Section C

Solution. 13 .
Let f(x) = asin

π/2

(x + 1), x ≤ 0

tan x - sin x/x³

, x > 0

is continuous at x = 0
∴ LHL = RHL = f(0)
Here, LHL = a sin

π/2

(x+1)
⇒ LHL = a sin

π/2

(-h+1)
[put x = 0 - h;when x →0^-, then h → 0]
a sin

π/2

= a
f(0) = a sin

π/2

= a
Now,we need to evaluate RHL at x = a .

[∵ LHL = f(0) = a and from this, we can't find the value of a]

Here,RHL =

tan x - sin x/x³

⇒ RHL =

tan h - sin h/h³

[put x = 0 + h = h; whenx → 0⁺,then h → 0]
=

sin h/cos h

- sin h/h³

=

sin h - sin h cos h/h³ cos h

=

sin h(1 - cos h)/h³ cos h

=

sin h/h

.

1 - cos h/h²

.

1/cos h

= 1 x

1 - cos h/h²

x 1
[∵

sin h/h

and

1/cos h

=

1/cos 0

=

1/1

= 1]
=

1 - cos h/h²

=

2 sin²(h/2)/h²

[∵ 1 - cos x = 2 sin² (x/2)]
=

2 x sin²(h/2)/(h²/4) x 4

= (2/4) x

sin²(h/2)/h²

= (1/2) x (

sin(h/2)/h/2

)² = (1/2) x 1

[∵

sin x/x

]
∴ RHL =

1/2

From Eq. (i) we have LHL = RHL
a =

1/2

---

Solution. 14 .
cos(cos^-1

1/√(x² + 1)

) = sin(sin^-1

4/5

)

1/√(x² + 1)

=

4/5

on squaring both sides, we get
16x² + 16 = 25
16x² = 9
x = ±

3/4

∴ cos(cos^-1x) = x and sin (sin^-1x) = x , x∈[-1,1]

---

Solution. 15 .
Given, f(x) = lxl + x and g(x) = lxl - x for all x ∈ R

⇒ f(x) = (x + x, x ≥ 0 )

(- x + x , x < 0)

and ⇒ g(x) = x - x, x ≥ 0

- x - x,x < 0

⇒ f(x) = 2x, x ≥ 0

0,x < 0

and g(x) = 0 , x ≥ 0

-2x , x < 0

Thus, for x ≥ 0, gof(X) = g(2x) = 0
and for x < 0, gof = g(0)
⇒ gof(x) = 0, all x ∈ R
Similarly , for x ≥ 0, fog(x) = f(0) = 0
and for x < 0 , fog (x) = f(-2x) = 2(-2x) = -4x
⇒ fog(x) = 0, x ≥ 0

-4x, x < 0

---

Solution. 16 .
= sin^-1(

1/3

) + sin^-1 (

2√2/3

)
= tan^-1

1/2√2

+ tan^-12√2
= tan^-1

1/2√2

+ 2√2/ 1 -

1/2√2

. 2√2

=

π/2

∴ tan^-1 x + tan^-1y = tan^-1

x+y/1 - xy

---

Solution . 17
Let P(x,y) be any point on the line joining A(1,3) and B(0,0)
Then area of (▵APB) = 0

1(y) - 3(x) + 1(0) = 0
y - 3x = 0
y = 3x
Since area of triangle ABC = 3 sq units
1(0) - 3(-k) + 1(0) = ± 6
3k = ±6
k = ± 2

---

Solution . 18
Solve this with the help of row and column transformation
So,Apply C₁ → C₁ + C₂ + C₃
=
On taking common (a+b+c) from C₁
=
Now apply R₂ → R₂ - R₁ and R₃ → R₃ - R₁
=
Now expand with respect to C₁
= (a+b+c)[(c-b)(b-c)-(a-b)(a-c)]
= (a+b+c)(-1)[(b-c)²+ (a-b)(a-c)]
= -(a+b+c)[b² + c² - 2bc + a² - ac - ba + bc]
= (a+b+c)[a² + b² + c² - ac - ab - bc]
multiply and divide by 2
=

-(a+b+c)/2

[2a² + 2b² + 2c² -2ac - 2ab - 2bc]
→ -

1/2

(a+b+c) [(a-c)² + (b-c)² + (c-a) ²] < 0
∴ a,b,c > 0 and a ≠ b ≠ c .

---

Solution . 19
This problem belongs to logarithmic differentiation, so take log both sides to get desired result
given x^y = e^x-y
On taking log both sides, we get
ylog_ex = (x - y)log_ee
⇒ ylog_ex = x - y [∵log_ee = 1]
⇒ y(1 + logx) = x ⇒ y =

x/1 + logx

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

dy/dx

=

(1 + logx)

d/dx

(x) -

d/dx

(1 + logx)/ (1 + logx)²

∵

d/dx

(

v/u

) =

u

dv/dx

- v

du/dx

/ u²

1 + logx - x.

