B) 144cm2 (C*) 150cm2 (D) 216cm2 [Sol. Equatingthe twovalues ofp 15 sin  = 16 cot   15 sin2 = 16 cos  solving cos  = 3/5 (cos  = – 5/3 rejected)  sin  = 4/5 x = 15 cos  = 9; p = 15 sin  = 12; Area = 2 12 25 = 150cm2] Q.6 Consider the eightdigit numberN =11115556.Which of the followingstatements are true? I. N isdivisible by11 II. N – 9 is a prime III N is a perfect square (A) I (B) II (C*) III (D) I, II and III [Hint: I SE = 12; So = 13  not divisibleby11 II N – 9 = 11115556 – 9 = 11115547 = 3331 × 3337 Note : N – 9 = (3334)2 – 32 = (3337) × (3331)  N – 9 can not be prime III N = 11115556 is a perfect square of 3334 ] Q.7 The valueoftheseries 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100, is (A) –100 (B) 0 (C) 1 (D*) 100 [Sol. 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... – 99 – 100 = (1 – 3) + (2 – 4) + (5 – 7) + (6 – 8) + ... = (–2) × 50 = – 100 ]
• 2. One or more than one is/are correct. Q.110/mod If x = t3 + t + 5 & y = sin t then d y dx 2 2 = (A*)      3 1 6 3 1 2 2 3 t t t t t    sin cos (B)     3 1 6 3 1 2 2 2 t t t t t    sin cos (C)      3 1 6 3 1 2 2 2 t t t t t    sin cos (D) cost t 3 1 2  Q.212/modIf y = 1 2 3 1 2 x x   then d y dx 2 2 at x =  2 is : (A*) 38 27 (B)  38 27 (C) 27 38 (D) none [Hint: y = 1 2 1 1 ( )( ) x x   = 2 2 1 2 2 ( )( ) x x   = 2 ( ) ( ) ( )( ) 2 2 2 1 2 1 2 2 x x x x      L N M O Q P ] Q.323/modIf y2 = P(x), is a polynomial of degree 3, then 2 d dx       y d y dx 3 2 2 .       equals : (A) P  (x) + P  (x) (B) P  (x) . P  (x) (C*) P (x) . P  (x) (D) a constant [Sol: 2 d dx (y3 y2) = 2 (y3.y3 + 3 y2 y1y2). Now differentiate y2 = P(x) thrice)] Q.425/modIf f (x) = x  2 & g (x) = f ( f (x)) then for x > 20, g  (x) = (A*) 1 (B)  1 (C) 0 (D) none [Hint: for x > 20, f (x) = x – 2 ; g () = f (x) – 2 = x – 4 ] Q.537/modPeoplelivingatMars,insteadoftheusualdefinition ofderivativeDf(x),defineanewkindofderivative, D*f(x)bytheformula D*f(x) = Limit h f x h f x h    0 2 2 ( ) ( ) where f(x) means [f(x)]2. If f(x) = x lnx then D f x x e * ( )  has the value (A) e (B) 2e (C*) 4e (D) none [Hint: D*f(x) = 2f(x).f (x) D*(x lnx) = 2x lnx (1 + lnx)] Q.6503/mod If f(x) = x . x, then its derivative is : (A) 2x (B) 2x (C*) 2x (D*) 2x sgn x
• 3. Q.1127/mod If y = 1 x 3 x 1 x x 2 2 4     and dx dy = ax + b then the value of a + b is equal to (A) cot 8 5 (B*) cot 12 5 (C) tan 12 5 (D) tan 8 5 [Sol. y = 1 x 3 x x 3 ) 1 x ( 2 2 2 2     = x 3 1 x ) x 3 1 x )( x 3 1 x ( 2 2 2       dx dy = 2x – 3  a = 2 & b = – 3 a + b = 2 – 3 = tan 12  = cot 12 5 Ans. ] Q.2 Valueofsin 705° is (A) 4 2 6  (B*) 4 6 2  (C) 4 2 3  (D) 2 3 1 Q.3 Findallrealfunctionsofonerealvariablethatsatisfythefollowingfunctionalequation, f (x) + 2 f       x 1 = x [Sol. f (x) + 2 f       x 1 = x ....(1) Replacingxby1/x and multiplyby2 yields 2 f       x 1 + 4 f (x) = x 2 ....(2) Andso,subtractingequation(1) from(2) gives 3 f (x) = x 2 – x  f (x) =        x x 2 3 1 ] Q.4 In the figure, each region T represents an equilateral triangle and each region S a semicircle. The complete figure is a semicircle of radius 6 withitscentreO.Thethreesmallersemicirclestouchthelargesemicircle at pointsA, B and C.What is the radius of a semicircle S? [Sol. If we cut out part of the diagram and label appropriately we find that DL = r and DO = 3 r . This gives OB = r(1 + 3 ) = 6 or r = 3 1 6  = 3( 1 3  ) Ans. ] Q.5 Solve 3 3 28 x 6 28 x 6    = 2. [Ans. x = 6] [Sol. let a = 3 28 x 6  and b = 3 28 x 6  a – b = 2 (a – b)3 = 8 a3 – b3 – 3ab(a – b) = 8 (using a – b = 2) 56 – 6 · 3 28 x 6  · 3 28 x 6  = 8  48 = 6 · 3 28 x 6  · 3 28 x 6   83 = (6x + 28)(6x – 28)  36x2 – 784 = 512  36x2 = 1296  x2 = 36  x = ± 6 ]
