Inverse Trigonometric Functions, Solved Paper Question: Jee 1 year MATH, Math - Meritnation

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Inverse Trigonometric Functions

All x satisfying the inequality (cot⁻¹ x)² − 7 (cot⁻¹ x) + 10 > 0, lie in the interval:

JEE Mains 2019

If x = sin^–1(sin10) and y = cos^–1(cos10), then y – x is equal to:

JEE Mains 2019

\cos^{- 1} (\frac{2}{3 x}) + \cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2} (x > \frac{3}{4}),

then x is equal to:

JEE Mains 2019

The number of real solutions of the equation

\sin^{- 1} (\sum_{i = 1}^{\infty} x^{i + 1} - x \sum_{i = 1}^{\infty} {(\frac{x}{2})}^{i}) = \frac{π}{2} - \cos^{- 1} (\sum_{i = 1}^{\infty} {(- \frac{x}{2})}^{i} - x \sum_{i = 1}^{\infty} {(- x)}^{i})

lying in the interval

(- \frac{1}{2}, \frac{1}{2})

is _____ .
(Here, the inverse trigonometric functions sin⁻¹𝑥 and cos⁻¹𝑥 assume values in [

- \frac{π}{2}, \frac{π}{2}

] and [0, 𝜋], respectively.)

JEE Advanced 2018

For any positive integer n, define f_n : (0, ∞) → ℝ as

f_{n} (x) = \sum_{j = 1}^{n} \tan^{- 1} (\frac{1}{1 + (x + j) (x + j - 1)})

for all x ∈ (0, ∞).

(Here, the inverse trigonometric function \tan^{- 1} x assumes values in (- \frac{π}{2}, \frac{π}{2}) .)

Then, which of the following statement(s) is (are) TRUE?

JEE Advanced 2018

α = 3 \sin^{- 1} (\frac{6}{11}) and β = 3 \cos^{- 1} (\frac{4}{9})

, where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)

JEE Advanced 2015

\tan^{- 1} y = \tan^{- 1} x + \tan^{- 1} (\frac{2 x}{1 - x^{2}}),

|x| < \frac{1}{\sqrt{3}}

. Then a value of y is:

JEE Mains 2015

f : [0, 4 π] \to [0, π]

be defined by f(x) = cos⁻¹ (cos x). The number of points x ∈

[0, 4 π]

satisfying the equation

f (x) = \frac{10 - x}{10}

JEE Advanced 2014

The value of cot is

JEE Advanced 2013

If x, y, z are in A.P. and tan⁻¹x, tan⁻¹y, and tan⁻¹z, are also in A.P., then :

JEE Mains 2013

Match List I with List II and select the correct answer using the code given below the lists:

	List I		List II
P.		1.
Q.	possible value of is	2.
R.		3.
S.	then possible value of x is	4.	1

JEE Advanced 2013

\cos^{- 1} (\cos (- \frac{17 π}{5}))

JEE Ranker's Test 0

If cos^–1x + cos^–1y + cos^–1z =

π

where –1 ≤ x, y ≤ 1 and x + y ≥ 0, then x² + y² + z² + 2xyz is equal to

JEE Ranker's Test 0

2(tan^–1 1 + tan^–1 2 + tan^–1 3) is equal to

JEE Ranker's Test 0

2 \cos^{- 1} x = \sin^{- 1} (2 x \sqrt{1 - x^{2}})

is valid for all values of x satisfying

JEE Ranker's Test 0

u = \cot^{- 1} \sqrt{\tan α} - \tan^{- 1} \sqrt{\tan α},

\tan (\frac{π}{4} - \frac{u}{2})

JEE Ranker's Test 0

If [x] denotes the integral part of x, then the domain of

f (x) = \cos^{- 1} (x + [x])

JEE Mains 0

The sum of infinite terms in the series $\cot^{- 1} (2) + \cot^{- 1} (\frac{9}{2}) + \cot^{- 1} (8) + \cot^{- 1} (\frac{25}{2}) + \cot^{- 1} (18) + . . .$ is

JEE Mains 0

Let the set of solution of the equation

\tan^{- 1} \sqrt{x (x + 1)} + \sin^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

is { α, β}. Then value of

\sqrt{α} + \sqrt{β}

JEE Mains 0

If 2sin⁻¹x + 2sin⁻¹y + 2sin⁻¹z + 3π = 0 then value of

\frac{x y z}{1 - x - y - z}

JEE Mains 0

If a, b, c and d ∈ [−1, 1] such that sin⁻¹a + sin⁻¹b + sin⁻¹ c + sin⁻¹ d = −2 $π$ , then what is the value of 1^a + 2^b + 3^c + 6^d?

