SOLUTIONS

1. (b) : Let θ=π ⇒ θ=π
   16

8

2

Ist and last gives tan^2 θ + cot^2 θ =^8
14

2

−

−
cos θ
Similarly other terms.

2. (c) : fx e()= sin()xx−π[]x cos
   sin(x – [x]) = sin{x}. Period is 1
   cosπx, Period is 2
   Hence f (x) is of period 2


3. (d) : f(x, y) is homogeneous function of degree
   n ∈ R in x, y if f (kx, ky) = kn f(x, y) ; where k > 0


4. (c) : (A)lim ln sin

/

x

x x

→

⎛⎝⎜ − ⎞⎠⎟

0

122 1

2

= ⎛⎝⎜ − ⎞⎠⎟

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

=

→

−

−

→

lim ln sin sin lim

sin

x

x

x

x

x

x

0

2

1

(^22)
(^22)
12
2
2
2
00
2 2
(^20)
−

sin x
x
(B)
lim
ln
lim
ln( )
lim
xh hln[( ) /h]
x
x
h
→→ →h h−−
− =
−

−
=−
10 011
1
1
1
1
1
(C)
lim
sin
x tan
xx
x
xx
x
→
−
0 − =− =−
3
3
1
6
1
3
1
2

5. (c)


6. We h a v e cos

cos

cos

y x

x

= 33

... (1)

=^43 − = 43 −

3

3

cos cos 2

cos

xx sec

x

x

⇒ cos^2 y = 4 – 3sec^2 x = 4 – 3(1 + tan^2 x)
= 1 – 3tan^2 x
⇒ sin^2 y = 3tan^2 x ⇒ sinyx=^3 tan

⇒ cosydy= sec

dx

3 2 x

∴ =

⋅

dy

dx x x

3

3

1

cos cos

[()]from

7. ∵ φ : R → R is continuous at x = a and satisfies
   f(x) – f(a) = φ(x)(x – a) ∀ x ∈ R

⇒ −

−

fx fa=

xa

() ()φ()x

⇒ −

−

=

→→

lim () () lim ( )

xa xa

fx fa

xa

φx

⇒ ′ ==

→

fa a x a

xa

()φφφ() [ lim ()∵ ()]

⇒ f is differentiable at x = a.

8. Given cosα + cos(α + β) + cos(α + β + γ) = 0
   sinα + sin(α + β) + sin(α + β + γ) = 0
   where β, γ ∈ (0, π)
   ⇒ [cosα + cos(α + β)]^2 + [sinα + sin(α + β)]^2 = 1
   ⇒ 2 + 2[cosβ] = 1
   ∴ cosβ=−^1
   2

.

Similarly, cosγ=−^1

2 ,

∴ βγ==^2 π

3

Now, fx x

x

() sin x

cos

= tan

+

(^2) =
12
and gx() tan= x 2
∴ f′⎛⎜⎝^23 ππ⎟⎠⎞=sec^223 = 4
and lim ( ) tan
x
gx
→
2 ==
3
π 3 3
π

9. Let O = (0, 0)

A

x

x x

=

⎡ −

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎛

⎝

⎜

⎜

⎜

⎞

⎠

⎟

⎟

⎟

=−

→

lim

cos

π ,(,)

π

2

(^2020)
B x
x x
x
= ⎛⎝⎜ ⎞⎠⎟
⎛
⎝
⎜
⎞
⎠
(^001) → 0 ⎟=
1
, lim tan (, )
/
∴ Area of ΔOAB=^1 −−=
2
20 1 sq. unit

10. We h a v e al + bm + cn = 0 ... (1)
    fmn + gnl + hlm = 0 ... (2)
    Eliminate n, we get

⇒ ag⎝⎜⎛l⎟⎠⎞ +++()⎜⎛⎝ ⎟⎠⎞+=

m

af bg ch l

m

bf

2

0

... (3)

Now, if l 1 , m 1 , n 1 and l 2 , m 2 , n 2 are d.c.’s of two lines

then roots of (3) are l

m

l

m

1

1

2

2

and.

∴ product of the roots =

l

m

l

m

bf

ag

1

1

2

2

. =

∴ ll =

fa

mm

gb

12 1 2

//

∴ ll ==

fa

mm

gb

nn

hc

12 1 2 1 2

///

∵ lines are perpendicular

∴ l 1 l 2 + m 1 m 2 + n 1 n 2 = 0

⇒ f++=

a

g

b

h

c

(^0) ””