Sample question paper 2 with solution
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Sample question paper 2 with solution

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CBSE Class XII Mathematics sample question paper with solurtion

CBSE Class XII Mathematics sample question paper with solurtion

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Sample question paper 2 with solution Sample question paper 2 with solution Document Transcript

  • Sample Question Paper MATHEMATICS Class XII Time: 3 Hours Max. Marks: 100 General Instructions 1. All questions are compulsory. 2. The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each, section B comprises of 12 questions of four marks each and section C comprises of 07 questions of six marks each. 3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 4. There is no overall choice. However, internal choice has been provided in 04 questions of four marks each and 02 questions of six marks each. You have to attempt only one of the alternatives in all such questions. 5. Use of calculators is not permitted. You may ask for logarithmic tables, if required. SECTION-A 1. If f(x) is an invertible function, find the inverse of f(x)=3x-2.  −1  2. find the principal value of sin −1    2  x + 3y 3. If   7−x y   4 −1 find the value of x and y. = 4  0 4     4. A matrix A of order 3X3 has determinant 5. what is the value of 3A . 5 .Find ∫ x + cos 6 x d ( x) 3 x 2 + sin 6 x 6. Evaluate ∫ xe x dx . Pratima Nayak,KV Teacher
  •         7. If p is a unit vector and x + p . x − p =. 80 )( ( ) 8.write the direction cosine of a line equally inclined the three co-ordinate axes. 9.write the value of the following determinant. a −b b−c c −a b−c c−a c−a c −a a −b b−c ( ) ( ) ˆ ˆ 0 ˆ ˆ 10. find the value of p, if 2i + 6 ˆ + 27 k X i + 3 ˆ + pk = j j SECTION-B 5 2  1 −1  1  11.Let the value of 2 tan −1   + sec −1   7  + 2 tan  8   5     OR π  x −1  −1  x + 1  Solve for x, tan −1  ,x≤  + tan  =1  x−2  x+2 2 13.Uasing properties of determinants prove that 2a 2a a −b−c 2b 2b b−c−a = 2c 2c c −a −b (a + b + c) 3 14.Show that the function f(x)= x + 2 is continuous at every x ∈ R but fails to be differentiable at x=-2. OR Verify lagrange`smean value theorem for the following function f(x)= x 2 + 2 x + 3 for [ 4, 6] 15.Differential following function w.r.to x  1 + x2 − 1 tan −1   x     OR Differentiate ( cos x ) sin x + ( sin x ) cos x with respect to x Pratima Nayak,KV Teacher
