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SOLVED PAPERS

MARCH -2024 (AP)

« MATHS 2B 2
AP-IPE 2024 SOLVED PAPER «

PREVIOUS PAPERS

IPE: MARCH-2024(AP)
Time : 3 Hours MATHS-2B Max.Marks : 75
SECTION-A
I. Answer ALL the following VSAQ: 10 ´ 2 = 20
1. Find the equation of the circle with (4, 2), (1, 5) as ends of a diameter.
2. Find the value of k if the length of the tangent from (2,5) to x2 + y2 − 5x + 4y + k = 0 is
37 .
3. 2 2 2 2
Find k if the pairs of circles x +y +4x +8=0, x +y –16y +k =0 are orthogonal.
4. Find the coordinates of the point on the parabola y2=8x, whose focal distance is 10.

-Q
5. If the angle between the asymptotes is 30º then find its eccentricity.


T
2
6. Evaluate ∫sec2x.csc2xdx 7. Find ∫ elog(1+tan x)dx . 8. Find ∫ cos 7 xsin 2 xdx

E
9. Find the area of the region enclosed by the given curves x = 4 – y2, x = 0.
0

L
6/5
⎛ d 2 y ⎛ dy ⎞3 ⎞
10. Find the order and degree of ⎜ 2 +⎜ ⎟ ⎟ = 6y

L
⎜ ⎝ dx ⎠ ⎠⎟
⎝ dx
SECTION-B
II. Answer any FIVE of the following SAQs:
U 5 ´ 4 = 20
11.
B
Find the length of the chord intercepted by the circle x2+y2–x+3y–22=0 onthe line y = x – 3
12. S.T the circles x2+y2–8x–2y+8 = 0,x2+y2–2x+6y+6 =0 touch each other and find the point of contact.

Y
13. Find the equation of tangent and normal to the ellipse 9x2+16y2=144 at the endof the
latus rectum in the first quadrant.
B
A 2 2
14. If P(x, y) is any point on the ellipse x2 + y2 = 1 with foci S & S' then SP + S'P is a constant.

B
a b
15. Find the centre, eccentricity, foci, length of latus rectum and equations of the directrices of x2 – 4y2 = 4

sinx 2 2
16. Evaluate ∫ sinx + cosx dx 17. Solve dy + 3x y = 1+ x

3 dx 3
1+ x 1+ x
SECTION-C
III. Answer any FIVE of the following LAQs: 5 ´ 7 = 35
18. Find the values of c if the points (2,0), (0,1), (4,5), (0,c) are concyclic.
19. Find the equation to the pair of transverse common tangents to the circles
x2 + y2 - 4x - 10y + 28 = 0 and x2 + y2 + 4x - 6y + 4 = 0
20. Define parabola. Derive its equation in the standard form.
21. Evaluate the reduction formula for In= òsinnxdx and hence find òsin4xdx
x +1
22. Evaluate ∫ 2 dx.
x + 3x + 12
π /4 x
sinx + cosx ⎛ x⎞
∫ 24. Solve ⎛⎜
x⎞ y
23. Evaluate 9 +16sin2x
dx
y⎟
dx + e ⎜1 − y ⎟ dy = 0
0 ⎝1+ e ⎠ ⎝ ⎠
« BABY BULLET-Q 3
AP-IPE 2024 SOLVED PAPER «

IPE AP MARCH-2024
SOLUTIONS
SECTION-A

1. Find the equation of the circle with (4, 2), (1, 5) as ends of a diameter.

Sol: Formula: The equation of the circle with A(x1,y1), B(x2,y2) as ends of a diameter is

(x−x1)(x−x2)+(y−y1)(y−y2)=0

∴ Equation of the required circle is (x−4)(x−1)+(y−2)(y−5)=0

⇒ (x2−x−4x+4)+(y2−5y−2y+10)=0⇒ x2+y2−5x−7y+14=0
-Q
T
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
2.
E
Find the value of k if the length of the tangent from (2,5) to x2+y2-5x+4y+k=0 is 37 .