JEE Mains 0

In a triangle ABC, if c² = a²+ b² and

\sin^{- 1} x = \sin^{- 1} \frac{ax}{c} + \sin^{- 1} \frac{bx}{c}

, where a, b, c are sides of the triangle,
then the total number of different value(s) of x that is(are) greater than or equal to zero is______

JEE Mains 0

If , then the value of x will be

JEE Mains 0

\sin^{- 1} x + \cos^{- 1} y + \cos^{- 1} 2 x y = \frac{π}{2}

then point (x, y) lies on-

JEE Mains 0

\tan^{- 1} a + \tan^{- 1} (1 - a) = \cot^{- 1} \frac{x}{y}; a \in (0, 1) then

JEE Mains 0

The difference between the highest and the lowest value of

{(\cos^{- 1} \sqrt{x})}^{3} + {(\sec^{- 1} \frac{1}{\sqrt{1 - x}})}^{3}

JEE Mains 0

If tan⁻¹x +cosec⁻¹

\sqrt{1 + y^{2}} = \sin^{- 1} \frac{3}{\sqrt{10}}

, where x, y ∈ z⁺, then the maximum value of

\frac{{(x + y)}^{2}}{27}

JEE Mains 0

Consider the following functions
Let f (x) = sin⁻¹x + cot⁻¹x + cos⁻¹x and let

g (x) = |\sin^{- 1} (\sin x)|

	List-I		List-II
P)	The domain of f is	1)	R
Q)	The domain of g is	2)	[−1, 1]
R)	The range of f is	3)	R-[−1,1]
S)	The range of g is	4)	$[0, \frac{π}{2}]$
		5)	$[0, \frac{π}{4}]$
		6)	$[\frac{3 π}{4}, \frac{5 π}{4}]$

The correct option is

JEE Advanced 0

Consider the following functions
Let f (x) =

[\tan^{- 1} |x|]

where [x] represents greatest integer less than or equal to x and let

g (x) = |\sin^{- 1} (\sin x)|

	List-I		List-II
P)	The domain of f is	1)	R-[−1,1]
Q)	The domain of g is	2)	[−1, 1]
R)	The range of f is	3)	R
S)	The range of g is	4)	{0, 1}
		5)	[0, $\frac{π}{2}$ ]
		6)	R-(0, 1)

The correct option is

JEE Advanced 0

Let the equation tan⁻¹y : tan⁻¹x = 6 : 1 is converted in the form of y = f (x).
Also,

\lim_{x \to 0}

f(x) tends to infinity if a⁶− 1 = (4k₁+ k₂) (a⁴− a²) where k₁, k₂∈ z⁺.
Then the possible value(s) of k₁k₂ is(are)

JEE Mains 0

The solution of the trigonometric equation is

\frac{{(\sin^{- 1} x)}^{3} + {(\cos^{- 1} x)}^{3}}{{(\tan^{- 1} x + \cot^{- 1} x)}^{3}} = 7

JEE Ranker's Test 0

Number of value of x satisfying the equation

\sin^{- 1} (\frac{1 + x^{2}}{2 x}) = \frac{π}{2} \sec (x - 1)

JEE Mains 0

For x ∈ (0, 1), f(x) = {x cos(cot^–1 x) + sin(cot^–1 x)}², then f′(x) equals

JEE Mains 0

\cos^{- 1} x = \tan^{- 1} x

JEE Advanced 0

If a, b and c are in GP and tan^–1 a, tan^–1 b and tan^–1 c are in AP, then

JEE Mains 0

Sum of the infinite series

\cot^{- 1} (1^{2} + \frac{3}{4}) + \cot^{- 1} (2^{2} + \frac{3}{4}) + \cot^{- 1} (3^{2} + \frac{3}{4}) + . . . . .

\tan^{- 1} k

, where the value of k is

JEE Advanced 0

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