1/x

/(1 + logx)²

=

1 + log x - 1 /(1 + logx)²

Hence,

dy/dx

=

log x/(1 + logx)²

Also,it can be written as

dy/dx

=

log x /(log_ee + logx)²

dy/dx

=

log x /(log(ex))²

---

Solution . 20
Given differential equation is
(x log lxl)

dy/dx

+ y = 2 loglxl
On dividing both sides by x log lxl, we get

dy/dx

+

y/x loglxl

=

2/x

which is a linear differential equation of the form

dy/dx

+ Py = Q
On comparing, we get
P =

1/x loglxl

and Q =

2/x

∴ IF = e^{∫(1/xloglxl)dx} = log^{logllog xl}

[put loglxl = t

1/x

dx = dt

∴ ∫

1/x loglxl

dx = ∫

dt/t

= logltl = log llog x l ]

= log x [∵ e^{log lxl} = x]
The solution of linear differential equation is given by
y x IF = ∫ Q x IF dx + C
∴ y x loglxl = ∫ 2/x loglxl dx
⇒ y log lxl = log lxl ∫2/x dx
- ∫ [d/dx (log lxl)∫2/x dx]dx
[using integration by parts]
⇒ y log lxl = log lxl.2log lxl - ∫ 1/x . 2log lxl dx
[∴ ∫ 1/x dx = log lxl + C]
⇒ y log lxl = 2(log lxl)² - 2 ∫ (log lxl)/x dx
⇒ y log lxl = 2(log lxl)² -

2(log lxl)²/2

+ C
[put log lxl = t ⇒

1/x

dx = dt]
∴ ∫

log lxl/x

dx = ∫ t dt =

t²/2

=

(log lxl)²/2

]
y = 2(log lxl) - (log lxl) +

C/log lxl

[dividing both sides by log lxl]
y = log lxl +

C/log lxl

which is the required solution .

---

Solution . 21
Let x and y denotes table and chair respectively
So, formulation of this problem
z = 20x + 30y
subject to: 4x + 5y ≥ 200
3x + 4y ≤ 240
y ≥ 20
x, y ≥ 20

Now, Solve this problem graphically. first of all draw the straight lines for all the given inequalities.

4x + 5y = 200
put x = 0 , y = 0
y = 0=40 , x = 50
(0,40) (50,0)
3x + 4y = 240
put x = 0 , y = 0
(0,60) y = 60 , x = 80 (80,0)

Solve equation 3 and 4, to get point A
Point A (30,20)
Solve equation 2 and 3, to get point B
3x + 4y = 240
x = 30
3(30) + 4y = 240
4y = 150
y = 37.5
Points B (30,37.5)
Solve equation 2 and 4, to get point C
3x + 4y = 240
y = 20
3x + 4(20) = 240
x = 53.4
Point C(53.4,20)

Now find z for all points
Z_A = 20(30) + 30(20) = 1200
Z_B = 20(30) + 30(37.5) = 1725
Z_C = 20(53.4) + 30(20) = 1668
Optimal solution is Rs.1200 at point A (30,20)

---

Solution . 22
Let P = probability of success and q = probability of failure and x be a random variable that denote the number of sucess in 6 trials .

x follows binomial distribution , having parameters n = 6, p

∴P(x = x) = 6C_x . P^x . q^n-x

x = 0,1,2,3,4,5,6

As per question P(x = 4) =

1/4

P(x = 2)

4P(x = 4) = P(x = 2)

4(6C₄P⁴q²) = 6C₂P²q⁴ (∵nc_x = nc_n-x)

4P² = q²

4P² = (1 - P)²

4P² = 1 + P² - 2P

3P² + 2P - 1 = 0

3P² + 3P - P - 1 = 0

3P(P + 1) - 1(P + 1) = 0

(3P - 1)(P + 1) = 0

P =

1/3

, -1

P =

1/3

( p = -1, Probability can't be negative.)

---

Solution . 23
Let us define the events
E₁ = a red and a black ball is transferred
E₂ = two red balls are transferred.
E₃ = two black balls are transferred.
A = ball drawn from bag B is red .
then P(E₁) =

4C₁ x 4C₁ / 8C₂

=

4/7

P(E₂) =

4C₂/8C₂

=

3/14

P(E₃) =

4C₂/8C₂

=

3/14

P(A/E₁) =

4/10

P(A/E₂) =

5/10

P(A/E₃) =

7/10

P(E₂/A) =

P(A/E₂).P(E₂)/ P(A/E₁).P(E₁) + P(A/E₂).P(E₂) + P(A/E₃) . P(E₃)

=

(5/10) x (3/14)/4/10 x 4/7 + 5/10 x 3/14 + 7/10 x 3/14

=

15/68