Sol: Length of the tangent from (2, 5) to

L
S = x2+y2−5x+4y+k = 0 is L
U
S11 = 37 ;

B
On squaring both sides, we get S11 = 37

⇒ (2) 2 + 52 − 5(2) + 4(5) + k = 37 ⇒ 4 + 25 − 10 + 20 + k = 37

⇒ 39 + k = 37 ⇒ k = −2 Y
B
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
3.
A
Find k if the pairs of circles x2+y2+4x+8 = 0, x2+y2–16y+k = 0 are orthogonal.

Sol:
B
From the given circles, we get

g = 2, f = 0, c = 8 and g' = 0, f ' = –8, c' = k

Orthogonal condition: 2gg' + 2ff' = c+c' ⇒ 2(2)(0) + 2(0)(–8) = 8 + k ⇒ k = –8

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
4. Find the coordinates of the point on parabola y2 = 8x, whose focal distance is 10.

Sol: Given parabola is y2 = 8x ⇒ 4a = 8 ⇒a=2

Given focal distance SP = 10

Formula: Focal distance SP = x1+a ⇒x1 + 2 =10 ⇒ x1= 8.

But, y12 = 8x1 ⇒ y12 = 8(8) ⇒ y1 = ±8

∴ P(x1,y1) = (8, ±8)

« MATHS 2B 4
AP-IPE 2024 SOLVED PAPER «

5. If the angle between the asymptotes is 30º then find its eccentricity.

Sol: The angle between the asymptotes of the hyperbola S = 0 is 2Sec–1e

∴ 2Sec–1e = 30º ⇒ Sec–1e = 15º ⇒ e = sec15º= 6 − 2

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
6. Evaluate òsec2x.csc2xdx

sin 2 x + cos 2 x ⎛ 1 1 ⎞
Sol: 2 2
∫ sec x.csc xdx = ∫
1
dx = ∫ dx = ∫ ⎜ + ⎟ dx
cos 2 x sin 2 x cos 2 x sin 2 x ⎝ cos x sin 2 x ⎠
2

= ∫ (sec 2 x + csc 2 x)dx

= ∫ sec 2 xdx + ∫ csc 2 xdx

-Q
= tan x − cot x + c
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
2
Find ∫ elog(1+tan x)dx . T
E
7.

L ⎡⎣' ∫ eloge f (x) = f (x) ⎤⎦

2
Sol: I = ∫ elog e (1+ tan x) dx = ∫ (1 + tan 2 x)dx = ∫ sec x dx = tan x + c
2

L
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

U


∫ cos xsin xdx

7 2
8. Find
0
B
Y

cos7 x sin 2 xdx = [
(6)(4)(2) ][(1) ]
∫
16
Sol: =

B
(9)(7)(5)(3) 315
0

A
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

B
9. Find the area of the region enclosed by the given curves x = 4 – y2, x = 0.

Sol : The given curve x = 4 – y2 is a horizontal (left) parabola.

P(0,2)
Put y = 0, then we get A = (4,0).

Put x = 0 then we get P(0, 2) and Q(0, –2). O A

x

From the diagram, Required area = 2 Area of OAP

2 Q(0,-2)
2 2
⎛ y3 ⎞ 16 32
A = 2∫ xdy = 2∫ (4 − y )dy = 2 ⎜ 4y −
2
⎟ = 2. = sq. units
0 0 ⎝ 3 ⎠0 3 3
« BABY BULLET-Q 5
AP-IPE 2024 SOLVED PAPER «

6
⎡ 2 3 ⎤5
10. Find the order and degree of the differential equation ⎢ d y + ⎛⎜ dy ⎞⎟ ⎥ = 6y
⎢⎣ dx2 ⎝ dx ⎠ ⎥⎦
6/5
⎛ d 2 y ⎛ dy ⎞3 ⎞
Sol: Given D.E is ⎜⎜ 2 + ⎜ dx ⎟ ⎟⎟ = 6y
⎝ dx ⎝ ⎠ ⎠

