Find the integral values of K for which the system of equations. arc cos x + arc sin y 2 = K π 2 4 arc sin y 2 arc cos x = π 2 16 possesses solution & find those solutions.

We have, cos − 1 x + sin − 1 y 2 = K π 2 4 Then, 0 ≤ cos − 1 x < π . . . i 0 ≤ sin − 1 y 2 ≤ π 2 4 . . . ii Adding i & ii , we get 0 ≤ cos − 1 x + sin − 1 y 2 ≤ π + π 2 4 ⇒ 0 ≤ K π 2 4 ≤ π + π 2 4 ⇒ 0 ≤ K ≤ 4 π + 1 ⇒ 0 ≤ K ≤ 2 . 27 Also, sin − 1 y 2 cos − 1 x = π 2 16 Let cos − 1 x = t , then ⇒ t K π 2 4 − t = π 4 16 ⇒ t 2 − K π 2 4 t + π 4 16 = 0 For solution D ≥ 0 ⇒ K 2 π 4 16 − π 4 4 ≥ 0 ⇒ K 2 1 − 4 ≥ 0 ⇒ K 2 ≥ 4 ⇒ K ≥ 2 or K ≤ - 2 ∴ only integral value of K = 2 .

Express β 3 2 cosec 2 1 2 tan - 1 β α + α 3 2 sec 2 1 2 tan - 1 α β as in integral polynomial in α & β .

We have, β 3 2 cosec 2 1 2 tan - 1 β α + α 3 2 sec 2 1 2 tan - 1 α β = β 3 2 sin 2 1 2 tan - 1 β α + α 3 2 cos 2 1 2 tan - 1 α β = β 3 1 − cos tan − 1 β α + α 3 1 + cos tan − 1 α β = β 3 1 − cos cos − 1 α α 2 + β 2 + α 3 1 + cos cos − 1 β α 2 + β 2 = β 3 1 − α α 2 + β 2 + α 3 1 + β α 2 + β 2 = α 2 + β 2 β 3 α 2 + β 2 - α + α 3 α 2 + β 2 + β = α 2 + β 2 β 3 α 2 + β 2 + α α 2 + β 2 - α α 2 + β 2 + α + α 3 α 2 + β 2 - β α 2 + β 2 + β α 2 + β 2 - β = α 2 + β 2 β 3 α 2 + β 2 + α α 2 + β 2 - α 2 + α 3 α 2 + β 2 - β α 2 + β 2 - β 2 = α 2 + β 2 β α 2 + β 2 + α + α α 2 + β 2 - β = α 2 + β 2 β α 2 + β 2 + β α + α α 2 + β 2 - α β = α 2 + β 2 α + β

Find the number of roots of the following equations: 1 + cos 2 x = 2 sin - 1 sin x

We have, 1 + cos 2 x = 2 sin - 1 sin x ⇒ 1 + 2 cos 2 x - 1 = 2 sin - 1 sin x ⇒ 2 cos x = 2 sin − 1 sin x ⇒ cos x = sin − 1 sin x cos x ∈ 0 , 1 ∀ x ∈ R Infinite points of intersections of cos x and sin − 1 sin x are possible.

Find the number of roots of the following equations: sin sin - 1 log 1 2 x + 2 cos sin - 1 x 2 - 1 = 0

We have, sin sin - 1 log 1 2 x + 2 cos sin - 1 x 2 - 1 = 0 ⇒ log 1 2 x + 2 cos sin - 1 x - 2 2 = 0 ⇒ log 1 2 x + 2 cos cos - 1 4 - x - 2 2 2 = 0 ⇒ log 1 2 x + 2 4 - x - 2 2 2 = 0 ⇒ log 1 2 x + 4 x - x 2 = 0 ⇒ 4 x - x 2 = - log 1 2 x Both the graphs are intersecting at one point, hence only one root is possible for the given equation.

E M B I B E

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

Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 4: EXERCISE-4

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 4: EXERCISE-4

Attempt the free practice questions on Chapter 15: Inverse Trigonometric Functions, Exercise 4: EXERCISE-4 with hints and solutions to strengthen your understanding. Beta Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 4: EXERCISE-4 with Hints & Solutions

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JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 35

Find the integral values of $K$ for which the system of equations.

$[\begin{matrix} arc \cos x + {(arc \sin y)}^{2} = \frac{K π^{2}}{4} \\ {(arc \sin y)}^{2} (arc \cos x) = \frac{π^{2}}{16} \end{matrix}$ possesses solution & find those solutions.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 36

Express $\frac{β^{3}}{2} {cosec}^{2} [\frac{1}{2} \tan^{- 1} \frac{β}{α}] + \frac{α^{3}}{2} \sec^{2} [\frac{1}{2} \tan^{- 1} \frac{α}{β}]$ as in integral polynomial in $α$ & $β$ .