Draw the graph of the following function: g x = cos - 1 x , where · denotes greatest integer function.

We have, g x = cos - 1 x Domain of cos - 1 x is - 1 , 1 and domain of greatest integer function is R , hence domain of g x is - 1 , 1 . Range of cos - 1 x is 0 , π . Now, when cos 1 < x ≤ 1 ⇒ cos - 1 1 ≤ cos - 1 x < cos - 1 cos 1 ⇒ 0 ≤ cos - 1 x < 1 ⇒ cos - 1 x = 0 When cos 2 < x ≤ cos 1 ⇒ cos - 1 cos 1 ≤ cos - 1 x < cos - 1 cos 2 ⇒ 1 ≤ cos - 1 x < 2 ⇒ cos - 1 x = 1 Similarly, cos 3 < x ≤ cos 2 ⇒ cos - 1 cos 2 ≤ cos - 1 x < cos - 1 cos 3 ⇒ 2 ≤ cos - 1 x < 3 ⇒ cos - 1 x = 2 Hence, required graph is as follows:

Draw the graph of the following function: g x = cos - 1 x , where · denotes greatest integer function.

We have, g x = cos - 1 x Domain of cos - 1 x is - 1 , 1 and domain of greatest integer function is R , hence domain of g x is - 1 , 1 . Range of cos - 1 x is 0 , π . Now, when cos 1 < x ≤ 1 ⇒ cos - 1 1 ≤ cos - 1 x < cos - 1 cos 1 ⇒ 0 ≤ cos - 1 x < 1 ⇒ cos - 1 x = 0 When cos 2 < x ≤ cos 1 ⇒ cos - 1 cos 1 ≤ cos - 1 x < cos - 1 cos 2 ⇒ 1 ≤ cos - 1 x < 2 ⇒ cos - 1 x = 1 Similarly, cos 3 < x ≤ cos 2 ⇒ cos - 1 cos 2 ≤ cos - 1 x < cos - 1 cos 3 ⇒ 2 ≤ cos - 1 x < 3 ⇒ cos - 1 x = 2 Hence, required graph is as follows:

Express f x = cos - 1 x + cos - 1 x 2 + 1 2 3 - 3 x 2 in simplest form and hence find the values of f 2 3

We have, f x = cos − 1 x + cos − 1 x 2 + 1 2 3 − 3 x 2 ⇒ f x = cos − 1 x + cos − 1 x 2 + 3 2 1 − x 2 ⇒ f x = cos − 1 x + cos − 1 x · 1 2 + 1 - 1 2 2 1 − x 2 ⇒ f x = cos − 1 x + cos − 1 1 2 - cos − 1 x ⇒ f x = cos − 1 1 2 = π 3 ∴ f 2 3 = π 3

Express f x = arc cos x + arc cos x 2 + 1 2 3 - 3 x 2 in simplest form and hence find the value of f 1 3 .

We have, f x = cos − 1 x + cos − 1 x 2 + 1 2 3 − 3 x 2 ⇒ f x = cos − 1 x + cos − 1 x 2 + 3 2 1 − x 2 ⇒ f x = cos − 1 x + cos − 1 x · 1 2 + 1 - 1 2 2 1 − x 2 ⇒ f x = cos − 1 x + cos − 1 x - cos − 1 1 2 ⇒ f x = 2 cos − 1 x - cos − 1 1 2 ⇒ f x = 2 cos − 1 x - π 3 ∴ f 1 3 = 2 cos − 1 1 3 - π 3

If α = 2 tan - 1 1 + x 1 - x & β = sin - 1 1 - x 2 1 + x 2 for 0 < x < 1 , then prove that α + β = π . What the value of α + β will be if x > 1 ?

We have α = 2 tan - 1 1 + x 1 - x & β = sin - 1 1 - x 2 1 + x 2 for 0 < x < 1 Let x = tan θ , where 0 < θ < π 4 Then, α = 2 tan - 1 tan π 4 + tan θ 1 - tan π 4 tan θ = 2 tan - 1 . tan θ + π 4 As we know, tan - 1 tan ( A ) = A ⇒ α = 2 θ + π 4 = 2 θ + π 2 . . . ( i ) Similarly, β = sin - 1 1 - tan 2 θ 1 + tan 2 θ = sin - 1 ( cos 2 θ ) = sin - 1 sin π 2 - 2 θ As we know, sin - 1 sin ( A ) = A ⇒ β = π 2 - 2 θ . . . ( ii ) From ( i ) & ( ii ) , we get α + β = 2 θ + π 2 + π 2 - 2 θ α + β = π Hence proved. Now, to find α + β for x > 1 Again, let x = tan θ , where θ = π 4 Then. α = 2 tan - 1 tan θ + π 4 - π = 2 θ + π 4 - π ⇒ α = 2 θ - 3 π 2 Similarly, β = sin - 1 sin π 2 - 2 θ ⇒ β = π 2 - 2 θ ( ∵ No change in β ) α + β = - π

If cos - 1 ⁡ x a + cos - 1 ⁡ y b = α , prove that x 2 a 2 - 2 x y a b cos ⁡ α + y 2 b 2 = sin 2 ⁡ α .