3
d 2 y ⎛ dy ⎞
⇒ + = (6y)5/6
2 ⎜⎝ dx ⎟⎠
dx

d2y
Here, the highest order derivative is ∴ order = 2
dx 2

d2y
∴ degree = 1 -Q
T
The exponent of 2 is 1
dx

E
L
L
U
B
Y
B
A
B
« MATHS 2B 6
AP-IPE 2024 SOLVED PAPER «

SECTION-B

11. Find the length of the chord intercepted by the circle x2 + y2 – x + 3y – 22 = 0 on the
line y = x – 3.
Sol: Given circle x2 + y2 – x + 3y – 22 = 0
It's Centre, C = (1/2,–3/2)
2
⎛1⎞ ⎛3⎞
2
1 9 1 + 9 + 88 98 49 7
radius, r = ⎜ ⎟ + ⎜ ⎟ + 22 = + + 22 = = = =
⎝2⎠ ⎝2⎠ 4 4 4 4 2 2

Perpendicular distance from the centre(1/2, –3/2) to the line y = x–3 = 0 ⇒ x – y – 3 = 0

1+3−6 −2
| 12 + 23 − 3|

-Q
2 2 1
is p = = = =
12 + 12 2 2 2

⎛ 49 ⎞ 1 48
T
E
∴ Length of the chord = 2 r 2 − p2 = 2 ⎜ ⎟ − = 2 = 2 24
⎝ 2 ⎠ 2 2

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
L
x2 y2-8x L
– 2y + 8 = 0, x2 + y2 - 2x + 6y + 6 = 0 touch each
U
12. Show that the circles +
other and find the point of contact.

Sol:
B
Given circle is S = x2+y2−8x–2y+8=0, centre C1=(4,1), radius r1 =

Y
16 + 1 − 8 = 9 =3

B
Other circle is S'=x2+y2−2x+6y+6=0, centre C2=(1,–3), radius r2 = 1 + 9 − 6 = 4=2

A
C1 C 2 =
B
(4 − 1) 2 + (1 + 3) 2 = 9 + 16 = 25 = 5

Also, r1 + r2 = 3 + 2 = 5 . Here, C1C 2 = r1 + r2

∴ the two circles touch each other externally.

Also the point of contact I divides C1C 2 in the ratio r1 : r2=3:2 internally.

⎛ 3(1) + 2(4) 3(−3) + 2(1) ⎞ ⎛ 11 −7 ⎞

∴ Point of contact I = ⎜ , ⎟=⎜ 5 , 5 ⎟
⎝ 3+ 2 3+ 2 ⎠ ⎝ ⎠
« BABY BULLET-Q 7
AP-IPE 2024 SOLVED PAPER «

13. Find the equation of tangent and normal to the ellipse 9x2 + 16y2 = 144 at the end
of the latus rectum in the first quadrant.

x 2 y2
Sol: Given Ellipse 9x2 + 16y2 = 144 ⇒ + = 1 ⇒ a 2 = 16, b 2 = 9 ⇒ a = 4, b = 3
16 9

a 2 − b2 16 − 9 7 7
e= = = ⇒e=
2 16 16 4
a

⎛ b2 ⎞ ⎛ 7 9⎞ ⎛ 9⎞

-Q
Positive end of latus rectum L = ⎜ ae, ⎟ = ⎜ 4 . , ⎟⎟ = ⎜ 7, ⎟
⎜ ⎜
a ⎟⎠ ⎝ 4 4⎠ ⎝ 4⎠
⎝

Equation of the tangent at L is S1 = 0

T
E
xx1 yy1
⇒ 2 + 2 =1⇒
x 7 y⎛9⎞
+ ⎜ ⎟ = 1 ⇒ 7x + 4y = 16
L
a b 16 9⎝4⎠
L
2 2
U
Equation of the normal at L is a x − b y = a 2 − b 2 ⇒ 16x − 9 y = 16 − 9

⇒
16x
x1

− 4y = 7 ⇒ 16x − 4 7y = 7 7
y1
B 7 9
4

7
Y
B
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

14. A x2 y 2

B
If P(x, y) is any point on the ellipse + = 1 with foci S & S' then SP + S'P is a constant.
a2 b2
Proof: Let N be the foot of the perpendicular from P(x, y) on to the x-axis.
from the diagram, PM = NZ = CZ − CN = a − x
e
a P(x , y)
and PM′ = NZ′ = CN + CZ′ = x + M' M
e
SP ⎛a ⎞ Z' S' C S N Z
Now, PM = e ⇒ SP = ePM = e ⎜ e − x ⎟ = a − ex
⎝ ⎠
S′P ⎛ a⎞
and PM′
= e ⇒ S′P = ePM′ = e ⎜ x + ⎟ = ex + a = a + ex
⎝ e⎠

∴ SP + S′ P = (a − ex) + (a + ex) = 2a which is a constant

« MATHS 2B 8
AP-IPE 2024 SOLVED PAPER «

15. Find the centre, eccentricity, foci, length of latus rectum and equations of the
directrices of the Hyperbola x2 – 4y2 = 4

x 2 y2
Sol: Given hyperbola is x2–4y2=4 ⇒ − = 1. Here a2=4, b2=1
4 1
(i) Centre C= (0,0)
a2 + b2 4 +1 5 5
(ii) Eccentricity e = 2
= = =
a 4 4 2
⎛ ⎛ 5⎞ ⎞
(iii) Foci = (±ae,0) = ⎜⎜ ±2 ⎜⎝ 2 ⎟⎠ , 0 ⎟⎟ = ± 5, 0 ( )
⎝ ⎠
a ⇒x=± 2 ⇒x=± 4
(iv) Equation of the directrices is x = ± 5 5
e 2

-Q
2b 2 2(1)
(v) Length of latusrectum = = =1

T
a 2
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

E
L
π /3
sin x
Evaluate ∫ sin x + cos x dx
L
16.
π /6

U
Sol: We know ∫a
b b
f ( x )dx = ∫ f (a + b − x )dx .
a B
Y ⎛π π ⎞

B
sin ⎜ + − x ⎟
π /3
π/3 ⎝ 6 3 ⎠
= ∫ dx

A
sin x
∴I = ∫ dx ........(1) π /6 sin ⎛ π π ⎞ ⎛ π π ⎞
sin x + cos x ⎜ + − x ⎟ + cos ⎜ + − x ⎟

B
π/6 ⎝6 3 ⎠ ⎝6 3 ⎠

⎛π ⎞
π /3 sin ⎜ − x ⎟ π /3
⎝2 ⎠ cos x
= ∫ dx = ∫ cos x + sin x
dx
............(2)
π /6 sin ⎛ π − x ⎞ + cos ⎛ π − x ⎞ π/6
⎜ ⎟ ⎜ ⎟
⎝2 ⎠ ⎝2 ⎠

π /3 π /3
sin x cos x
From (1) and (2), I + I = ∫ sin x + cos x dx + ∫ sin x + cos x dx
π /6 π /6

π /3 π/3
sin x + cos x π /3 π π π π π
⇒ 2I = ∫ sin x + cos x
dx = ∫ 1dx = [x ]π/6 = 3 − 6 = 6 ⇒ 2I = 6 ⇒ I = 12
π/6 π /6
« BABY BULLET-Q 9
AP-IPE 2024 SOLVED PAPER «

dy 3x2 1 + x2
17. Solve + y=
dx 1 + x3 1 + x3

dy ⎛ 3x 2 ⎞ 1 + x 2
Sol: Given D.E is + y⎜ ⎟= 3
. This is a linear D.E in y.
dx ⎝ 1 + x3 ⎠ 1 + x

2
dy
+ yP(x) = Q(x) where P(x) = 3x 1 + x2
It is in the form and Q(x) =
dx 1 + x3 1 + x3

3x 2 3x 2 ⎡ f ′(x) ⎤
Here, P(x) = ⇒ ∫ P(x) dx = ∫ 1 + x3 dx = log(1 + x 3 ) ⎢' ∫ = log f (x) ⎥

