1. No category

Solutions to CBSE Pariksha-2015 (Mathematics–XII)
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RSPL/1
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1. a = – b ⇒ vectors have same magnitude, but opposite directions. Hence | a | = | b | is
true.
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Converse may not be true, e.g. a = it + tj – 2kt and b = 2it + tj + kt have same magnitude, i.e.
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| a | = | b | but a ! b or a ! –b.
2.
Consider dy
dx
dy
=– a sin x
+ y tan x = – a sin x + a cos x. tan x
dx
Hence, solution=– a sin x + a sin x = 0
3 2 6
3. Direction cosines of perpendicular from origin are ± < , - , >
7 7 7
3 2
6
\ Equation is ± ( x– y + z ) = 7 ⇒ ± (3x – 2y + 6z) = 49.
7 7
7
4. Given equation is c
dy 4
d2 y
m + 3y 2 = 0
dx
dx
Order = 2, degree = 1
5. Given vector a = it + tj + 2kt
Vector of magnitude
6 $ at =
6 units in the direction of a is
6⋅
it + tj + 2kt
1+1+4
= it + tj + 2kt
6. Let point (x, y) lies on the line joining the points (3, –5) and (0, 4). As points lie on the
line so, these points are collinear.
x
y 1
1
1
⇒ 3 - 5 1 = 0 ⇒ [x(–9) – y(3) + 1(12)] = 0
2
2
0
4 1
⇒ –9x – 3y + 12 = 0 ⇒ 3x + y – 4 = 0
7. Number of students for thinking skills = x
Number of students for attitude towards school programmes = y
Number of students for participation in sports = z
According to the question
y + 2z=4 ⇒
0x + y + 2z=4
⇒
x + 2z=5
⇒
x + 0y + 2z=5
⇒
x + 2y =7 ⇒
x + 2y + 0z=7
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 1
Corresponding matrix equation is
0 1
>1 0
1 2
i.e.
4
2 x
2 H > y H = >5 H
7
0 z
AX=B, Its solution is X = A–1B
For A : –1
0 1
|A|= 1 0
1 2
-4
Adj A= > 4
2
\
A–1=
...(i)
2
2 = 0 – 1(–2) + 2(2) = 6 ≠ 0
0
2l
-4
1H = > 2
2
-1
4
-2
1
-4
1
2
>
6
2
2
2H
-1
2
-2
2
Adj A
|A |
=
4
-2
1
2
2H
-1
Substituting in (i), we get
⇒
X=
-4
1
> 2
6
2
4
-2
1
2 4
- 16 + 20 + 14
1
2H>5 H = > 8 - 10 + 14 H
6
8+5 -7
-1 7
18
x
3
1
> yH= 6 >12H = >2H
6
z
1
⇒ x = 3, y = 2, z = 1
\ Number of students awarded for thinking skills = 3
Number of students awarded for attitude towards school programme = 2
Number of students awarded for participation in sports = 1
Participation in physical activities and sports is of utmost importance as it inculcates
team spirit, fitness, decision making, sportsmanship. 8. Consider sin–1 x + sin–1 2x =
⇒
⇒
⇒
⇒
⇒
π
3
π
– sin–1x
3
π
2x =sin c - sin - 1 xm
3
π
π
2x =sin cos(sin–1x) – cos sin (sin–1x)
3
3
1
3
⋅ 1 - x2 – ⋅ x
2x =
2
2
sin–1 2x =
3
5x
=
2
2
1 - x2
2 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
Squaring both sides, we get
25x2 = 3(1 – x2)⇒28x2 = 3
⇒
x2 =
⇒
x =
3
⇒x=±
28
3
28
3
as x = 28
3
does not satisfy the given equation.
28
OR
Let x =a tan q ⇒ q = tan–1 x a
3a3 tan q – a3 tan3 q
3a 2 x – x 3 –1
3
tan )
= tan–1 )
3
a (a2 – 3a2 tan2 q)
a (a2 – 3x2)
= tan–1 )
a3 (3 tan q – tan3 q)
= tan–1 )
3 tan q – tan3 q
a3 (1 – 3 tan2 q)
2
1 – 3 tan q
x
= 3q = 3tan–1 a
ax + 1, x # 3
9. Consider function f(x) = )
bx + 3, x > 3
For continuity at x = 3
⇒
...(i)
3
3 = tan–1(tan 3q)
[from (i)]
Lim f(x) = Lim f(x) = f(3)
x → 3+
x → 3–
Lim f(3 – h) = Lim f(3 + h) = 3a + 1
h→0
h→0
⇒ Lim {a(3 – h) + 1} = Lim {b(3 + h) + 3} = 3a + 1
h→0
h→0
⇒
3a + 1 =3b + 3 = 3a + 1
⇒
3a + 1 =3b + 3 ⇒ 3a – 3b = 2
For values of a and b satisfying the relation 3a – 3b = 2, function is continuous at x = 3
10. Consider xy =yx
Taking log on both sides, we get
ylogx =xlogy
Differentiating both sides w.r.t. x, we get
⇒
⇒
y$
1
1
+ log x $ yl = x $ $ yl + log y $ 1
x
y
y
x
yl= log x– G = log y –
y
x
y′ =
y (x log y – y)
x (y log x – x)
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 3
11. Consider curves
y2 =4ax
and
xy =c2
...(i)
...(ii)
Differentiating (i) and (ii), with respect to x, we get
2y
dy
dy 2a
=4a ⇒
=
dx
dx
y
...(iii)
dy
dy
y
+ y.1 =0 ⇒
=– dx
dx
x
If curves cut at right angles, then from (iii) and (iv), we get
and
...(iv)
x
2a
y
× c- m =–1 ⇒ x = 2a
y
x
From (i) we get
...(v)
y2 =4a × 2a = 8a2
...(vi)
From (ii) squaring both sides, we get
x2y2 =c4
⇒
(2a) (8a2) =c4
⇒
32a4 =c4
2
12. Consider I =
cos x
dx
y cos
3x
=
y
cos x
3
4 cos x - 3 cos x
[from (v) and (vi)]
dx =
y
1
4 cos2 x - 3
dx
Dividing numerator and denominator by cos2x, we get
I= y
= y
sec2 x
1 - 3 tan2 x
y
1
3 1-t
2
dx =
y
sec2 x
4 - 3 - 3 tan2 x
dx
dx Let
dt ⇒
3 tan x = t
3 sec2x dx = dt
1
1+t
log
H+ c
1–t
3 2 ×1
1
=
>
1
2 3
13. Consider I= y
Using property
4 - 3 sec2 x
1
=
=
sec2 x
π
2
0
log
1 + 3 tan x
1 - 3 tan x
+c
x
dx sin x + cos x
y0a f (x) dx = y0a f (a - x) dx , we get
4 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
...(i)
I= y
π
-x
2
π
2
π
π
sin c - x m + cos c - x m
2
2
π
π
-x
2
2
= y
dx 0 cos x + sin x
Adding (i) and (ii), we get
π
π
2
2
2I= y
dx
0 sin x + cos x
0
π
=
2
=
=
=
⇒
y0
π
2
π
2 2
π
2 2
π
2 2
dx
...(ii)
1
dx
sin x + cos x
y0
y0
y0
π
2
1
1
2
π
2
π
2
sin x +
1
π
sin c x + m
4
1
2
dx
cos x
dx
π
cosec c x + m dx
4
π
π
π 2
I=
= log cosec c x + m - cot c x + m G
4
4
0
4 2
=
=
=
=
=
π
π
π π
π π
π
π
; log cosec c + m - cot c + m - log cosec - cot E
2 4
2 4
4
4
4 2
π
4 2
π
4 2
π
4 2
π
2 2
; log sec
2 +1
log
2 -1
$ 2 log
log
π
π
+ tan - log
4
4
=
π
4 2
2 -1 E
^ 2 + 1h
2
log
2-1
2 +1
2 +1
14. General equation of a plane through the intersection of planes x + 3y + 6 = 0 and
3x – y – 4z = 0 is
x + 3y + 6 + l(3x – y – 4z) = 0
⇒ (1 + 3l)x + (3 – l)y – 4lz + 6 = 0
...(i)
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 5
If distance of (i) from origin is one unit, then
(1 + 3l) × 0 + (3 – l) × 0 – 4l × 0 + 6
(1 + 3l) 2 + (3 – l) 2 + 16l2
=1
Squaring both sides, we get
⇒
36=1 + 9l2 + 6l + 9 + l2 – 6l + 16l2
26l2=26 ⇒ l2 = 1 ⇒ l = ±1.
