GURU HARKRISHAN PUBLIC SCHOOL, LONI ROAD
PRE-MOCK EXAMINATION -2014
CLASS XII
SUBJECT-MATHEMATIC
ANSWER KEY
Section (A)
1. 3*4 = 2 (3)+ 4-3×4 = -2
2. tan-12sin2cos-132=tan-12sin2π6
= tan -1[2 sin π /3 ]
= tan -12.32
= π3
3. | adj A| = lAl3-1
= l A l2 =64
= l A l= ±8
4. If singular 5-xx+124= 0
20– 4x- 2x- 2 = 0
ð 6 x – 18 = 0
x = 3
5. Degree = 1
6. f(-x) = Sin 5(-x) = -Sin 5x
odd function
-π2π2sin5xdx=0
SECTION-B
7. Sin -1x + Sin – 1(1-x) = cos-1 x
= > Sin -1[ x1-(1-x)2+(1-x)1-x2=cos-1x [1]
· = > x2x-x2 +1-x1-x2=sincos-1x=1-x2 [1]
= > x2x-x2+ 1-x2(1-x-1)=0
= > x [2x-x2- 1-x2] = 0 [1]
= > x = 0 or 2x-x2 = 1-x2
= > x = ½ or 0 [1]
8 Given (a,b) * (c,d) = [a+c,b+d]
For commutative
Let (a,b), (c,d) ∈A
(a,b) * (c,d) = [a+c, b+d]
= [c+a, d+b]
= [c,d] * (ab) [1]
Hence. Commutative
For Associative let (a,b) , (c,d) (e,f) ∈ A
Then [ (a,b) * (c,d)] * (e,f) = [a+c, b+c]*(e.f)
= [(a + c) +e, (b+d) +f)
= [a+(c+e), b + (d+f)]
= (a,b) * [ c+e, , d+f)
= (a,b) *[ (c,d) * (e,f)]
Hence Associative [1 / ½]
For a,b,∈ N X N ,if (c,d) ∈ NXN is identify element, then
(a,b) * (c,d)= (c,d) *(a,b) = (a,b)
(a+c, b+d) = (c+a, d+b) = (a,b)
a+c= a, b+d = b, =>c=0 and d = 0
ð (0,0)∈ N X N is identify element but (0,0) ∈ NXN
No identify element [1 / ½]
Or
We have the given relation
R= { a,b) : l a- b l is a multiply of 4}
Where (a,b) ∈A and A=[0,1,2 …..12]
Reflexivity for any a ∈ A we have | a-a| = 0
Which is multiple of 4
ð (a,a) ∈R for all a∈R
So , R is reflexive [1]
For Symmetric fora,b∈A
(a,b) ∈R = ⇒ a-bisdivisible by 4⇒|- (b-a)| is divisible by 4, true
⇒|b-a| is divisible by 4
(a,b) ∈ R ⇒(b,a)∈R hence symmetric [1]
For transitive : For a,b,c∈A
(a,b) ∈R and b,c∈R
⇒ a-bisdivisible by 4and b-c is divisible by 4
⇒ a-c is divisible by 4
⇒(a,c) ∈ R Hence transitive.
Hence relation R is an equivalence relation. [2]
9.
Let p(n) = (a I + bA)n = a nI +nan-1 bA
P (1) :aI + bA = aI + bA [1]
Let P(k) be true ⇒(a I + bA)k = a nI +nak-1 bA
To show P(k+1 ) is true i.e (a I + bA)k+1 = a k+1 I+ (k+1) akbA [1]
LHS = (aI+ bA)k(aI+ bA)
= (akI + kak – 1bA) (aI+ bA)
= ak + 1 I2+ akbIA + kakbAI + kak – 1 b2 A2 A2 = 01000100=0000 [2]
OR
We have A = 2-334
A2=2-3342-334=-5-18187
6A=62-334=12-181824 [1]
17I=171001=170017
A2-6A+17I=-5-18187-12-181824+170017 [1]
=0000=0
A2-6A=-17I
17A-1=6I-A [1]
A-1=1176I-A
=1176006-2-334=11743-32 [1]
10.1+a2-b22ab-2b2ab1-a2+b22a2b-2a1-a2-b2
C1→C1-bC3,C2→C2+aC3
=1+a2+b20-2b01+a2+b22ab+a2b+b3-a-a3-ab21-a2-b2 [2]
=1+a2+b2210-2b012ab-a1-a2-b2 [1]
R3→R3-bR1
=1+a2+b2210-2b012a0-a1-a2+b2=1+a2+b23 [1]
11.