  • 16.Evalute π x sin x ∫ 1 + (cos x) 2 dx 0 17.Solve the following differential equation ( x 2 − y 2 )dx + 2 xydx = 0, y (1) = 1 18.The volume of a spherical balloon is increasing at the rate of 25 cm3 / sec. Find the rate of change of its surface area at the instant when its radius is 5cm. 19.Form the differential equation representing the family of parabola having vertex at origin and axis along positive of X-axis.      ˆ ˆ 20.Find the projection of b + c on a when a = i − 2 ˆ + k , b = + 2 ˆ − 2k and 2ˆ j ˆ i j  ˆ ˆ j c = 2i − ˆ + 4k . 21.Find the equation of the plan passing through the points (0,-1,0),(1,1,1) and (3,3,0). 22.The probability of A solving a problem is 3/7 and that B solving it is 1/3 what is the probability (i) at least one of them solve the problem? (ii)only one of them will solve the problem? SECTION-C (23) Using matrices solve the following system of linear equations2x-y+z=3, -x+2y-z=-4, x-y+2z=1 OR Using elementary transformations, find the inverse of the following matrix 1 3 − 2 − 3 0 − 5   2 5 0   (24)Determine the points on the curve y = x2 which are nearest to point (0, 5) 4 OR Pratima Nayak,KV Teacher
  • Show that the surface area of a closed cuboid with square base and given volume is maximum when it is a cube. (25) Find the area of the region included between parabola y 2 =x and the line x +y =2. a  a−x  (26) Prove that- ∫   a + x  dx = π a − a  (27) Mona wants to invest at most Rs.12000 in Saving certificate (SC) and National saving bonds (NSB).She has to invest at least Rs.2000 in SC and at least Rs.4000 in NSB. If the rate of interest on SC is 8 pa. and the rate of interest on NSB 10pa, how much money should she invest to earn maximum yearly income? Also find the maximum income. (28) Find the foot of perpendicular drawn from the point A (1, 0, 3) to the join of the points B (4, 7, 1) and C (3, 5, 3) and also find the perpendicular distance. (29) Urn A contains 1 white, 2 black, 3 red balls; Urn B Contains 2 white ,1 black ,1red ball; Urn C contains 4 white ,5 black and 3 red balls. One urn is chosen at random and two balls are drawn, these happens to be one white and one red. What is the probability that they come from urn C. Solution SECTION-A 1. f(x)=3x-2 Let y =f(x) y =3x-2 3x+2 =y f −1 ( x) =3x+2 2. y = sin −1 (−1/ 2) = −1 (1/ 2) = 6 − sin −π / 3. x + 3y  7−x  4. 3 A = 33 A = 27 × 5 = 135 y   4 −1 7, − = ⇒ x + 3 y = y = 1, 7 − x = ⇒ x = y = 1 4, − 0 4  0 4     Pratima Nayak,KV Teacher
  • 5. I = 1 x + cos 6 x log ( 3 x 2 + sin 6 x ) +c dx = 2 6 + sin 6 x ∫ 3x ∫ xe dx = x ∫ e x dx − ∫ 1.e x dx = xe x − e x + c 6. I= 7. ( x + p ) . ( x − p ) = 80 ⇒ 8. l= m= n= ± 9. Taking operation C1 → C1 + C2 + C3  x    2 2 2  x − p = 80 ⇒ x = 81 ⇒ x = 9 1 3 a −b b−c c −a b−c c−a c−a c −a a −b b−c 10 = 0 2 0 ( 2iˆ + 6 ˆj + 27kˆ ) X (iˆ + 3 ˆj + pkˆ ) =⇒ 1 = 6 = 27 ⇒ P = 27 3 P 2 SECTION-B 5 2  1 −1  1  11. 2 tan −1   + sec −1   7  + 2 tan  8  =  5      1 1  5 2   +  1 1  2  tan −1 + tan −1  + sec −1 = 2 tan −1  5 8  + tan −1  7   1 1 5 8    1 − .   5 8 2 5 2    7  −1     2   3 1  +   3  π −1  1  −1  1  −1 −1  1  −1  4 −1 7 . = 2 tan   + tan  = tan   =  + tan   = tan  = tan (1)  3 1 1 4 3 7 7 1 − ×  1 −   4 7  9 Value= π 4 OR  x −1   x +1  π tan −1  + tan −1  =  x − 2   x + 2 4  Pratima Nayak,KV Teacher
  •  x −1 x +1  +   π −1 x−2 x+2  = ⇒ tan  1 −  x − 1   x + 1   4   x − 2  x + 2       ( x − 1)( x + 2 ) + ( x − 2 )( x + 1)  π ⇒ tan −1  =  ( x − 2 )( x + 2 ) − ( x − 1)( x + 1)  4 ⇒ 1 x2 + x − 2 + x2 − x − 2 2x2 − 4 = 1⇒ == 1⇒ x ± 2 2 x − 4 − x +1 −3 2 12. R= {(T1 , T2 ) : T1 ≅ T2 } (i) R is reflexive:T1 ≅ T1 ⇒ (T1 , T1 ) ∈ R ∴ R is reflexive (ii) R is symmetric:Let (T1 , T2 ) ∈ R ⇒ T1 ≅ T2 ⇒ T2 ≅ T1 ⇒ (T2 , T1 ) ∈ R ⇒ R is symmetric. (iii) R is Transitive:Let (T1 , T2 ) ∈ R & (T2 , T3 ) ∈ R ⇒ T1 ≅ T2 , T2 ≅ T3 ⇒ T1 ≅ T3 ⇒ (T1 , T3 ) ∈ R ∴ R is Transitive. ∴ R is equivalence relation. a −b−c 2a 2a 13. = ∆ 2b b−c−a 2b 2c 2c c−a −b R1 → R1 + R2 + R3 a+b+c a+b+c a+b+c = ∆ b−c−a 2b 2b 2c 2c c −a −b Taking (a + b + c) common 1 1 ∆ = (a + b + c) 2b b − c − a 2c 2c 1 2b c−a−b Pratima Nayak,KV Teacher
  • C1 → C1 − C 2 , C 2 → C 2 − C 3 0 0 1 ∆ = (a + b + c) a + b + c − (a + b + c) 2b c−a−b a+b+c 0 1 0 0  1 −1 ∆ = (a + b + c) 1 − 1 2b = (a + b + c) 3 0 − 0 + 1  0 1   0 1 c−b−a 3 ∆ = (a + b + c) 3 14. f ( x) = x + 2 f (x) =x+2 if x>-2 f(x) =0 x=-2 if f(x)=-(x+2) if x<-2 For continuity Case(i) When c>-2 lim x →c = lim x →c ( x + 2) = c + 2 = f (c) ∴ f ( x)is continuous for all c > -2 Case(ii) When c <- 2 then lim x →c f ( x) = lim x →c − ( x + 2) = −(c + 2) = f (c) ∴ f(x) is continuous for all c <- 2 Case(iii) When c=-2 lim x →−2+ f ( x) = lim h→0 f (−2 + h) = lim h→0 (−2 + h + 2) = 0 lim x →−2− f ( x) = lim h→0 f (−2 − h) = lim h→0 − (−2 − h + 2) = 0 ∴ lim x →−2 f ( x) = f (−2) ∴ f ( x)is continuous at x=-2 ∴ f ( x)is continuous for all x ∈ R Pratima Nayak,KV Teacher
  • For differentiability at x=-2 Rf ` (−2) = lim h→0 Lf (−2) = lim h→0 f (−2 + h) − f (−2) = lim h→0 h (−2 + h − 2) − 0 =1 h (2 + h − 2) − 0 f (−2 − h) − f (−2) = −1 = lim h→0 −h −h Lf ` (−2) ≠ Rf ` (−2) ⇒ f(x) is not diff. at x=-2 OR f(x)= x 2 + 2 x + 3 for [ 4, 6] ⇒ f ` ( x) = 2 x + 2  F(x) is a polynomial function ∴ f(x) is continuous for x ∈ [4,6] and differentiabl for x ∈ (4,6) Now ∃x = c ∈ (4,6) s.t. f ` (c ) = f (b) − f (a ) ⇒ 2c + 2 = 12 ⇒ c = 5 ∈ (4,6) b−a ∴ Lagranges theorem satisfied.  1 + x 2 − 1 15.y= tan −1   x     Let x= tan θ Y=  1 + tan 2 θ − 1 θ  θ 1 −1 −1  sec θ − 1  −1 1 − cos θ  −1  tan −1   = tan   = tan  sin θ  = tan  tan 2  = 2 = 2 tan x tan θ  tan θ          y= 1 dy 1 tan −1 x ⇒ = 2 dx 2(1 + x 2 ) OR Let y= ( cos x ) sin x + ( sin x ) cos x y=u+v Pratima Nayak,KV Teacher