-Q
1 + x3 ⎣ f (x) ⎦

T
3
∴I.F = e∫ P(x)dx = elog(1+x ) = 1 + x3

\ The solution is y(I.F) = ∫ (I.F)Q(x)dx

E
L
L
⎛ 1 + x2 ⎞ x3
∫
⇒ y(1 + x 3 ) = (1 + x 3 ) ⎜
⎜ 1 + x3
⎟ dx = ∫ (1 + x 2 )dx = x +
⎟
+c

U
⎝ ⎠ 3

3
∴ y(1 + x ) = x +
x3
3
+c B
Y
B
A
B
« MATHS 2B 10
AP-IPE 2024 SOLVED PAPER «

SECTION-C
18. Find the values of c if the points (2, 0), (0, 1), (4, 5), (0, c) are concyclic.

Sol: Let A = (2, 0), B = (0, 1), C = (4, 5), D = (0, c)

We take S(x1, y1) as the centre of the circle ⇒ SA = SB = SC

Now, SA = SB ⇒ SA 2 = SB2 ⇒ (x1 − 2)2 + (y1 − 0)2 = (x1 − 0)2 + (y1 − 1)2

⇒ ( x12 − 4x1 + 4) + ( y12 ) = ( x12 ) + ( y12 − 2y1 + 1)

⇒ 4x1 − 2y1 +1− 4 = 0 ⇒ 4x1 − 2y1 − 3 = 0.........(1)

Also, SB = SC ⇒ SB2 = SC2 ⇒ (x1 − 0) 2 + (y1 − 1)2 = (x1 − 4) 2 + (y1 − 5) 2

⇒ ( x12 ) + ( y12 − 2y1 + 1) = ( x12 − 8x1 + 16) + ( y12 − 10y1 + 25)

-Q
T
⇒ 8x1 − 2y1 + 10y1 + 1 − 16 − 25 = 0 ⇒ 8x1 + 8y1 − 40 = 0 ⇒ 8(x1 + y1 − 5) = 0 ⇒ x1 + y1 − 5 = 0......(2)

E
L
Solving (1) & (2) we get the centre S(x1,y1)

2 × (2) ⇒2x1+2y1–10=0...........(3)
L
U
B
(1) + (3) ⇒6x1–13=0.⇒6x1=13⇒x1=13/6

13 30 − 13 17 17
(2) ⇒ y1 = 5 − x1 = 5 − = = ⇒ y1 =
6 6
Y 6 6

B ⎛ 13 17 ⎞
∴ Centre of the circle is S(x1 , y1 ) = ⎜ , ⎟

A
⎝6 6⎠

B
Also, we have A = (2,0) Hence, radius r = SA ⇒ r 2 = SA 2
2 2 2 2
⎛ 13 ⎞ ⎛ 17 ⎞ ⎛ 12 − 13 ⎞ ⎛ 17 ⎞ ⎛ 1 ⎞ ⎛ 289 ⎞ 290
∴ r 2 = SA 2 = ⎜ 2 − ⎟ + ⎜ 0 − ⎟ = ⎜ ⎟ +⎜ ⎟ =⎜ ⎟+⎜ ⎟=
⎝ 6⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 36 ⎠ ⎝ 36 ⎠ 36
2 2
⎛ 13 17 ⎞ 2 290 ⎛ 13 ⎞ ⎛ 17 ⎞ 290
∴ Circle with Centre, ⎜ , ⎟ and r = is ⎜ x − ⎟ + ⎜ y − ⎟ =
⎝6 6⎠ 36 ⎝ 6⎠ ⎝ 6⎠ 36

2 2
⎛ 13 ⎞ ⎛ 17 ⎞ 290 ⎛ 17 ⎞2 290 169 121
Put D(0, c) in the above equation ⇒ ⎜ 0 − ⎟ + ⎜c − ⎟ = ⇒ ⎜c − ⎟ = − =
⎝ 6⎠ ⎝ 6⎠ 36 ⎝ 6⎠ 36 36 36

2
⎛ 6c − 17 ⎞ 121 (6c − 17)2 112
⇒⎜ ⎟ = ⇒ = ⇒ 6c − 17 = ±11
⎝ 6 ⎠ 36 36 36
28 14
⇒ 6c = ±11 + 17 ⇒ 6c = 28 ⇒ c = = (or) 6 c = 6 ⇒ c = 1
6 3
∴ c= 14/3 (or) 1
« BABY BULLET-Q 11
AP-IPE 2024 SOLVED PAPER «

19. Find the equation to the pair of transverse common tangents to the circles
x2+y2-4x-10y+28=0 and x2+y2+4x-6y+4=0

Sol: For the circle x2 + y2 − 4x − 10y + 28 = 0, centre C1 = (2, 5),

radius r1 = (−2)2 + (−5)2 − 28 = 1 = 1

For the circle x2 + y2 + 4x − 6y + 4 = 0, centre C2 = (−2, 3),

radius r2 = 2 2 + ( −3) 2 − 4 = 9 = 3

-Q
The internal centre of similitude, I divides C1C2 internally in the ratio r1 : r2 = 1:3

T
⎛ 1( −2) + 3(2) 1(3) + 3(5) ⎞ ⎛ 4 18 ⎞ ⎛ 9 ⎞
E
L
∴I = ⎜ , ⎟ = ⎜ , ⎟ = ⎜ 1, ⎟
⎝ 1+ 3 1+ 3 ⎠ ⎝ 4 4 ⎠ ⎝ 2 ⎠

L
U
The equation to the pair of transverse common tangents is S12 = S11 (S)

⎡ 9 ⎛ 9⎞ ⎤
2
B
⎛ 81 ⎞ 2 2
⇒ ⎢ x + y − 2(x + 1) − 5 ⎜ y + ⎟ + 28⎥ = ⎜1 + − 4 − 45 + 28 ⎟ (x + y − 4x − 10y + 28)

Y
⎣ 2 ⎝ 2 ⎠ ⎦ ⎝ 4 ⎠

⎛ y 7⎞ 1
2
B (2x − y + 7)2 1 2 2

A
⇒ ⎜ −x − + ⎟ = (x2 + y2 − 4x − 10y + 28) ⇒ = (x + y − 4x − 10y + 28)
⎝ 2 2⎠ 4 4 4
2
B
⇒ (2x + y − 7) = x + y 2 − 4x − 10y + 28
2

⇒ 4x 2 + y2 + 49 + 4xy − 28x − 14y = x 2 + y2 − 4x − 10y + 28 ⇒ 3x 2 + 4xy − 24x − 4y + 21 = 0

« MATHS 2B 12
AP-IPE 2024 SOLVED PAPER «

20. Derive the standard form of the parabola.

Sol: Let S be the focus and L=0 be the directrix of the parabola.
Y-axis

>
Let Z be the projection of S on to the directrix P
M N
.
Let A be the mid point of SZ
SA
>
A(0,0)
.S(a,0) >
⇒ SA = AZ ⇒ =1 Z X-axis
AZ L=0
⇒ A is a point on the parabola.

>
Take AS, the principle axis of the parabola as X-axis and

-Q
the line perpendicular to AS through A as the Y-axis ⇒ A=(0,0)

Let AS = a ⇒ S = (a,0), Z = (−a,0)

T
⇒ the equation of the directrix is x = −a ⇒ x + a = 0
E
L
L
Let P(x1, y1) be any point on the parabola.

U
N be the projection of P on to the Y-axis.

Here PM=PN+NM = x1+a

B
M be the projection of P on to the directrix.

('PN = x − coordinate of P and NM = AZ = AS = a)

Y
Now, by the focus directrix property of the parabola,

B
A
SP
we have = 1 ⇒ SP = PM ⇒ SP2 = PM2
PM

B
⇒ (x1 − a) 2 + (y1 − 0) 2 = (x1 + a) 2

⇒ y12 = (x1 + a) 2 − (x1 − a) 2

⇒ y12 = 4ax1 [' (a + b)2 − (a − b)2 = 4ab]

∴ The equation of locus of P(x1, y1) is y2 = 4ax

« BABY BULLET-Q 13
AP-IPE 2024 SOLVED PAPER «

21. Evaluate the reduction formula for In= òsinnxdx and hence find òsin4xdx

Sol: Given, In = ∫sinnxdx=∫sinn−1x(sinx)dx.