When l = 1 from (i)
4x + 2y – 4z + 6 = 0, i.e. 2x + y – 2z + 3 = 0 is the required equation of the plane
When l = – 1. from (i)
– 2x + 4y + 4z + 6 = 0, i.e. x – 2y – 2z – 3 = 0 is the required equation of the plane
OR
Given line is 3x = 2y = – z
y
x
z
= =
2
3 -6
Line parallel to the given line and passing
or
A (1, –2, 3)
3x = 2y = – z
B
through A(1, –2, 3) is
y+2
x-1
z-3
=
=
=l
2
3
-6
General point on the line is B(2l + 1, 3l – 2, – 6l + 3)
If this point lies on the plane x – y + z = 5
1
7
3
–6
9 11 15
2
Point of intersection is B c + 1, - 2,
+ 3m , i.e. B c , – ,
m
7
7
7
7
7 7
then 2l + 1 – 3l + 2 – 6l + 3 = 5 ⇒ –7l = – 1 ⇒ l =
\ Distance AB =
2
2
2
9
15
- 11
+ 2m + c
- 3m
c - 1m + c
7
7
7
=
9 36
4
=
+
+
49 49 49
49
= 1 unit
49
15. Given vectors are a = 2x2 it + 4x tj + kt and b = 7it – 2 tj + x kt
cos q=
7 × 2x 2 + 4x × (- 2) + x × 1
4x 4 + 16x2 + 1 49 + 4 + x2
If angle between two vectors is obtuse then
14x2 – 8x + x < 0 ⇒ 14x2 – 7x < 0
⇒ 7 (2x – 1) < 0
6 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
x–y+z=5
⇒ x < 0, 2x – 1 > 0 or x > 0, 2x – 1 < 0
1
2
⇒ no solution ⇒ x < 0, x >
\ For 0 < x <
or x > 0, x <
or 0 < x <
1
2
1
2
1
angle between vectors a and b is obtuse.
2
16. A : girl students
n(A) = 430
n(B) = number of girls studying in class XII = 10% of 430 = 43
P(B/A) =
P (A + B)
43/1000
1
=
=
P (A)
430/1000 10
OR
Given integers are 1 to 11.
There are 6 odd numbers and 5 even numbers.
E : sum of two integers choosen is even
A : both even;B: both odd
5
6
C
C
P(E/A) = 11 2 ; P(E/B) = 11 2
C2
C2
Probability that both numbers are odd
6
P^ E/Bh
=
=
-7
17. Let A = =
4
P (E/A) + P (E/B)
6
5
1
G
-1
We have A = IA ⇒ =
C2
6
C2 + C2
=
=
C2
11
5
C2
6
C2
+
11
11
C2
C2
C2
15
15 3
=
=
10 + 15 25 5
1
1 0
-7
G==
GA
4 -1
0 1
1 -1
1 2
G= =
GA
4 -1
0 1
[By performing R1 → R1 + 2R2]
⇒ =
1 -1
1
2
G= =
GA
0
3
-4 - 7
[By performing R2 → R2 – 4R1]
⇒ =
1
2
1 -1
G= =
GA
0
1
- 4/3 - 7/3
⇒ =
⇒ \
1 0
- 1/3
G= =
=
0 1
- 4/3
A–1 = =
- 1/3
- 4/3
- 1/3
GA
- 7/3
[By performing R2 →
1
R]
3 2
[By performing R1 → R1 + R2]
- 1/3
G
- 7/3
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 7
OR
Consider equations 2X + 3Y = =
3X + 2Y = =
and
2
4
-2
1
3
G
0
...(i)
