JEE Main 2022 (Online) 24th June Morning Shift | Inverse Trigonometric Functions Question 39 | Mathematics

The domain of the function

$$f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}}$$ is :

$$( - \infty ,1) \cup (2,\infty )$$

$$(2,\infty )$$

$$\left[ { - {1 \over 2},1} \right) \cup (2,\infty )$$

$$\left[ { - {1 \over 2},1} \right) \cup (2,\infty ) - \left\{ 3,{{{3 + \sqrt 5 } \over 2},{{3 - \sqrt 5 } \over 2}} \right\}$$

JEE Main 2021 (Online) 1st September Evening Shift

MCQ (Single Correct Answer)

-1

$${\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))$$ is equal to :

(The inverse trigonometric functions take the principal values)

3$$\pi$$ $$-$$ 11

4$$\pi$$ $$-$$ 9

4$$\pi$$ $$-$$ 11

3$$\pi$$ + 1

JEE Main 2021 (Online) 31st August Evening Shift

MCQ (Single Correct Answer)

-1

The domain of the function

$$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$$ is :

$$\left[ {0,{1 \over 4}} \right]$$

$$[ - 2,0] \cup \left[ {{1 \over 4},{1 \over 2}} \right]$$

$$\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\} $$

$$\left[ {0,{1 \over 2}} \right]$$

JEE Main 2021 (Online) 27th August Evening Shift

MCQ (Single Correct Answer)

-1

Let M and m respectively be the maximum and minimum values of the function
f(x) = tan^$$-$$1 (sin x + cos x) in $$\left[ {0,{\pi \over 2}} \right]$$, then the value of tan(M $$-$$ m) is equal to :

$$2 + \sqrt 3 $$

$$2 - \sqrt 3 $$

$$3 + 2\sqrt 2 $$

$$3 - 2\sqrt 2 $$

PREVIOUS NEXT

Questions Asked from Inverse Trigonometric Functions (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions