Section D

Solution.24
Let I = (2 log lsin xl - loglsin 2xl)dx

⇒ I = (2 log lsin xl - logl2sinx cos xl)dx

[∵ sin x > 0 for 0 < x <

π/2

and sin 2x = 2sinx cosx]

⇒ I = [2 log(sin x) - (log 2 + log(sin x) + log(cos x)]dx

[∵ log(mnp) = log m + log n + log p]

⇒ I = [2 log(sinx) - log 2 - log(sin x) - log(cos x)]dx

⇒ I = (log(sin x) - log 2 -log(cos x))dx

⇒ I = log(sin x)dx - log 2 dx - log(cos x) dx

⇒ I = log cos x dx - log 2 [

π/2

- 0] - log cos x dx

∴ I = -

π/2

log 2

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Solution.25
Let C denotes the curved surface area , r be the radius of base, h be the height aand V be the volume of right circular cone.
To show, h = √2r
We know that, volume of cone is given by
V=

1/3

πr²h ⇒ h =

3V/πr²

Also,the curved surface area of cone is given by C = πrl, where l = √(r² + h²) is the slant height of cone.
∴ C = πr√(r² + h²)
On squaring both sides, we get
C² = π²r² (r² + h²)⇒ C² = π²r⁴ + π²r²h²
Let C² = Z
Then Z = π²r⁴ + π²r²h²
⇒ Z = π²r⁴ + π²r²(

3V/πr²

)²
⇒ Z = π²r⁴ + π²r²

9V²/ π²r⁴

⇒ Z = π²r⁴ +

9V²/r⁴

On differentiating both sides w.r.t.r, we get

dZ/dr

= 4π²r³ -

18V²/r³

For maxima or minima put

dZ/dr

= 0
⇒ 4π²r³ -

18V²/r³

= 0
⇒ 4π²r³ =

18V²/r³

⇒ 4π²r³ = 18 (

1/3

πr²h )² [∵V =

1/3

πr²h ]
⇒ 4π²r⁶ = 18 x

1/9

π²r⁴h²
⇒ 4π²r⁶ = 2π²r⁴h² ⇒ 2r² = h²
∴ h = √2r
Hence , height = √2 x (radius of base)
Also,

d²Z/dr²

=

d/dr

(

dZ/dr

) =

d/dr

(4π²r³ -

18V²/r³

)
= 12π²r² +

54V²/r⁴

)

d²Z/dr²

= 12π²r² +

54V²/r⁴

> 0

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Solution.26
The region has a circle with equation
x² + y² = 16 (circle)
x + y = 4 (line)
Circle has center (0,0) and radius is 4, wide line has equal intercept on x and y axis that is 4.
Point of intersection is calculated as follows
x² + y² = 16
x² + + (4 - x)² = 16
x² + 16 + x² - 8x = 16
2x² - 8x = 0
2x(x - 4) = 0
x = 0 or 4
when x = 0, then y = 4
and when x = 4, then y = 0
So the points of intersection are (0,4) and (4,0) on drawing the graph, area of shaded region is to be calculated

So, required area = (y circle - y line) dx
= [ - (4 - x)]dx
= ()dx - (4 - x)dx
= [

x/2

+

16/2

sin^-1

x/4

] - [4x -

x² /2

]
= [0 + 8sin^-1(1) - 0 - 8sin^-1(0)] - [16 -

16/2

- 0]
= = 8 .

π/2

- 8
= 8(

π /2

- 1)
= 8 (

π - 2/2

= 4(π - 2)sq. unit

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Solution.27
Given differential equation is
x

dy/dx

sin(

y/x

) = y sin (

y/x

) - x

⇒

dy/dx

=

y/x

-

1/sin

y/x

[dividing both sides by x sin (

y/x

)]

Let F(x,y) =

y/x

-

1/sin

y/x

On replacing x by kx and y by ky both sides, we get
F (kx,ky) =

ky/kx

-

1/sin

ky/kx

= k⁰ (

y/x

-

1/sin

y/x

) = k⁰ F(x,y)

So,given differential equation is homogeneous. On putting y = vx
⇒

dy/dx

= v + x

dy/dx

In equation (1) we get
V + x

dy/dx

= V -

1/sin v

⇒ x

dy/dx

= -

1/sin v

⇒ sin v dv = -

dx/x

On integrating both sides, we get
∫ sin v dv = ∫

dx/x

⇒ - cos v = - log lxl + C
⇒ - cos

y/x

= - log lxl + C [put v =

y/x

]
Also, given that x = 1, when y =

π/2

On putting x = 1and y =

π/2

In equation (ii), we get
- cos (

π/2

) = log l1l + C
⇒ -0 = 0 + C ⇒ C = 0

On putting the value of C in equation (ii), we get
cos

y/x

= log lxl
which is the required solution .

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Solution.28
a→ = î + ĵ +k̂,b→ = 2î + 4ĵ -5k̂ and c→ = μî + 2ĵ +3k̂
Now, b → + c→ = 2î + 4ĵ - 5k̂ + μî + 2ĵ + 3k̂
= (2 + μ)î + 6ĵ - 2k̂

∴ lb→ + c→ l = √[(2 + μ)² + (6) ² + (-2)²]

= √[4 + μ² + 4μ + 36 + 4]

= √(μ² + 4μ + 44)

The unit vector along b→ + c→

b→ + c→/ lb→ + c→l

=

(2 + μ)î + 6ĵ - 2k̂/ √(μ² + 4μ + 44)

Now as given in question, scalar product of (î + ĵ +k̂) with unit vector b→ + c→ is 1.

(î + ĵ +k̂).

b→ + c→/ lb→ + c→l

= 1

î + ĵ +k̂.

(2 + μ)î + 6ĵ - 2k̂/ √(μ² + 4μ + 44)

= 1

(2 + μ)1 + (6)1 + (-2)1/ √(μ² + 4μ + 44)

= 1

(2 + μ) + 6 - 2/ √(μ² + 4μ + 44)

= 1

μ + 6 = √(μ² + 4μ + 44)

(μ + 6)² = μ² + 4μ + 44, squaring both sides we get,

μ² + 36 + 12μ = μ² + 4μ + 44

8μ = 8

μ = 1

So, the value of μ is 1.

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Solution.29
Given equation of planes are
r→.(î - 2 ĵ + 3 k̂) = 4 and

r→.(-2î - ĵ + k̂) = - 5

On comparing these with r→. n→= d, now we get

n₁→ = î-2ĵ+3k̂, d₁ = 4,

n₂→ = -2î + ĵ + k̂ and d₂ = -5

Now, the equation of the plane which contains the intersection of the given plane is

r→.(n₁→ + μn₂→) = d₁ + μd₂

r→ .[î - 2ĵ + 3ĵ + μ(-2î + ĵ + k̂)] = 4 - 5μ

r→ .[(1-2μ)î + (-2 + μ)ĵ + (3 + μ)k̂] = 4 - 5μ -----(1)

Now it is given that intercept on X and Y axis are same

4 - 5μ/1 - 2μ

=

4 - 5μ/-2 + μ

On putting μ = 1 in equation (1), we get

r→. [(1-2)î + (-2 + 1)ĵ + (3 + 1)k̂] = 4 - 5 x 1

r→ .(-î - ĵ + 4k̂) = -1

equation of plane.

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