2
G
-5
...(ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting, we get
3(2X + 3Y) – 2(3X + 2Y) = 3 =
2
4
3
-2
G – 2=
0
1
⇒
6 9
-4
5Y= =
G–=
12 0
2
⇒
10
5Y= =
10
4
6+4
G==
12 - 2
- 10
5
G
10
1 10
Y= =
5 10
⇒
2
G
-5
5
2
G==
10
2
1
G
2
Substituting Y in (i) we get
2X= =
2
4
3
2
G – 3=
0
2
= =
2
4
3
6
G–=
0
6
= =
-4
-2
1 -4
=
2 -2
⇒
X=
Hence,
X= =
-2
-1
1
G
2
3
2-6
G==
6
4-6
3-3
G
0-6
0
G
-6
0
-2
G==
-6
-1
0
2
G; Y = =
2
-3
0
G
-3
1
G
2
18. Let A be first term and R be common ratio of a GP.
Then a = ARp–1, b = ARq–1, c = ARr–1
log a
Consider D= log b
log c
p 1
q 1
r 1
log AR p - 1
= log AR q - 1
log AR r - 1
p 1
q 1
r 1
8 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
9-4
G
0 + 10
log A + (p - 1) log R
= log A + (q - 1) log R
log A + (r - 1) log R
p 1
q 1
r 1
log A
= log A
log A
p 1 (p - 1) log R
q 1 + (q - 1) log R
r 1 (r - 1) log R
p 1
q 1
r 1
1
=log A 1
1
p 1
p-1
q 1 + log R q - 1
r 1
r-1
p 1
q 1
r 1
p 1
q 1
r 1
p
=log A × 0 + log R q
r
=0 + log R × 0 = 0
y
19. Consider
1
x2
(x2 + a2) (x2 + b2)
=
x2
1
=
dx =
y
a2 - b2 x2 + b2
=
=
=
=
a2 –b
1
a2 - b
1
a2 - b
1
a -b
2
y
(x2 + b2) - b2
(x2 + b2)
= y e1 2
a2 - b2
2
(x2 + a2) (x2 + b2)
>y
2
1
dx
x2 [(x2 + a2) - (x2 + b2)
y
2
=x -
b2
x2 + b
[Performing C1 → C1 + C3]
dx
x2
x2 + a2
dx -
y
dx G
(x2 + a2) - a2
(x2 + a2)
o dx - y e1 2
a2
x2 + a2
dx H
o dx G
a2
x
b2
x
tan - 1 - x +
tan - 1 G + c
a
a
b
b
= a tan - 1
x
x
- b tan - 1 G + c
a
b
OR
Consider
y
1
4
x +1
=
dx
1
2
1 (1 + x2) + (1 – x2)
dx = y
dx
y
2 x4 + 1
2
x4 + 1
1 x2 + 1
= = y
dx –
2 x4 + 1
y
x2 – 1
x4 + 1
dx G ...(i)
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 9
y
Consider
2
x +1
x4 + 1
1+
dx= y
2
x +
Consider
y
2
x –1
x4 + 1
dx =
y
2
x dx =
1
2
t +2
1
2
2
x +
= y
t
=
2
1
2
x dx =
1
x2
1
2
1+
1
x2
1 2
cx – m + 2
x
dx
dt tan –1
1–
y
x2
1
= y
=
1
y
1
2
tan –1
1–
1
=t
x
1
⇒ e1 + o dx = dt
x2
Let x –
1
2
x = 1 tan - 1 x - 1 2
2
2x
x–
1
x2
1 2
cx + m – 2
x
dx
dt 1
=t
x
1
⇒ e1 – o dx = dt
x2
Let x +
t –2
1
t– 2
=
log
2 2
t+ 2
1
x+ – 2
1
x2 + 1 - 2 x
1
x
=
log
=
log
1
2 2
2 2
x2 + 1 + 2 x
x+ + 2
x
Substituting from (ii) and (iii) in (i), we get
y
...(ii)
...(iii)
1 1
x2 – 1
1
x2 + 1 – 2 x
dx = =
tan –1
–
log
G+ c
2 2
2x 2 2
x4 + 1
x2 + 1 + 2 x
1
20. Given relation is R = {(P1, P2) ∈ A × A : P1 and P2 have same number of sides}
For reflexive: For P ∈ A
(P, P) ∈ R ⇒ P, P have same number of sides, true.
Hence, reflexive.
For symmetric: For P, Q ∈ A
(P, Q) ∈ R ⇒ P and Q have same number of sides
⇒ Q and P have same number of sides
⇒ (Q, P) ∈ R
As (P, Q) ∈ R⇒ (Q, P) ∈ R for P, Q ∈ A
Hence, symmetric.
For transitive: For P, Q, T ∈ A
Let (P, Q) ∈ R and (Q, T) ∈ R
10 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
⇒ P and Q have same number of sides and Q and T have same number
of sides
⇒ P and T have same number of sides
⇒ (P, T) ∈ R
As (P, Q) ∈ R, (Q, T) ∈ R ⇒ (P, T) ∈ R for P, Q, T ∈ A .
Hence, transitive.
As relation R is reflexive, symmetric and transitive.
Hence, relation R is an equivalence relation.
Set of all elements in set A related to right triangle T with sides 3, 4 and 5.
⇒ These elements should have same number of sides as right angled triangle has, i.e. 3
\ Set of all element in set A related to right triangle T is set of all triangles.
OR
Given set {0, 1, 2, 3, 4, 5} and binary operation
a * b = (a + b) mod 6, where (a + b) mod 6 is remainder obtained on dividing (a + b) by 6.
Operation table for *
*
0
1
2
3
4
5
0
0
1
2
3
4
5
1
1
2
3
4
5
0
2
2
3
4
5
0
1
3
3
4
5
0
1
2
4
4
5
0
1
2
3
5
5
0
1
2
3
4
From table we notice that a * 0 = a and 0 * a = a
Hence, 0 is identity for this operation.
For inverse, we notice.
5 * 1 = 0 ⇒ 5 * (6 – 5) = 0
4 * 2 = 0 ⇒ 4 * (6 – 4) = 0
3 * 3 = 0 ⇒ 3 * (6 – 3) = 0
2 * 4 = 0 ⇒ 2 * (6 – 2) = 0
1 * 5 = 0 ⇒ 1 * (6 – 1) = 0
In general
a * (6 – a) = 0
⇒ 6 – a is inverse of a.
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 11
21. Let square of side x cm be cut off from each corner to make a box.
Then
l=(45 – 2x) cm, b = (24 – 2x) cm, h = x cm ...(i)
x
x
x
x
24 cm
Volume of the box (V)= x(45 – 2x) (24 – 2x)
V= 2x(45 – 2x)(12 – x)
x
x
x
Differentiating both sides w.r.t. x, we get
x
45 cm
dV
=2[x(45 – 2x)(–1) + x(12 – x) (–2) + (45 – 2x) (12 – x) (1)]
dx
=2[–45x + 2x2 – 24x + 2x2 + 540 – 69x + 2x­2]
=2[6x2 – 138x + 540] = 12[x2 – 23x + 90]
=12(x – 18) (x – 5)
For maximum V,
⇒
d2 V
dV
= 0 ⇒ x – 18 = 0 or x – 5 = 0
dx
x=18 (rejected) [from (i)] or x = 5
dx2
d2 V
dx2
G
=12[(x – 18) ⋅ 1 + (x – 5) ⋅ 1] = 12 [2x – 23]
=12[10 – 23] < 0
x=5
Hence, for x = 5, volume is maximum.
Hence, a square of side 5 cm must be cut off from each corner to make a box of maximum
volume.
22. Given reigon {(x, y) : x2 ≤ y ≤ |x|}
Corresponding inequations are y ≥ x2, y ≤ |x|
Plotting the graph of inequations we notice we have to find the shaded area.
we notice both the curves y = x2 and y = |x|
are symmetrical with respect to y-axis as both the
functions are even with respect to x.
Y
\ Area = 2 × area in first quadrant.
For first quadrant, equations are y = x2, y = x,
(x > 0)
y=
y=
–x
x2
y=
Eliminating y, we get x2 = x ⇒ x(x – 1) = 0
⇒ x = 0, 1
\
Area=2 y (x - x2) dx = 2 =
1
0
1
x 2 x3
- G 2 3 0
1 1
1 1
= 2 ;c - m - 0E = 2 × = sq units.
2 3
6 3
12 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
–1
0
1
X
x
x2
23. Let ellipse be
y2
= 1, (a > b)
a2 b2
Differentiating both sides w.r.t. x, we get
⇒
1
a2
+
1
(2x) +
b
c 2y $
2
...(i) Y
F2
dy
m=0
dx
F1
0
X
y dy
x
y dy
b2
⇒ $
=–
=–
x dx
b2 dx
a2
a2
Again differentiating both sides w.r.t. x, we get
x) y $
d2 y
dx2
+c
dy 2
dy
m 3 – cy $
m1
dx
dx
x2
⇒ xy
d2 y
dx2
+ xc
=0
dy 2
dy
= 0 is the required differential equation.
m –y
dx
dx
OR
Consider equation (1 + y + x2y)dx + (x + x3)dy = 0
⇒ {1 + y(1 + x2)}dx + x(1 + x2)dy = 0
⇒ x(1 + x2)
⇒
dy
dx
=
dy
dx
= –{1 + y(1 + x2)}
-1
2
-
y
x
x (1 + x )
dy y
-1
⇒ + =
x
dx
x (1 + x2)
Here, P
(x) =
1
-1
, Q(x) =
x
x (1 + x2)
Integrating factor = e y
1
dx
x
= elog x = x
Solution is, (I.F)y= y {(I $ F) Q (x)} dx
-1
y x$
x.y=
xy= – tan–1x + c
x (1 + x2)
1
= - y
dx
1 + x2
dx
...(i)
When x = 1, y = 0
⇒ 0 = – t an–11 + c ⇒ c =
π
4
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 13
Substituting in (i), we get
xy = – tan–1x +
π
is the required solution. 4
x-1 3-y z+1
=
=
5
2
4
x-1 y-3 z+1
or
=
=
5
4
-2
Vector equation of the line is
24. Given line is
r =^it + 3tj - kth + l^5it - 2tj + 4kth ...(i)
As lines are parallel, so direction vector of required line is ^5it - 2tj + 4kth and line passes
through the point (3, 0, –4), i.e. point with position vector ^3it - 4kth
\ Vector equation of the line is
r = ^3it - 4kth + µ^5it - 2tj + 4kth ...(ii)
From line (i), a1 = it + 3tj - kt , b = 5it - 2tj + 4kt
From line (ii), a2 = 3it - 4kt , b = 5it - 2tj + 4kt
Distance between parallel lines =
(a2 - a1) × b
b
...(iii)
a2 - a1 = 3it - 4kt - it - 3tj + kt = 2it - 3tj - 3kt
it
^ a2 - a1h × b = 2
5
kt
- 3 = it (- 18) - tj (23) + kt (11) = - 18it - 23tj + 11kt
4
tj
-3
-2
^ a2 - a1h × b =
324 + 529 + 121 =
25 + 4 + 16 =
b =
45
974
units
45
\ Distance between parallel lines =
25.
974
{from (iii)}
1 ball
I
6R+5B
II
5R+8B
Case I: When ball transferred is red.
P` R II j =
5
13
Number of balls in bag I = (6 + 1) R + 5 B
P` B I j =
5
12
\ Probability in this case =
5
5
× 13 12
14 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
...(i)
Case II: When ball transferred is blue.
P` B II j =
8
13
Number of balls in bag I = 6 R + ( 5 + 1) B = 6 R + 6 B
P` B I j =
6
12
\ Probability in this case =
8
6
× 13 12
...(ii)
\Probability of drawing a blue ball from bag I when one ball is transferred from
5
5
8
6
[From (i) & (ii)]
bag II to bag I is
×
× +
13 12 13 12
25 + 48
73
=
=
.
156
156
26.
Flour
Fat
Cake I (x)
200 g
25 g
Cake II (y)
100 g
50 g
≤ 5 kg
≤ 1 kg
Let x cakes of type I and y cakes of type II are made.
Y
Then LPP is
50
40
To maximise Z = x + y
Subject to constraints
30
x ≥ 0, y ≥ 0
20
200 x + 100 y ≤ 5000 ⇒ 2x + y ≤ 50
10
25 x + 50 y ≤ 1000 ⇒ x + 2y ≤ 40
0
20
A(25, 0)
30 40
x 50
+
2y
y
=
X
=
40
50
Z=x+y
10
+
Point
B(20, 10)
2x
Plotting the graph of inequations we notice shaded portion
is feasible solution. Possible points for maximum Z are
A(25, 0), B(20, 10), C(0, 20)
C(0, 20)
Value
A(25, 0)
25 + 0
25
B(20, 10)
20 + 10
30
C(0, 20)
0 + 20
20
← Maximum
We notice for B(20, 10), i.e. x = 20, y = 10, Z is maximum.
Hence, 20 cakes of type I and 10 cakes of type II must be made for maximum number of
cakes within given constraints.
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 15
Solutions to CBSE Pariksha-2015 (Mathematics–XII)
RSPL/2
1. Given relation R = {(x, y) : x ≤ y3, x, y ∈ R}
Let a = 27, b = 4 and c = 2
(27, 4) ∈ R as 27 ≤ (4)3 ⇒ 27 ≤ 64, true
(4, 2) ∈ R as 4 ≤ (2)3 ⇒ 4 ≤ 8, true
Consider (27, 2) ∈ R ⇒ 27 ≤ (2)3 ⇒ 27 ≤ 8, false
\ (27, 4), (4, 2) ∈ R ⇒ (27, 2) ∈ R.
Hence, not transitive.
5p
5p
5p
as
∉ [0, p]
m!
4
4
4
5p
3p
cos–1 ecos o = cos–1 'cos c2p –
m1
4
4
2. Consider, cos–1 ccos
\
= cos–1 'cos
3.
P(A ∪ B)= P(A) + P(B) – P(A ∩ B)
⇒
⇒
3p
3p
3p
[Q
∈ (0, p)]
1=
4
4
4
0.6= 0.4 + 0.2 – P(A ∩ B)
P(A ∩ B)= 0 ≠ P(A)P(B)
Hence, events A and B are not independent.
4. For inequation 2x – 3y ≤ 6
For the point (1, –2), 2 + 6 ≤ 6 ⇒ 8 ≤ 6, false.
Therefore, we shade the portion which does not contain (1, –2).