y=sin-1x1-x2
⇒1-x2 y=sin-1x
⇒1-x2dydx - x1-x2y = 11-x2 [2]
⇒(1-x2) dydx – xy =1
Again differentiating we get,
(1-x2)d2ydx2 – 2xdydx- xdydx – y =0
⇒(1-x2)d2ydx2 – 3xdydx– y =0 [2]
OR
log (x2+y2)=2tan-1yx
1x2+y22x+2ydydx=21+y2/x2[xdydx - yx2 ] [2]
⇒x+ydydx=xdydx-y
⇒x-ydydx=x+y
⇒dydx=(x+y)/x-y [2]
12.
x=acost+logtant2
dxdt=a-sint+1tant2sec2t2.12
=a-sint+1sint
=acos2tsint [1]
y=a(1+sint)
dydt=accost [1]
dydx=acostsintacos2t= tant [1]
d2ydx2=sec2tdtdx
=sec2tsintacos2t
=1asec4t sint [1]
13.
L.H.L=limx→0-f(x)=limh→0f(0-h)=limh→0f(-h)
=limh→01+k(-h)-1-k(-h)-h
=limh→01-kh-1+kh(1-kh+1+kh)-h(1-kh-1+kh)
=limh→0-2kh-h1-kh-1+kh =-2k2=k [2]
R.H.L= limh→0+f(x)=-1
f(0)=-1 [1]
Since f(x) is conts
...L.H.L=R.H.L=f(0)
...k=-1 [1]
14.
y=4sinθ2+cosθ-θ
dydθ=2+cosθ4cosθ-4sinθ(-sinθ)2+cosθ2-1 [1]
=8cosθ+42+cosθ2-1
=8cosθ+4-4-cos2θ-4cosθ2+cosθ2 [1]
=4cosθ-cos2θ2+cosθ2=cosθ(4-cosθ)2+cosθ2 [1]
In [0,π/2] cosθ>0
Also use -1≤cosθ≤1, dydx>0
...function is increasing [1]
15.
y=4x3-2x5
dydx=12x2-10x4 [1]
Let [x1,y1], tangent passes through origin
dydx]x1,y1, =12x12-10x14
Eq. of tangent
y-y1=(12x12-10x14)(x-x1) [1]
Passing through (0,0)
(0-y1)= (12x12-10x14)(0-x1)+( 4x13-2x15)
=8x15-8x13=0 [1]
... (0,0) (1.2)(-1,-2) [1]
16.
sinx+cosx9+16sin2xdx
Let Sinx-Cosx = t
(Cosx + Sinx)dx = dt
Squaring
1-sin2x = t2
1-t2=sin2x [1]
=dt9+16(1-t2)
=dt25-16t2) [1]
=14sin-14t5+C [1]
=14sin-14(sinx-cosx)5+C [1]
17.
dxcosx-acos(x-b)=1sin(a-b)sin(a-b)dxcosx-acos(x-b) [1]
=1sin(a-b)sin[x-b-x-a]cosx-acos(x-b)dx [1]
=1sin(a-b)tanx-b-tanx-adx [1]
=1sin(a-b)(log sec(x-b)-log sec(x-a)) + C [1]
18.
01cot-11-x+x2dx
=01tan-11-x+x1-x(1+x)dx [1]
=01(tan-11-x+tan-1x)dx
=01(tan-11-(1-x)dx+01tan-1xdx [1]
=201tan-1xdx
=2xtan-1x01-201x1+x2dx [1]
=2xtan-1x-22log(1+x2)01
=π2-log2 [1]
OR
01log(1+x)1+x2dx
Put x= tanθ, we get
I= 0π/4log(1+tanθ)dθ ……(i) [1]
Using property, 0afxdx=0afa-xdx
I= 0π/4log1+tanπ4-θdθ [1]
= 0π/4log1+1-tanθ1+tanθdθ
= 0π/4log21+tanθdθ …….(ii) [1]
Adding (i) and (ii), we get
2I= 0π/4log2dθ = π4log2
⇒ I= π8log2 [1]
19.
Given (x-y)dydx= x + 2y
dydx = x+2yx-y - (i)
Let y = vx, dydx= v+ xdvdx [1]
From (i), v+ xdvdx = 1+2v1-v
1-vv2+v+1dv = dxx [1]
Solving integration,
-12log(x2+xy+ y2) + 3tan-12y+x3x= C [2]
SECTION-C
20.