We take First function u = sinn−1x and Second function v = sinx ⇒ ∫ v = –cosx

From By parts rule, we have

In = sinn−1x (−cosx) −∫(n−1)sinn−2x cosx (−cosx)dx

= −sinn−1xcosx+(n−1)∫sinn−2xcos2xdx

= −sinn−1xcosx+(n-1)∫sinn−2x(1−sin2x)dx

-Q
T
n −2
= − sin n −1 x cos x + (n − 1)[ ∫ sin xdx − ∫ sin n xdx]

= − sin n −1 x cos x + (n − 1)[I n − 2 − I n ] E

L
I n = − sin n −1 x cos x + (n − 1)I n − 2 − (n − 1)I n
L
U
= − sin n −1 x cos x + (n − 1)I n − 2 − nI n + I n
B
Y
n−1
⇒ nIn = −sin xcosx + (n −1)In−2 + In − In

− sinn−1 xcosx ⎛ n −1 ⎞ B
⇒ In =
n
+⎜
⎝ n ⎠
A⎟ In−2 ....(1)

B
Put n = 4, 2,0 successively in (1), we get

sin 3 x cos x 3
I4 = − + I2
4 4

sin 3 x cos x 3 ⎡ sin x cos x 1 ⎤

=− + ⎢− + I0⎥
4 4⎣ 2 2 ⎦

− sin 3 x cos x 3 3
= − sin x cos x + I 0
4 8 8

sin3 xcosx 3 3
=− − sin xcosx + x + c [' I0 = x]
4 8 8
« MATHS 2B 14
AP-IPE 2024 SOLVED PAPER «

x +1
22. Evaluate ∫ x2 + 3x + 12 dx
d 2
Sol: Let x + 1=A (x +3x+12) + B .......(i)
dx

∴ x + 1 = A(2x + 3) + B ⇒ x + 1 = 2Ax + (3A + B)

Equating the coefficients of 'x', we get 2A=1 ⇒ A=1/2

3 1
Equating the constant terms, we get 3A+B=1 ⇒ B=1–3A=1– ⇒ B= −
2 2

−1
-Q
T
1 1 d 2 1
Putting A = , B = in (i), we get Nr. x + 1 = (x + 3x + 12) −
2 2 2 dx 2

E
L
L
d (x 2 + 3x + 12)
x +1 1 1 dx
∴I = ∫
2∫ ∫
dx = dx dx −
2 2 2
x + 3x + 12 x + 3x + 12 2 x + 3x + 12

U
1 1
= log x 2 + 3x + 12 − ∫
dx B
Y
2 2 2 2 2 2
⎛ 3 ⎞ ⎛ 39 ⎞ ⎛3⎞ ⎛3⎞
⎜x + ⎟ +⎜ ⎟ ' x 2 + 3x + 12 = x 2 + 3x + ⎜ ⎟ − ⎜ ⎟ + 12

B
⎝ 2⎠ ⎝ 2 ⎠ ⎝2⎠ ⎝2⎠

A
⎛ ⎛3⎞ ⎞ 9
2
⎛ 3⎞
2
9 + 48
= ⎜ x 2 + 3x + ⎜ ⎟ ⎟ − + 12 = ⎜ x + ⎟ −
⎜ ⎝ ⎠ ⎠⎟ ⎝ ⎠

B
2 4 2 4
⎝
⎛ ⎞
−1 ⎜ x + 2
3
1 1 2
= log x 2 + 3x + 12 − Tan ⎟+c ⎛ 3⎞ 39
2
2 2 39 ⎜ 39 ⎟ =⎜x + ⎟ +
⎝ 2 ⎠ ⎝ 2⎠ 4
dx 1 −1 x
' ∫ x 2 + a 2 = a Tan a
1 1 ⎛ 2x + 3 ⎞
= log x 2 + 3x + 12 − Tan −1 ⎜ ⎟+c
2 39 ⎝ 39 ⎠
« BABY BULLET-Q 15
AP-IPE 2024 SOLVED PAPER «


sinx + cosx
23. Evaluate ∫ 9 + 16sin2x
dx
0

Sol: Here, we take the substitution sinx–cosx=t. Then (cosx+sinx)dx=dt.