  •  u = (sin x) cos x ⇒ log u = cos x log(sin x) ⇒ d log u cos x cos x + log(sin x)(− sin x) = sin x dx 1 du du sin x cos x ] = cos x. cot x − sin x. log sin x ⇒ = (sin x) cos x [ ( cos x ) + ( sin x ) u dx dx v = (cos x) sin x ⇒ log v = sin x log(cos x) ⇒ 1 dv 1 (− sin x ) + cos x log(cos x) = sin x × cos x v dx dv = (cos x) sin x [− sin x tan x + cos x log cos x ] dx dy du dv = + dx dx dx ⇒ dy sin x cos x ] + (cos x) sin x [− sin x tan x + cos x log cos x ] = (sin x) cos x [ ( cos x ) + ( sin x ) dx π π π x sin x (π − x) sin(π − x) (π − x) sin x dx ⇒ I = ∫ dx ⇒ I = ∫ dx 2 2 2 0 1 + cos (π − x ) 0 1 + cos x 0 1 + cos x 16. I= ∫ π I= ∫ 0 π sin x 1 + cos x 2 π x sin( x) dx 1 + cos 2 ( x) 0 π π sin( x) ∫ ∫ 1 + cos 2I= π dx − ∫ 2 0 ( x) dx ⇒ I = 1 π sin x dx 2 ∫ 1 + cos 2 x 0 Let cosx=t ⇒ − sin xdx = dt −1 [ ] [ ] 1 −1 dt ⇒ π tan −1 t −1 = π tan −1 (1) − tan −1 (− 1) 2 1 1+ t I= π ∫ I= 17. π2 4 dy − ( x 2 − y 2 ) = dx 2 xy Let y=vx ⇒ ⇒v+x dy dv =v+x dx dx 2v dv − x 2 (1 − v 2 ) dx = ⇒ dv = − 2 2 dx x 2x v 1+ v Pratima Nayak,KV Teacher
  • ⇒∫ dx 2v dv = − ∫ ⇒ log(1 + v 2 ) = − log x + c ⇒ log( x 2 + y 2 ) − log x = c …….(1) 2 x 1+ v Now put x=1,y=1 C=log2 From (1)  x2 + y2  2 2 log   = log 2 ⇒ ( x + y ) = 2 x x   18.Let any time t,radius=r,volume=V,surface area=S g.t. dV ds = 25cm 3 / sec, = ? dt dt 4 dV dr 1  dr   V = πr 3 ⇒ = 4πr 2 ⇒  = 3 dt dt  dt  r =5 4π S = 4πr 2 ⇒ ds dr = 8πr dt dt  ds  ⇒   = 10cm 2 / sec  dt  r =5 19. Equ. Of parabola y 2 = 4ax ………(i) d.w.r.to x dy 4a y dy = ⇒a= dx 2 y 2 dx From equ. (i) y = 2x dy dx   ˆ  ˆ ˆ ˆ j 20. a = i − 2 ˆ + k , b = + 2 ˆ − 2k , c = 2i − ˆ + 4k . 2ˆ j ˆ i j   ˆ ˆ j ⇒ b + c = 3i + ˆ + 2k      2  b + c .a Projection of b + c on a = = Ans.  a 14 ( ) 21. points (o,-1,0) ,(1,1,1) ,(3,3,0). Pratima Nayak,KV Teacher
  • Equ. Of plan passing through point (0,-1,0) A(x-0) + B(y+1) + C(z-0) = 0 ……….(i) Point (1, 1, 1)and (3,3,0) satisfy equ.(i) we get A + 2B + C = 0……….(ii) 3A + 4B + 0.C = 0…….(iii) Solve eque. (i)@ (ii) A B C = = =λ⇒ A −4 3 −2 A = - 4λ , B = 3λ , C = - 2λ Putting in (1) -4 λ x + 3 λ (y + 1) - 2 λ z = 0 -4x + 3 (y + 1) - 2z = 0 4x + 3y + 2z-3 = 0 Ans. 22. Given that probability of A solving the question P(A) = Probability of B solving the question P (B) = − ∴ P( A ) =1 - 1 3 3 4 = 7 7 − and 3 7 P( B ) = 1 - 1 2 = 3 3 − − (1) Probability that at least one of them solve the problem = P(A B ) + P( A B) + P(AB) = 3 2 4 1 3 1 . + . + . 7 3 7 3 7 3 = 6 + 4 + 3 13 = 21 21 Pratima Nayak,KV Teacher
  • − − (2) Probability that only one of them will solve the problem = P(A B ) + P( A B) = 3 2 4 1 . + . 7 3 7 3 = 6+4 21 = 10 Ans. 21 SECTION-C (23) Given equations can be written as  2 −1 1   x  3  − 1 2 − 1  y  = − 4       1 −1 2  z  1       ⇒ AX=B → (1) A =4 ≠ 0 ⇒ A −1 exits. A 11 =(-1) 1+1 (4-1)=33 Similarly other cofactors can be obtained.  3 1 − 1 adjA 1  = 1 3 1 Adj. A=  4 A − 1 1 3    From (1) X= A −1 B Pratima Nayak,KV Teacher
