Marking Scheme (1St Pre-Board 2017-18 )

MARKING SCHEME (1ST PRE-BOARD 2017-18 )

SUBJECT: MATHEMATICS

SECTION : A

(1)  17 (2) 1 (3) 1 (4) X-22=Y-3-3= Z+5-2

SECTION : B

(5) ffx= f 3-x313 (1 mark) = x (1 mark)

(6) x=2 & y =9 (1mark for each value)

(7) R.H.L = 10 (1 mark) & k = 95(1 mark)

(8) xy= asin-1t+cos-1t= aπ2 (1 mark) &dydx= -yx (1 mark)

(9) I .F =ecotx dx (1 mark) = elogsinx= sinx (1 mark)

(10) x2=y3=z4 (1 mark) &r= μ2i+3j+4k (1 mark)

(11) For graph(1 mark) & Z = 16 at (0,4) (1 mark)

(12) P( exactly one of them solve ) = P(A)P(B )P(C ) + P(A ) P(B) P(C ) + P(A ) P(B ) P(C) (1 mark)

= 2556(1 mark)

SECTION : C

(13) R1→R1+R2+R3

L.H.S = 2(a+b+c)2(p+q+r)2(x+y+z)c+ar+pz+xa+bp+qx+y(1 mark)

= 2 (a+b+c)(p+q+r)(x+y+z)c+ar+pz+xa+bp+qx+y (1 mark)

R2→R1-R2 ; R3 → R1- R3

= 2 (a+b+c)(p+q+r)(x+y+z)bqycrz (1 mark)

R1→R1-R2-R3

=2apxbqycrz = RHS (1 mark )

(14) RHL = 2 (1 mark)

LHL = 2 ( 1 mark )

f(0) = 2 (1 mark)

∴RHL = LHL = f(0) (1mark )

(14) OR dxdy = sina+ycosy-siny cos(a+y)sin2a+y( 2 marks )

dxdy = sin(a+y-y)sin2a+y (1 mark)

dydx = sin2a+ysina (1 mark )

(15) Point of intersection of given curves = k23 ,k13 (1 mark )

For 1stcurve ,dydx =12k13 = m1say ( 1 mark )

For 2ndcurve ,dydx = -k13k23 = m2say ( 1 mark )

m1m2=-1

=>12k13-k13k23= -1

=> 8 k2=1 (1 mark)

(16) Let I=0πxa2cos2x+b2sin2x dx

I= 0ππ-xa2cos2(π-x)+b2sin2(π-x) dx

2 I=0ππa2cos2x+b2sin2x dx( 1 mark )

I=π0π21a2cos2x+b2sin2x dx( 1 mark )

I= π 0π2sec2xa2+b2tan2x dx (1mark)

Put b tanx =t

I = π22ab(1 mark)

(16) OR Let ,I = π6π3dx1+tanx

I =π6π3cosxcosx+sinx dx

I = π6π3cosπ3+π6-xcosπ3+π6-x+sinπ3+π6-x dx(1 mark )

I =π6π3sinxsinx+cosx dx

2I =π6π31 dx( 1 mark )

I =12xπ6π3( 1 mark)

I = π12 (1 mark)

(17) Figure 1/2

Equation of sides of∆ABC are

Y = 2 (x-1 ) ; y = (4-x) ; y = 12x-1 (12+12+1 2mark)

Area (∆ABC)= 12 2x-1dx+ 234-xdx-13x-12dx ( 1 mark )

= 32unit2 (1 mark )

(18) dydx = yx -cosec yx

Put yx=vdydx =v+x dvdx (1 mark )

1xdx = sinv dv

=> logx= -cos yx + C ( 1 mark )

Put x = 1 and y = 0 , then C = 1 (1mark)

∴particular solution : logx+cosyx= 1 (1mark)

(19) AB= i+x-2j+4 k ; BC=2-xj-7k ; CD=2i+3j+k ( 12+12+12mark )

[AB ,BC ,CD ]=0 (Coplanar) ( 12mark)

=>1x-2402-x-7231=0 (1mark)

=> x = 5 (1mark)

(20) a1= i+2j+k , b1= i-j+k (12mark )

a2=2 i-j-k , b2=2 i+j+2k (12mark )

a2-a1= i-3j-2k (12mark )

b1 Xb2= -3i+3k (12mark )

b1 X b2= 32 (12mark )

a2-a1 . b1 X b2 = - 9 (12mark )