π π π 1 1
Now x = 0 ⇒ t = sin0–cos0 = 0–1= –1 and x = ⇒ t = sin − cos = − =0
4 4 4 2 2

Also,(sinx–cosx)2=t2 ⇒ sin2x+cos2x–2sinxcosx=t2 ⇒ 1–sin2x=t2 ⇒sin2x=1–t2

∴ 9+16sin2x=9+16(1–t2)=9+16–16t2=25–16t2
-Q
T
π /4
sin x + cos x
0 0
E
L
dt dt
∴I = ∫ 9 + 16sin 2x
dx = ∫ 25 − 16t 2 = ∫ 52 − (4t)2
L
0 −1 −1

U
B
0
1 1 ⎡ 5 + 4t ⎤
= . .log ⎢ ⎥
4 2(5) ⎣ 5 − 4t ⎦ −1

Y
=
1 ⎡ ⎡5 + 0⎤
⎢ log ⎢ − log ⎢ ⎥ B
⎡ 5 − 4 ⎤⎤ 1 ⎡
=
1⎤
log 1 − log ⎥

A
⎥
40 ⎣ ⎣ 5 − 0 ⎦ ⎥ ⎢
⎣ 5 + 4 ⎦ ⎦ 40 ⎣ 9⎦

=
1 ⎡ B1
0 − log 9 −1 ⎤ = [log 9 ] =
log 32 2 log 3 log 3
= =
40 ⎣ ⎦ 40 40 40 20
« MATHS 2B 16
AP-IPE 2024 SOLVED PAPER «

x
⎛ x⎞
Solve ⎛⎜ ⎞ x y
24. ⎟ dx + e ⎜ 1  ⎟ dy = 0
y
⎝1 + e ⎠ ⎝ y⎠

⎛ ⎞ ⎛ ⎞
Sol: Given that (1 + e x/y )dx + e x/y ⎜1 − x ⎟ dy = 0 ⇒ (1 + ex/y )dx = −e x/y ⎜1 − x ⎟ dy
⎝ y⎠ ⎝ y⎠

dy (1 + e x/ y ) dx ⎡ e x/ y (1 − (x / y)) ⎤
⇒ =− ⇒ = −⎢ ⎥ .....(1)
dx ⎛ x⎞ dy ⎢⎣ 1 + e x/ y ⎥⎦
e x/ y ⎜1 − ⎟
⎝ y ⎠

x dx dv
Put y = v ⇒ x = vy differentiating w.r.to y, we have dy = v + y. dy

-Q
⎛ dv ⎞ ⎡ e (1 − v) ⎤
∴ (1) ⇒ v + y ⎜ ⎟ = − ⎢
v
⎥
T
E
v
⎝ ⎠
dy ⎣⎢ 1 + e ⎦⎥

L
⎛ dv ⎞ ⎡ e v (1 − v) ⎤ ⎡ e v − v.e v + v + ve v ⎤
L
⎡ v + ev ⎤

U
⇒ y⎜ ⎟ = − ⎢ + v⎥ = − ⎢ ⎥ = −⎢ ⎥
v
⎝ dy ⎠ ⎣⎢ 1 + e ⎦⎥ ⎣⎢ 1 + ev ⎦⎥
v
⎣⎢ 1 + e ⎦⎥

⎛ 1 + ev ⎞
B
Y
⎛1⎞ 1 + ev 1
⇒⎜ ⎟ dv = − ⎜ ⎟ dy ⇒ ∫ dv = − ∫ dy
⎜ v + ev ⎟ ⎝y⎠ v+e v y

B
⎝ ⎠

A
B
⇒ log | v + e v |= − log | y | + log | c |⇒ log | v + e v | + log | y |= log | c |

⎡x ⎤
⇒ log | (v + e v )y |= log | c |⇒ ⎢ + e x/ y ⎥ y = c ⇒ x + ye x/ y = c
⎣y ⎦