5. Given lines are
x−2 y+1
x−1 y+3 z+5
; z = 2 and
=
=
=
3
–2
1
3
2
⇒ x–2
y+1 z–2 x−1 y+3 z+5
=
;
=
=
=
3
0
1
3
2
−2
\
cosq=
6. Consider
9+4+0 1+9+4
q= cos–1 e
\ 3 ×1 − 2 × 3 + 0 × 2
2x + 3
2x + 1
−3
182
=
−3
13 14
=
−3
182
o
x−3
=0
x+2
⇒ (2x + 3) (x + 2) – (2x + 1) (x – 3) = 0
⇒ 2x2 + 4x + 3x + 6 – 2x2 + 6x – x + 3 = 0
⇒ 12x + 9 = 0 ⇒ x = –
9
3
=–
12
4
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 1
7. Given a * b = a + b – ab ∀ a, b ∈ Q – {1}
For closure property: a + b – ab ∈ Q – {1}
i.e., a + b – ab ≠ 1 ⇒ a + b – ab – 1 ≠ 0
⇒ (a – 1) – b(a – 1) ≠ 0 ⇒ (a – 1) (1 – b) ≠ 0
⇒ a – 1 ≠ 0 and 1 – b ≠ 0 ⇒ a ≠ 1, b ≠ 1, true.
Hence, closure property holds.
Let e ∈ Q – {1} is identity element.
\ a * e = e * a = a
⇒ a + e – ae = e + a – ae = a
⇒ e – ae = 0 ⇒ e(1 – a) = 0
⇒ e = 0 as a ≠ 1.
\ Identity element is 0.
For inverse of a ∈ Q – {1}, Let b is its inverse.
\ a * b = b * a = 0
⇒ a + b – ab = b + a – ab = 0
⇒ a + b(1 – a) = 0
⇒ b =
a
is inverse of a.
a−1
3 sin 2a
1
m + tan–1 c tan a m
5 + 3 cos 2a
4
2 tan a
3$
3 sin 2a
1 + tan2 a
First, consider
=
5 + 3 cos 2a
1 − tan2 a
5 + 3e
o
1 + tan2 a
8. LHS = tan–1 c
=
=
6 tan a
1 + tan2 a
×
6 tan a
8 + 2 tan2 a
1 + tan2 a
5 + 5 tan2 a + 3 − 3 tan2 a
=
3 tan a
4 + tan2 a
3 tan a
1
o + tan–1 c tan a m
4
4 + tan a
R
V
3 tan a
1
S
+ tan a W
2
S
W
4
=tan–1 S 4 + tan a
W
S1 − 3 tan a c tan a m W
e
o
S
W
4
4 + tan2 a
T
X
3
–1 12 tan a + 4 tan a + tan a
=tan =
G
4 (4 + tan2 a) − 3 tan2 a
tan a (16 + tan2 a)
=tan–1 =
G
16 + tan2 a
LHS=tan­–1 e
2
=tan–1[tan a] = a = RHS
2 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
OR
Consider, cos –1
⇒
x
y
+ cos –1 = a
a
b
x y
cos –1 = . −
a b
1−
xy
−
ab
⇒
x2
a2
1−
x2
a2
Squaring, we get
x2 y2
a2 b2
x2 y2
⇒
a2 b2
1−
G= a
y2
b2
= cos a
1−
x2
a2
1−
y2
b2
2xy
x2
y2
cos a= e1 − oe1 − o
ab
a2
b2
+ cos2 a –
2xy
x2
y2
x2 y2
cos a= 1 –
–
+
ab
a2 b2
a2 b2
a2
x2
⇒
b2
+ cos2 a –
x2
⇒
y2
xy
− cos a =
ab
⇒
1−
a2
–
2xy
y2
cos a +
= 1 – cos2 a
ab
b2
–
2xy
y2
cos a +
= sin2 a.
ab
b2
9. Consider function f(x) = (x – 3) (x – 6) (x – 9); [3, 5]
(i) f is continuous in [3, 5] as polynomial function is continuous and product of continuous
functions is continuous.
f ′(x)= (x – 3) (x – 6) + (x – 6) (x – 9) + (x – 3) (x – 9)
(ii)
= x2 – 9x + 18 + x2 – 15x + 54 + x2 – 12x + 27
= 3x2 – 36x + 99
Q f ′(x) exists in (3, 5), h
ence, f is derivable in (3, 5)
Therefore, LMV Theorem is satisfied. Hence, there exists at least a point c ∈ (3, 5)
f (5) − f (3)
=f ′(c)
5−3
^5 − 3h (5 − 6) (5 − 9) − 0
=3c2 – 36c + 99
2
8
=3c2 – 36c + 99
2
such that
⇒
⇒
⇒
3c2 – 36c + 95=0
⇒
c=
⇒
c=
\
36 ! 1296 − 1140 36 ! 156
=
6
6
36 ! 2 39
39
= 6!
6
3
39
c=6 –
∈ (3, 5)
3
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 3
OR
Consider, f(x) =
x , then
f(x + Dx)=
f ′(x)=
Also,
Using approximations,
x + Dx , where Dx is small increment in x.
1
2 x
f (x + Dx)= f(x) + f ′(x) . Dx
⇒
x + Dx =
x+
Let x = 0.04, Dx = – 0.003
1
2 x
. Dx ...(i)
Substituting in (i), we get
0.04 – 0.003 = 0.04 +
⇒ 0.037 = 0.2 –
= 0.2 –
⇒
1
2 0.04
× – 0.003
1
× 0.003
2 × 0.2
0.03
= 0.2 – 0.0075
4
0.037 = 0.1925.
10. Consider, y = (x cos x)x + ^ x sin xhx
1
1
x
⇒ y= e log (x cos x) + e log (x sin x) x
1
y= e x log (x cos x) + e x
log (x sin x)
dy
1
=ex log (x cos x) ⋅ ; x $
" x $ ^− sin xh + cos x $ 1, + log (x cos x)E
dx
x cos x
1
+ ex
log (x sin x)
1
1
−1
$ (x cos x + sin x) + log (x sin x) e o3
) $
x x sin x
x2
1
=(x cos x)x {–x tanx + 1 + log(x cos x)} + (x sin x) x )
11. n =10
p = probability of defective item =
q =1 –
5
1
=
100 20
1 19
=
20 20
Probability of r successes, using binomial theorem,
P(r)= Cr c
10
19 10 − r 1 r
m
c m
20
20
4 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
cot x 1 1
+ − log (x sin x)3
x
x2 x2
P(not more than 1 defective) = P(0) + P(1)
10
0
9
=10C0 c 19 m c 1 m + 10C1 c 19 m c 1 m
20
20
20 20
= c
12. Consider,
y
=
y
=
y
sin − 1 x
(1 − x2) 3/2
dx
q
2
(1 − sin q)
q
3
cos q
19 9 19 10
29 19 9
m ; + E=
c m
20 20 20 20 20
3/2
Let x = sin q
$ cos q dq ⇒ dx = cos q dq
$ cos q dq = y q sec2 q dq I
=q ⋅ tan q –
II
1
y 1 $ tan q dq
x
q
=q tan q – log |sec q| + c
1−x
x
=sin–1x ⋅
1
1 − x2
+c
1
$ sin − 1 x + log 1 − x2 + c
2
1−x
x
=
2
13.