• 4. Q.1130/mod If f (x) = x5 + 2x3 + 2x and g is the inverse of f then g'(–5) is equal to (A*) 13 1 (B) 1 (C) 7 1 (D) none [Sol. y = x5 + 2x3 + 2x dx dy = 5x4 + 6x2 + 2  dy dx = 2 x 6 x 5 1 2 4   = g ' (y) when y = – 5, x5 + 2x3 + 2x + 5 = 0 (x + 1)(x4 – x3 + 3x2 – 3x + 5) = 0  x = – 1  1 x 5 y dy dx     = 2 6 5 1   = 13 1 ] Q.2126/mod Suppose the function f (x) – f (2x) has the derivative 5 at x = 1 and derivative 7 at x = 2. The derivative of the function f (x) – f (4x) at x = 1, has the value equal to (A*) 19 (B) 9 (C) 17 (D) 14 [Sol. y = f (x) – f (2x) y' = f ' (x) – 2 f ' (2x) y'(1) = f ' (1) – 2 f ' (2) = 5 ....(1) and y'(2) = f ' (2) – 2 f ' (4) = 7 ....(2) now let y = f (x) – f (4x) y' = f ' (x) – 4 f ' (4x) y ' (1) = f ' (1) – 4 f ' (4) ....(3) substituting the value of f ' (2) = 7 + 2 f ' (4) in (1) f ' (1) – 2 [7 + 2 f ' (4)] = 5 f ' (1) – 4 f ' (4) = 19  (A) ] Q.3 Findthenumberoftriangles withsides of integerlength whoseperimeteris 10. (A) 4 (B) 3 (C*) 2 (D) 0 [Sol. Let the sides are x, y, 10 – (x + y), x, y  N Now x + y > 10 – (x + y) x + y > 5 ....(1) also 10 – x > x  0 < x < 5 ....(2) |||ly 0 < y < 5 ....(3) If x = 1, y can be 5, 6, 7..... not possible x = 2, y = 4, z = 4; x = 3, y = 3, z = 4  (2, 4, 4) & (3, 3, 4) ] Q.4 The value of the parameter m for whichthe equation 1 x 3 2 x   = m x 6 1 x 2   has no solution for x, is (A) 12 (B) 18 (C) 14 (D*) 13 [Sol. x = m 13 1 m 2   , for no solution m = 13 ] Q.5 A circle of area 20 is centered at the point indicated bya solid dot in the accompanying figure. Suppose that ABC is inscribed in that circle and has area 8. The central angles ,  and  are as shown. What is the value of sin  + sin  + sin ? (A*) 5 4 (B) 4 3 (C) 3 2 (D) 2  [Hint: 2 1 r2(sin  + sin  + sin ) = 8 and r2 = 20 ]
• 5. Q.195/mod Which of the following could be the sketch graph of y= d dx (x ln x)? (A) (B) (C*) (D) Q.278/mod If y = (A + Bx) emx + (m  1)2 ex then d y dx 2 2  2m dy dx + m2y is equal to : (A*) ex (B) emx (C) emx (D) e(1  m) x [Hint: multiplygivenequationby emx &then differentiatetwice] Q.3 Number ofsides in aregularpolygonifthesumofits interioranglesis 2520°, is (A) 14 (B*) 16 (C) 18 (D) 20 [Hint (n – 2) = 2520° external angle = n          n 2 = 2520°  n = 16 n – 2 = 2520°  (n – 2)180 = 2520  n – 2 = 14 n = 6] Q.4 The value of x such that 4(1 + y)x2 – 4x + 1 – y = 0 is true for all real values of y, is (A) – 1/2 (B) – 2/3 (C) 2/3 (D*) 1/2 [Sol. x = ) y 1 ( 8 ) y 1 ( 4   or ) y 1 ( 8 ) y 1 ( 4   ; hence x = 2 1 Ans. ] Q.5 The lengths of the sides of triangleABC are 60, 80 and 100 with A=90°.The lineAD divides triangleABC into two triangles ofequal perimeter. Calculatethelength ofAD. [Ans. 24 5 ] [Sol. 60 + x = 100 – x + 80 2x = 120  x = 60 now usingcosinelaw inACD y2 = (60)2 + x2 – 2x · 60 cos C (cosC = 3/5) = 3600 + 3600 – 2 · 60 · 60 · 5 3 = 7200 – 24 · 60 · 3 = 2880 y = 2880 = 24 5 Ans. Alternatively: sin 2 C = 60 · 2 y = 120 y y = 120 sin 2 C ....(1) now cos C = 5 3 1 – 2 sin2 2 C = 5 3  2 sin2 2 C = 5 2  sin 2 C = 5 1 . Hence from (1), y = 5 120 = 5 24 Ans. ]
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