The given differential equation can be written as :
dxdy + (coty)x = cosy , which is a linear differential equation [1]
I.F = ecoty dy= siny [1]
=> The general equation is
xsiny = cosy.sinydy + C [1]
xsiny = - 14cos(2y) + C [1]
It is given that y = 0, when x = 0
ð C = ¼
ð The required equation is : x siny = 14(1 - cos(2y)) = 12sin2y [2]
21.
LHS=tanπ4+12cos-1ab+tanπ4-12cos-1ab
=1+tan12cos-1ab1-tan12cos-1ab + 1-tan12cos-1ab1+tan12cos-1ab [2]
=1+tan12cos-1ab2+1-tan12cos-1ab21-tan2(cos-1ab)=21+tan12cos-1ab21-tan2(cos-1ab) [2]
=2cos212cos-1ab = 2coscos-1ab=2ab= 2ba= RHS [2]
OR
tan-12tanα2.tanπ4-β21-tan2α2tan2π4-β2
= tan-12tanα2.tan1-tanβ21+tanβ21-tan2α2tan21-tanβ21+tanβ22 [112]
= tan-12tanα2.1-tan2β21-tan2α2+tan2β2+2tanβ2-tan2α2tan2β2+2tan2α2tan2β2 [112]
=tan-12tanα2.1-tan2β21-tan2α21+tan2β2+2tanβ21+tan2α2
=tan-12tanα21+tan2α2.1-tan2β21+tan2β21-tan2α21+tan2α2+2tanβ21+tan2β2 [112]
= tan-1sin α cos βcos α+sinβ [112]
22. Graph - [2]
Max Z = 22x + 18y
Sub to x + y ≤ 20
And, 360x + 240y ≤ 5760 or 3x + 2y ≤ 48 [2]
x ≥ 0, y ≥ 0
Feasible regions are O(0,0), A(0,20), P (8,12), B(16,0)
Corner Points / Z = 22x + 18y(0,0) / 0
(0,20) / 360
(8,12) / 392
(16,0) / 352
[2]
23.
Let award money for Honesty,Regularity and Homework be Rs x, Rsy,Rs z resp.
Acc. to ques,
x+y+Z=6000
x+3z=11000
x+z=2y
x-2y+z=0 [1]
11110-31-21xyz=6000110000
AX=B
X=A-1B
|A|=6≠0 [1]
... A-1 exist
Adj A=62-2-3033-2-1=6-3320-2-23-1
A-1=166-3320-2-23-1 [2]
X=166-3320-2-23-16000110000
xyz=50020003500
Award for Hardwork = Rs 3500
Award for Honesty = Rs 500
Award for Regularity = Rs 2000 [1]
The Value which should be included is Empathy. [1]
24.
Hypotenuse = H = AB = AP + PB ………..(i)
Now, cosecθ=APa⇒AP=a cosecθ
And BPb=secθ ⇒ BP=bsecθ
From (i), h = a cosecθ+ bsecθ [1]
dhdθ=a(-cosecθcotθ)+ bsecθtanθ [1]
For minimum h,dhdθ=0⇒ a cosecθcotθ= bsecθtanθ
⇒ab=tan3θ . ⇒ tanθ=ab13⇒sinθ=a13a23+b23, cosθ=b13a23+b23 [2]
Show that d2hdθ2> 0, for tanθ= ab13
Put in (ii) h = a.a23+b23a13+b.a23+b23b13 = a23+b233/2 [2]
OR
Profit P =x5-x100-x5+500 [1]
= 5x-x2100-x5 + 500
dpdx = 5-2x100-15 [2]
= -2x100+245=0
⇒2x100=245
x = 240 [1]
d2pdx2= -150<0 for x=240 [2]
25.
013x2+2x+1dx
a=0 ,b=1 ,nh=1
013x2+2x+1dx=limh→0hf0+fh+f2h+...…...fx-1h [1]
=limh→0h1+3h2+2h+1+3(2h)2+22h+1+... ….3x-1h2+2x-1h+1 [1]
=limh→0hx+3h2(x-1)x(2x-1)6+2hx(x-1)2 [2]
=limh→0nh+2nh-hnh2nh-h2+nhnh-h
=1 + 1 + 1 = 3 [2]
26.
Given circles are x2+y2=4 and x-22+y2=4.
Eliminating y from the two equations,we get
4-x2=4-x-22
⇒4-x2=4-x2+4x-4
⇒ x=1 [1]
Area = 2014-x-22dx+124-x2dx [2]
=2x-224-x-22+2sin-1x-2201+x24-x2+2sin-1x212 [1]
=2-32-2.π6+2.π2+2.π2-32-2.π6
= 22π-2π3-3=8π3-23sq units. [1]
11