1−x
– log
2
2
–1
0
1
2
|x + 1|
(x + 1)
(x + 1)
(x + 1)
|x|
–x
x
x
|x – 1|
– (x – 1)
–(x – 1)
(x – 1)
2
y− 1 ( x + 1 +| x |+| x − 1 |) dx
0
1
2
=
y− 1 (x + 1 − x − x + 1) dx + y0 (x + 1 + x − x + 1) dx + y1
=
y–1 (2 − x) dx + y0 (x + 2) dx + y1
0
= =2x −
1
0
2
1
^ x + 1 + x + x − 1h dx
3xdx
2
x2
x2
3x2
G + = + 2x G + =
G
2 −1 2
2 1
0
1
1
4 1
9 19
=(0) – c− 2 − m + c + 2m − (0) + 3 c − m = 5 + =
2
2
2 2
2 2
y
y
+ x – y sin = 0
x
x
y
y
xy′ sin = y sin – x
x
x
14. Consider, equation xy′ sin
⇒
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 5
y
−x
x
⇒
y′ =
y
x sin
x
y
y sin − x
x
F(x, y) =
y
x sin
x
ly
ly $ sin − lx
lx
Consider, F(lx, ly) =
ly
lx $ sin
lx
y
l ; y sin − x E
x
=
= F[x, y]
y
l ; x sin E
x
\ Function is homogeneous, so, corresponding equation is homogeneous.
y sin
y = nx ⇒ y′ = n ⋅ 1 + x ⋅ n′
Let
⇒
n + xn′ =
⇒
xn′ =
⇒
⇒
...(i)
x
nx sin n − x n sin n − 1
=
x sin n
sin n
n sin n − 1
n sin n − 1 − n sin n
−n =
sin n
sin n
dn
1
= –
dx
sin n
sin n dn = –
dx
x
Integrating both sides, we get
⇒
⇒
y sin n dn = – y
– cos n = –log |x| – c
cos
When
⇒
dx
x
y
= log |x| + c
x
x = 1, y =
cos
...(i)
p
2
p
= log |1| + c ⇒ c = 0
2
Substituting is (i), we get
y
cos = log|x| is particular solution.
x
6 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
15. Given line L : r = 2it − 2tj + 3kt + l (it − tj + 4kt ) and plane p : r $ ^it + 5tj + kth = 5
If line L is parallel to the plane p, the line is perpendicular to normal to plane,
i.e. ^it − tj + 4kth ⋅ (it + 5tj + kt ) = 1 – 5 + 4 = 0, true
Hence, line is parallel to the plane.
Distance between line and a plane is the distance of point on line with position vector
(2it − 2tj + 3kt ) , from plane p.
(2it − 2tj + 3kt ) $ (it + 5tj + kt ) – 5
Distance = =
=
1 + 25 + 1
=
2 − 10 + 3 – 5
27
=
− 10
3 3
=
3
units
9
OR
General equation of the plane through point with position vector 2it – kt or point
(2, 0, –1) is
a(x – 2) + b(y – 0) + c(z + 1)=0 ...(i)
x
y−2
1−y
= z + 1 and x – 4 =
= 2z
=
4
2
−3
x
y−2 z+1
x−4 y−1 z
or
and
=
=
=
= , then
4
1
2
1
−3
−4
–3a + 4b + c=0
If plane (i) is parallel to the lines
and
⇒ i.e.
2a – 4b + c=0
a
c
a b c
−b
=
=
⇒ = =
4+4
8 5 4
− 3 − 2 12 − 8
a : b : c=8 : 5 : 4
Substituting in (i), we get
8(x – 2) + 5(y) + 4(z + 1) = 0 ⇒ 8x – 16 + 5y + 4z + 4 = 0
⇒ 8x + 5y + 4z – 12 = 0 is the required equation of the plane.
16. Let A : knows answer, B : guesses the answer
3
1
, P(B) =
4
4
E : answer correctly
P(A) =
1
4
Using Bayes' Theorem, probability that student knows the answer given that he answered
correctly, is
P(E/A) = 1; P(E/B) =
P (A) P (E/A)
P (A) P (E/A) + P^ Bh P (E/B)
3
×1
12
12
4
=
=
=
3
1 1 12 + 1 13
×1 + ×
4
4 4
P(A/E) =
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 7
OR
A : person suffers from TB. B : person is free of disease
P(A) =
1
999
; P(B) =
1000
1000
E : person is diagonised to have TB.
P(E/A) = 0.99 ; P(E/B) = 0.001
Using Baye’s Theorem,
P(A/E)=
P (A) P (E/A)
P (A) P (E/A) + P (B) P (E/B)
1
× 0.99
1000
=
1
999
× 0.99 +
× 0.001
1000
1000
=
17. Given
3 2
6 7
A = =
G and B = =
G
7 5
8 9
3 2 6 7
18 + 16
AB = =
G=
G==
7 5 8 9
42 + 40
= =
|A| =
adj A = =
\
34
82
21 + 18
G
49 + 45
39
G
94
3 2
= 15 – 14 = 1
7 5
\
990
990
110
=
=
990 + 999 1989 221
5
−7
−2
G
3
5 −2
1 5 −2
A–1 = =
G==
G
7
3
7
3
−
−
1
6 7
= 54 – 56 = –2
|B| =
8 9
adj B = =
9
−8
−7
G
6
1 9
B–1 = – =
2 −8
|AB| =
34
82
adj (AB) = =
94
− 82
...(i)
−7
G
6
39
= 3196 – 3198 = –2
94
− 39
G
34
8 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
...(ii)
\ (AB)–1 = –
− 39
G
34
1 94
=
2 − 82
...(iii)
From (i) and (ii)
1 9
B–1 A–1 = – =
2 −8
5
−7
G=
6 −7
1 45 + 49
= – =
2 − 40 − 42
1 94
= – =
2 − 82
−2
G
3
− 18 − 21
G
16 + 18
− 39
G = (AB)–1­­
34
[from (iii)]
Hence, (AB)–1 = B–1 A–1
OR
LetP(n) : (aI + bA)n = anI + nan–1bA
P(1) : aI + bA = aI + 1.a°.bA = aI + bA
Hence, P(1) is true
Let P(k) be true
i.e. P(k): (aI + bA)k = akI + kak–1bA
...(i)
To show P(k + 1) is true
i.e. P(k + 1) : (aI + bA)k + 1 = ak + 1I + (k + 1)akbA
Consider (aI + bA)k+1 = (aI + bA)k (aI + bA)
=(akI + kak–1bA) (aI + bA)
[from (i)]
=ak + 1 I2 + akbIA + kakbA I + kak – 1 b2A2
I2 = I, IA = AI = A, A2 = =
0
0
1 0
G=
0 0
1
0 0
G==
G
0
0 0
(aI + bA)k+1= ak + 1 I + akbA + kakbA + O
= ak + 1 I + (k + 1) akbA
Hence, P(k) true ⇒ P(k + 1) is ture, P(1) is also true.