We have cos - 1 x a = α - cos - 1 y b ⇒ x a = cos α - cos - 1 y b ∵ cos A - B = cos A cos B + sin A sin B ⇒ x a = cos α cos cos - 1 y b + sin α sin cos - 1 y b Let cos - 1 y b = θ ⇒ cos θ = y b ⇒ sin θ = ± 1 - y b 2 ∵ sin 2 α + cos 2 α = 1 ⇒ θ = sin - 1 ± 1 - y b 2 Now, x a = cos α cos cos - 1 y b + sin α sin sin - 1 ± 1 - y 2 b 2 ⇒ x a = cos α · y b + sin α ± 1 - y 2 b 2 ⇒ x a - cos α · y b = ± sin α 1 - y 2 b 2 Squaring both sides, we get x 2 a 2 + y 2 b 2 cos 2 α - 2 x y a b cos α = sin 2 α 1 - y 2 b 2 . ⇒ x 2 a 2 - 2 x y a b cos α + y 2 b 2 cos 2 α + sin 2 α = sin 2 α ⇒ x 2 a 2 - 2 x y a b cos α + y 2 b 2 = sin 2 α

Prove that cos - 1 x = 2 sin - 1 1 - x 2 = 2 cos - 1 1 + x 2 .

We need to prove that, cos - 1 x = 2 sin - 1 1 - x 2 = 2 cos - 1 1 + x 2 Taking R . H . S . = 2 sin - 1 1 - x 2 = 2 cos - 1 1 + x 2 Let x = cos θ ⇒ θ = cos - 1 x ⇒ 2 sin - 1 1 - cos θ 2 = 2 cos - 1 1 + cos θ 2 ⇒ 2 sin - 1 2 sin 2 θ 2 2 = 2 cos - 1 2 cos 2 θ 2 2 ∵ 1 - cos x = 2 sin 2 x 2 ; 1 + cos x = 2 cos 2 x 2 ⇒ 2 sin - 1 sin 2 θ 2 = 2 cos - 1 cos 2 θ 2 ⇒ 2 sin - 1 sin θ 2 = 2 cos - 1 cos θ 2 ⇒ 2 × θ 2 = 2 × θ 2 ∵ sin - 1 sin x = x ; cos - 1 cos x = x ⇒ θ = θ ⇒ cos - 1 x = cos - 1 x [ Substituting value from θ ] = R . H . S . Hence proved.

If sin 2 x + sin 2 y < 1 for all x , y ∈ R , then prove that sin - 1 tan x tan y ∈ - π 2 , π 2

We have, sin 2 x + sin 2 y < 1 Let tan x = a ⇒ sin x = a 1 + a 2 and tan y = b ⇒ sin y = b 1 + b 2 . Now, sin 2 x + sin 2 y < 1 ⇒ a 2 1 + a 2 + b 2 1 + b 2 < 1 ⇒ a 2 1 + b 2 + b 2 1 + a 2 < 1 + a 2 1 + b 2 ⇒ a 2 + a 2 b 2 + b 2 + b 2 a 2 < 1 + a 2 + b 2 + a 2 b 2 ⇒ a 2 b 2 < 1 ⇒ a b < 1 ⇒ tan x tan y < 1 ⇒ sin − 1 tan x tan y ∈ − π 2 , π 2

Solve the following equation: sin - 1 x + sin - 1 2 x = π 3

We have, sin - 1 x + sin - 1 2 x = π 3 ⇒ s i n - 1 x 1 - ( 2 x ) 2 + 2 x 1 - x 2 = π 3 ∵ sin - 1 x + sin - 1 y = sin - 1 ( x 1 - y 2 + y 1 - x 2 ⇒ s i n - 1 x 1 - 4 x 2 + 2 x 1 - x 2 = π 3 ⇒ x 1 - 4 x 2 + 2 x 1 - x 2 = s i n π 3 = 3 2 ⇒ 2 x 1 - 4 x 2 + 4 x 1 - x 2 = 3 ⇒ 2 x 1 - 4 x 2 = 3 - 4 x 1 - x 2 Squaring both sides 4 x 2 1 - 4 x 2 = 3 + 16 x 2 1 - x 2 - 8 3 x 1 - x 2 ⇒ 8 3 x 1 - x 2 = 16 x 2 1 - x 2 + 3 - 4 x 2 ( 1 - 4 x 2 ) ⇒ 8 3 x 1 - x 2 = 16 x 2 - 16 x 4 + 3 - 4 x 2 + 16 x 4 ⇒ 8 3 x 1 - x 2 = 12 x 2 + 3 Again squaring both sides 64 · 3 · x 2 ( 1 - x 2 ) = 12 x 2 + 3 2 ⇒ 192 x 2 1 - x 2 = 144 x 4 + 9 + 72 x 2 ⇒ 192 x 2 - 192 x 4 = 144 x 4 + 9 + 72 x 2 ⇒ 336 x 4 - 120 x 2 + 9 = 0 ⇒ 112 x 4 - 40 x 2 + 3 = 0 ⇒ 28 x 2 - 3 4 x 2 - 1 = 0 ⇒ x 2 = 3 28 or x 2 = 1 4 ⇒ x = ± 3 28 , ± 1 2

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 11

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Draw the graph of the following function:

$g (x) = [\cos^{- 1} x]$ , where $[\cdot]$ denotes greatest integer function.

Important Questions on Inverse Trigonometric Functions

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 12

Draw the graph of the following function:

$h (x) = - |\tan^{- 1} (3 x)|$

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 13

Express $f (x) = \cos^{- 1} x + \cos^{- 1} (\frac{x}{2} + \frac{1}{2} \sqrt{3 - 3 x^{2}})$ in simplest form and hence find the values of

$f (\frac{2}{3})$

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 14

Express $f (x) = arc \cos x + arc \cos (\frac{x}{2} + \frac{1}{2} \sqrt{3 - 3 x^{2}})$ in simplest form and hence find the value of $f (\frac{1}{3})$ .