  •  3 1 − 1  3  X   Y  = 1  1 3 1   − 4 ⇒    4  − 1 1 3   1  Z       1  = − 2 ⇒ x=1, y=-2, z=-1    − 1   OR A =AI R 1 → R 1 +3R 1 1 3 − 2  1 0 0 0 9 − 11 = 3 1 0 A      2 5 0  0 0 1      R 1 → R 1 +3R 3 , R 2 → Applying R 3 → R 3 -2R 1 1 9 11 R 2 R 3 → R 3 +R 2 R 3 → R 3R 2 →R 2 + R3 9 25 9 R 1 → R 1 -10 R 3 1 0 0 ⇒ 0 1 0  =   0 0 1    −2 5 4 25 1 25   1 − 2   5 −3  5  − 3 5  11   25  9  25   ⇒ I=BA ⇒ B is inverse of A (24)let (x,y) be the foot of perpendicular on the given curve which is nearest to (0,5) [ D= ( x − 0) + ( y − 5) 2 2 ] 1 2 Pratima Nayak,KV Teacher
  • If D is minimum then D 2 = x 2 +(y-5) 2 =E is also minimum dE = 2y-6 dY For minimum distance At y=3 dE = 0 ⇒ y=3 dY d 2E =2 〉 0 dy 2 ⇒ for y=3 ,distance is minimum ⇒ x= ± 2 3 Reqd. point is ( ± 2 3 ,3 ) OR Let x be the side of the square base & y be the height of cuboid V=x 2 y ⇒ y = V x2 Surface area (S)=2x 2 +4 xv / x 2 =ds/dx=4x-4v/x 2 1 For minimum ds/dx=0 => x 3 =v ,x=v 3 1 3 3 S’’ =4+8v/x when x =v =4+8 >0 therefore S is ,minimum 3 When x 3 =v=>x =x 2 y =>x =y There when S is mini mum cuboid is cube. Q25 given curves are y 2 =x and x+y =2 solving points of intersection aare (1,1),(4,-2) 1 The area of shaded region ∫ ( x 2 − x1 )dy −2 1 = ∫ (2 − y − y 2 )dy −2 =9/2 sq units. Pratima Nayak,KV Teacher
  • (26) a Given integral = ∫ −a a a (a − x ) 2 2 dx − ∫ −a xdx (a 2 − x ) 2 Putting x=asin θ ,dx=acos θdθ ,and changing limits in second integral Π Π Π Given integral= a  − (−  +a [cos θ ] 2 =a Π 2 2 −Π 2 (27) Let she invests Rs. x in saving certificates and Rs. y in National saving bonds Then LPPis To maximize Z=0.08x+0.1y Subject to constraints x ≥ 2000, y ≥ 4000, x + y ≤ 12000 corner points of feasible region ABC are A(2000,4000), B(8000,4000),c(2000,10000) at A, Z=160+400=560 at B, Z=640+400=1040 at C,Z=160+1000=1160 thus Rs. 2000 should be invested in saving certificates and Rs.10000 in National saving bonds. Maximum yearly income is Rs. 1160. Pratima Nayak,KV Teacher
  • (28)Let A(1,0,3),B(4,7,1),C(3,5,3) be the given points . Let P be the foot of perpendicular from A on BC. If P divides BC in k:1 then coordinates of P are ( 3k + 4 5k + 7 3k + 1 ) , , k +1 k +1 k +1 d.r.’s of BC are 1,2,-2 d.r.’s of AP are 2 k + 3 5k + 7 − 2 , , k +1 k +1 k +1 since AP ⊥ BC therefore k= −7 4 5 7 17 thus coordinates of P are ( , , ) 3 3 3 reqd. ⊥ dis tan ce = AP = 117 7 17  5 2 2 2 ( 3 − 1) + ( 3 − 0) + ( 3 − 3)  = 3   (29) Let E1,E2 ,E3 be the events that the balls are drawn from urn A, urn B, urn C respectively and let E be the event that balls drawn are one white and one black Then P(E1)=P(E2)=P(E3)= P(E/E1)= 1 3 C (1,1) * C (3,1) 1 = C (6,2) 3 Similarly P(E/E2)= 1 2 , P(E/E3)= 11 3 Using Bayes’ Theorem reqd. probability= P(E3/E) 2 15 33 = ( )= 1 1 2 59 + + 15 9 33 Pratima Nayak,KV Teacher