Hence, statement is true for all natural numbers by the principle of mathematical
induction.
18.
Note Books
Pencils
Woollen Clothes
Group A:
2 dozen
1 dozen
4 dozen
Group B:
3 dozen
2 dozen
2 dozen
Group C:
1 dozen
5 dozen
6 dozen
Cost of each note book, pencil, woollen cloth is ` 20, ` 5 and ` 120.
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 9
Using matrices, we have
24 12 48
A
20
> 36 24 24 H > 5 H = > B
12 60 72 120
C
480 + 60 + 5760
A
⇒
=
720
B
120
2880
+
+
H >
>
240 + 300 + 8640
C
6300
A
⇒
> 3720 H = > B
9180
C
H
H
H
\Money spent by Group A, Group B and Group C is ` 6300, ` 3720 and ` 9180
respectively and total money spent = ` (6300 + 3720 + 9180) = ` 19,200.
Values reflected: helping needy, concern for others, kindhearted.
a
a+b
a + 2b
19. Consider a + 2b
a
a+b
a+b
a + 2b
a
3 (a + b) 3 (a + b) 3 (a + b)
=
a + 2b
a
a+b a+b
a + 2b
a
[Using R1 → R1 + (R2 + R3)]
1
1
1
= 3(a + b) a + 2b a
a+b a + b a + 2b
a
0
= 3(a + b) b
b
0
−b
2b
[By taking 3(a + b) common from R1]
1
a+b a
[By performing C1 → C1 – C3 and C2 → C2 – C3]
= 3(a + b) [0 – 0 +1(2b2 + b2)
[By expanding along R1]
= 9b2(a + b)
4−1
⇒ nh = 3
n
4 2
y (x + 3x) dx = Lim h 6 f (1) + f (1 + h) + f (1 + 2h) + ... + f "1 + ^n − 1h h,@ 20. Here, f(x) = x2 + 3x, a = 1, b = 4, h =
h→0
1
...(i)
f(1)=1 + 3 = 4
f(1 + h)=(1 + h)2 + 3(1 + h) = h2 + 5h + 4
f(1 + 2h)=(1 + 2h)2 + 3(1 + 2h) = 4h2 + 10h + 4
. .
. .
. .
f{1 + (n – 1)h}={1 + (n – 1)h}2 + 3{1 + (n – 1)h}
=(n – 1)2 h2 + 5(n – 1)h + 4
Substituting in (i), we get
y1
4
(x2 + 3x) dx = Lim h[4 + (h2 + 5h + 4) + (4h2 + 10h + 4) + ... + {(n – 1)2h2 + 5(n – 1) h + 4}]
h→0
10 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
= Lim h[h2 {1 + 4 + ... + (n – 1)2} + 5h{1 + 2 + ... + (n – 1)} + 4n]
h→0
(n − 1) n ((2n − 1)
^n − 1h n
+ 5h
+ 4n]
6
2
(nh − h) (nh) (2nh − h) 5
= Lim ;
+ (nh − h) (nh) + 4nhE
h→0
6
2
(3 − h) (3) (6 − h) 5
= Lim <
+ (3 − h) (3) + 4 × 3F
h→0
6
2
= Lim h[h2 ⋅
h→0
=
3×3×6 5
+ (3) (3) + 12
2
6
45
45 42 + 45 87
+ 12 = 21 +
=
=
2
2
2
2
21. Consider f(x) = sin x (1 + cos x)
= 9 +
\
f ′(x)=sinx(–sin x) + (1 + cos x) cos x
=–sin2 x + cos x + cos2 x
=2cos2 x + cos x – 1
For a point of local maximum or minimum
f ′(x)=0 ⇒ 2cos2 x + cos x – 1 = 0
⇒ (2cos x – 1)(cos x + 1) = 0
⇒
2cos x – 1=0 or cos x + 1 = 0
⇒
⇒
⇒
2cos x=1 or cos x = –1 ⇒ cos x =
p
cos x=cos c! m
3
p
x=± ∈ (–p, p)
3
f ′′(x)=–2 sin2x – sin x
1
or x = ± p (rejected) ∉ (–p, p)
2
p
p
2p
p
, f ′′ c m = –2 ⋅ sin
– sin < 0
3
3
3
3
p
\ Function attain local maximum at x =
3
p
p
3 3
p
Local maximum value = f c m = sin c1 + cos m =
3
3
4
3
p
For x = –
3
p
2p
p
f ′′ c – m =–2 sin c – m – sin c – m
3
3
3
2p
p
=2 sin
+ sin > 0
3
3
p
\ Function attains local minimum at x = –
3
p
p
p
\ Local minimum value = f c – m = sin c – m '1 + cos c− m1
3
3
3
For x =
= e –
3
3 3
3
o × = –e
o
2
4
2
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 11
OR
Let hypotenuse and other side be h and x respectively.
Given
then
\
x + h=S (say) ...(i)
h
third side= h2 − x2
1
1
Area (A)= x h2 − x2 = x (h + x) (h − x)
2
2
A=
x
q
S
S
x. S−x−x =
x S − 2x
2
2
Differentiating both sides w.r.t. x, we get
dA
S
1
=
(− 2) +
=x .
dx
2
2 S − 2x
=
For maximum area A,
⇒
S − 2x . 1G
S − x + S − 2x
S (S − 3x)
.
G=
=
2
2
S − 2x
S − 2x
dA
=0
dx
S – 3x=0 ⇒ S = 3x ⇒ x =
1
S
3
(− 2)
S>
H
2
S
2
x
−
=
2
2
(S − 2x)
dx
R
V
2S
S
W
S
(
3
)
0
−
−
−
d2 A
SS
W
3
=
G
S
W< 0
2
S
2
2
S
dx x =
S
W
cS −
m
3
S
W
3
T
X
S
\
for x= , area is maximum
3
⇒
3x=S ⇒ 3x = x + h
[from (i)]
x 1
1
p
⇒
2x=h ⇒ = ⇒ cos q = ⇒ q = .
h 2
2
3
p
Hence, the area of the triangle is maximum when the angle between them is .