S.D = a2-a1 . b1 X b2b1 X b2 (12mark )

=322unit (12mark )

(21) Let , no of Desktop model = x units & no of Portable model = y units

x≥0 , y ≥0 , x+y ≤250 (12mark )

25 000 x + 40 000 y ≤ 70 00 000

=>5x+8y ≤1 400 (12mark )

Maximise profit , Z=4 500 x+5 000 y (12mark ) For correct graph (1mark)

By Corner Point Method

Maximum , Z = 11 50 000 at ( 200 , 50 ) (112mark)

(22) Let E be the event that the man reports that six occurs in the throwing of the die and let S1 be the event that six occurs and S2 be the event that six does not occur .

Then , PS1= 16 , PS2= 56 ; P ES1= 34 ; P ES2=1- 34=14 ; P S1E= ?(2marks )

By Bayes Theorem , PS1E= PS1 . P ES1PS1 . P ES1+ PS2 P ES2 (1mark)

= 16 X 3416 X 34 + 56 X 14= 38 (1 mark)

(23) (i) PX=1 => 10 k2+9k-1=0 => k=-2neglecting& k=110(1 mark)

(ii) PX<3=3k= 310=0.3 (1mark)

(iii) P( X> 6 ) = 7 k2 + k = 17100=0.17 (1mark)

(iv) P( 0 < X < 3 ) = 3k =310 = 0.3 (1mark)

SECTION : D

(24) (i) For commutative : (2 marks)

(ii) For associative : (2marks)

(iii) For no Identity : (1mark)

(iv) (2,3)* (5,7) = (2+5 , 3+7) = (7,10) (1mark)

(25) 1-1202-33-24–20192-361-2= 1000100o1 (3 marks)

1-1202-33-24xyz=112 (1mark)

=>xyz= 1-1202-33-24-1112 =–20192-361-2112=053 (1mark)

=> x=0 , y=5 and z=3 (1mark)

(26) x logx dydx+y= 2x logx

=>dydx + 1xlogx y= 2x2 (1mark)

P = 1xlogx& Q = 2x2 (1mark)

I.F = ePdx = e1xlogxdx = logx (1mark)

General Solution : y × I .F = Q × I .Fdx +C (1mark)

Applying Product Rule of Integration (1mark)

Solution : y logx = -2x1+logx+C (1mark)

(27) Let , x = radius of the cone & y = height of the cone

For figure : (1mark )

y-R2 + x2= R2=>x2=2Ry-y2 (1 mark)

Let z = the volume of the cone

∴ z= 13πx2y =π32Ry-y2y = π32Ry2-y3 (1mark)

dzdy= π34Ry-3 y2= 0=> y =4R3 (1mark)

d2zdy2 = π34R-6y , at y =4R3 the value of d2zdy2=-4R3 (Negative) (1mark)

Maximum volume z of the cone =8274π3R3= 827Volume of the sphere(1mark)

(28) Let , I = 0π2tanx+cotxdx

= 0π2sinx+cosxsinx cosx dx (1mark)

= 20π2sinx+cosx1-(1-2sinx cosx) dx (1mark)

= 20π2sinx+cosx1-sinx-cosx2 dx (1mark)

Put sinx-cosx=t

∴ I =2-11dt1-t2 (1mark)

=2sin-1t-11 (1mark)

=2π (1mark)

(29) Given line in Cartesian form : x+12 = y-33= z-1-1

Let the pointQ (α,β,γ ) be the foot of perpendicular drawn from given point P(5,4,2) to the given line

∴α+12 = β-33= γ-1-1= μ say=>α=2μ-1; β=3π+3 ; γ=-μ+1 (1mark)

Now ,PQ=α-5i+β-4j+γ-2k perpendicular to 2i+3j-k

=>α-52+β-43+γ-2-1=0

Putting the value α=2μ-1; β=3π+3 ; γ=-μ+1 in this equation ,we get

μ=1 (1mark)

Coordinate of the foot of perpendicular Q =Q( 1,6,0) (1mark)

Length of perpendicular = 5-12+4-62+2-02 = 26 units (1 mark )

Let R ( p , q ,r ) be the image of the point P (5 ,4,2 ) w.r.t.the given line.

∴ 1= p+52=> p = -3

6 = q+42=>q= 8

0 = r+22=> r = -2

∴ Required image = (-3 , 8 , -2 ) (2mark)

Alternative methods are accepted. Proportional marks are to be awarded.