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Inverse Trigonometric Functions - 1 - JEE MCQ


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30 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Inverse Trigonometric Functions - 1

Inverse Trigonometric Functions - 1 for JEE 2025 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Inverse Trigonometric Functions - 1 questions and answers have been prepared according to the JEE exam syllabus.The Inverse Trigonometric Functions - 1 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Inverse Trigonometric Functions - 1 below.
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Inverse Trigonometric Functions - 1 - Question 1

If ƒ(x) = √(2tan(x)),then f-1(√(2)) =

Detailed Solution for Inverse Trigonometric Functions - 1 - Question 1

Correct Answer :- b

Explanation :

Step 1: Write down the equation you need to solve
You want to find x such that
f(x) = √(2)
⟹ √(2 tan x) = √(2)

Step 2: Eliminate the square root
Square both sides (remembering (√A)2 = A):
(√(2 tan x))2 = (√(2))2
⟹ 2 tan x = 2

Step 3: Solve for tan x
Divide both sides by 2:
tan x = 2 / 2 = 1

Step 4: Find x from tan x = 1
On the principal branch (–π/2 < x < π/2), tan x = 1 gives
x = tan-1(1) = π/4

Step 5: State the inverse value
Therefore
f⁻¹(√(2)) = π/4

Inverse Trigonometric Functions - 1 - Question 2

The value of cos15º− sin15º is

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The maximum value of sin x + cos x is

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Inverse Trigonometric Functions - 1 - Question 4

The value of tan15+ cot15is

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The value of tan150 + cot 150 

Inverse Trigonometric Functions - 1 - Question 5

The number of solutions of the equation sin-1 x - cos-1 x = sin-1(1/2) is

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Hence , the given equation has only one solution.

Inverse Trigonometric Functions - 1 - Question 6

What is the maximum and minimum value of sin x +cos x?

Detailed Solution for Inverse Trigonometric Functions - 1 - Question 6

Let y= sin x + cos x

dy/dx=cos x- sin x

For maximum or minimum dy/dx=0

Setting cosx- sin x=0

We get cos x = sin x

x= π/4, 5π/4———-

Whether these correspond to maximum or minimum, can be found from the sign of second derivative.

d^2y/dx^2=-sin x - cos x=-1/√2–1/√2 (for x=π/4) which is negative. Hence x=π/4 corresponds to maximum.For x=5π/4

d^2y/dx^2=-(-1/√2)-(-1/√2)=2/√2 a positive quantity. Hence 5π/4 corresponds to minimum

Maximum value of the function

y= sin π/4 + cos π/4= 2/√2=√2

Minimum value is

Sin(5π/4)+cos (5π/4)=-2/√2=-√2

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*Multiple options can be correct
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Inverse Trigonometric Functions - 1 - Question 15

Equation of the image of the line x + y = sin–1(a3 + 1) + cos–1(a2 + 1) – tan–1(a + 1), a ∈ R about y-axis is given by-

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a = 0


Inverse Trigonometric Functions - 1 - Question 16

Number of solutions of the equation 2 cot–12 + cos–1(3/5) = cosec–1 x is

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*Multiple options can be correct
Inverse Trigonometric Functions - 1 - Question 18

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Inverse Trigonometric Functions - 1 - Question 19

Which one of the following can best represent the graph of the function, (x) = cos–1(2x2 – 1)?

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Inverse Trigonometric Functions - 1 - Question 26

The set of values of x, satisfying the equation tan2(sin–1x) > 1 is

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Inverse Trigonometric Functions - 1 - Question 30

cos−1[cos(2cot−1(√2−1)] is equal to

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