3
Y
22. Given curves are x2 + y2 = 8x and y2 = 4x
y2 = 4x
S − 2x (− 3) − (S − 3x) .
2
d A
i.e. (x – 4)2 + y2 = 16
and
y2 = 4x
(x – 4)2 + y2 = 16
Circle: centre is (4,0); radius is 4;
On plotting the equations we notice we have to find
the shaded area.
Eliminating y from two equations, we get
x2 + 4x = 8x ⇒ x2 – 4x = 0
⇒ x(x – 4) = 0 ⇒ x = 0, 4
12 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
0
4
8
X
\
Area =
4
y0
" 16 − (x − 4) 2 − 4x , dx
4
x−4
16
x−4
2
16 − (x − 4) 2 + sin − 1 d
n − 2 $ x3/2E
2
2
4
3
0
0
4
4
−
= ; 16 + 8 sin − 1 0 − (4) 3/2E − ;
0 + 8 sin − 1 (− 1) − 0E
2
3
2
32
32
= – + 4p = c4p − m sq units.
3
3
= ;
a=ait + tj + kt
b=it + bjt + kt
and
g =it + tj + ckt
23. Given vectors are
If vectors are coplanar, then
a 1 1
1 b 1 =0
1 1 c
Preforming C1 → C1 – C2 and C2 → C2 – C3, we get
a−1
0
1
1 − b b − 1 1 =0
0
1−c c
⇒ (a – 1) [c(b – 1) – (1 – c)] – 0 + 1[(1 – b) (1 – c) – 0] = 0
⇒ (a – 1) (b – 1)c – (a – 1) (1 – c) + (1 – b) (1 – c) = 0
⇒ (1 – a) (1 – b)c + (1 – a) (1 – c) + (1 – b) (1 – c) = 0
Dividing both sides by (1 – a) (1 – b) (1 – c), we get
c
1
1
= 0
+
+
1−c 1−b 1−a
⇒ 1 − (1 − c)
1
1
= 0
+
+
1−c
1−b 1−a
⇒ 1
1
1
= 0
−1+
+
1−c
1−b 1−a
⇒
1
1
1
= 1
+
+
1−a 1−b 1−c
24. Let Q(a, b, g) be the image of the point P(2, 3, 7) in the plane 3x – y – z = 7.
P(2, 3, 7)
Let PQ meets the plane at R.
Then
(i)R is mid-point of PQ
(ii) R is foot of the perpendicular from the point P to the
plane.
R
As R is mid-point of PQ, therefore, coordinates of R are
c
a+2 b+3 g+7
,
,
m
2
2
2
...(i)
3x – y – z = 7
Q(a, b, g)
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 13
Also, as PR is perpendicular the plane 3x – y – z = 7
\ Direction ratios of PR are 3, –1, –1
x−2
y−3
z−7
=
=
= l (say)
3
−1
−1
General point on the line is R(3l + 2, –l + 3, –l + 7)
\ Equation of PQ is
...(ii)
If this lies on the plane 3x – y – z = 7,
then 3(3l + 2) – 1(–l + 3) – 1(–l + 7) = 7
⇒ 9l + 6 + l – 3 + l – 7 = 7
⇒ 11l = 11 ⇒ l = 1
Substituting in (ii), we get
Foot of perpendicular R(5, 2, 6)
\ c
a+2 b+3 g+7
,
,
m = (5, 2, 6)
2
2
2
[from (i)]
g+7
a+2
b+3
= 2;
=6
= 5,
2
2
2
⇒ a = 8, b = 1, g = 5
⇒
\ Image is Q(8, 1, 5)
25.
Vitamin A
Vitamin C
Cost (in `)
Food I (x)
2
1
50
Food II (y)
1
2
≥10
70
≥8
Y
Let x kg of food I and y kg of food II are mixed. Then LPP is
14
To minmise Z = 50x + 70y
12
Subject to constraints
10
x ≥ 0, y ≥0
2x + y ≥8
x + 2y ≥10
8
6
B(2, 4)
x+
2
4
y=
2
2
4
6
8
A(10, 0)
10 12 14
X
+
y=
8
Z = 50x + 70y
0
10
2x
On plotting the graph of inequations, we notice shaded
portion is optimum solution. Possible points for minimum
Z are A(10, 0), B(2, 4) and C(0, 8)
Point
C(0, 8)
Value
A(10, 0)
500 + 0
500
B(2, 4)
100 + 280
380
C(0, 8)
0 + 560
560
← Minimum
Z is minimum for B(2, 4), i.e. x = 2, y = 4. Hence, 2kg of food I and 4kg of food II must be
mixed for a minimum cost of ` 380, to meet the requirments.
14 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
26. Consider equation (1 + y2) + (x – etan
⇒ (1 + y2)
–1y
)
dy
=0
dx
dx
–1
+ (x – etan y) = 0
dy
−1
⇒
dx
x
e tan y
+
=
dy 1 + y2
1 + y2
Here, P(y) =
1
1 + y2
, Q(y) =
e tan
− 1y
1 + y2
Integrating factor (I.F.) = e
y
1
1 + y2
dy
= etan
–1y
Solution is,
(I.F.)x =
x etan
–1y
=
=
y {(I.F.) $ Q (y)} dy
y
e tan
− 1y
$
e tan
− 1y
1 + y2
dy
Let etan
y t.dt t2
= + c
2
⇒
–1y
etan
⇒ x =
⋅ x =
⇒
e tan
–1y
− 1y
1 + y2
=t
dy = dt
1 tan–1y 2
(e
) +c
2
1 tan–1y
–1
e
+ ce–tan y is the required solution.
2
OR
Consider equation
Let x + y = t dy
= sin(x + y) + cos(x + y)
dx
dy
dt
dy
dt
⇒ 1 +
= ⇒ =
−1
dx dx
dx dx
...(i)
Substituting in (i), we get
⇒
⇒
dt
− 1 =sin t + cos t
dx
dt
=1 + sin t + cos t
dx
dt
=dx
1 + sin t + cos t
Integrating both sides, we get
y 1 + sindtt + cos t = y dx
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 15
⇒
⇒
⇒
⇒
⇒
⇒
y
1
1+
2z
1−z
2
+
1 + z2 1 + z2
y
×
2
1+z
2
dz = y dx 2
dz = y dx
1 + z + 2z + 1 − z2
y 1 1+ z dz = y dx
2
log |1 + z|=x + c
t
log|1 + tan |=x + c
2
x+y
=x + c is the required solution.
log 1 + tan
2
16 Solutions to CBSE Pariksha-2015 (Mathematics-XII)
Let tan
t
=z
2
⇒ t = 2 tan–1 z
2
⇒ dt =
.dz
1 + z2
2z
sin t =
1 + z2
cos t =
1 − z2
1 + z2