JEE Main Previous Year Questions (2016- 2024): Inverse Trigonometric Functions | Mathematics for GRE Paper II PDF Download

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Q.1.then x is equal to:    (2019)




Ans. (1)
Solution.





 

Q.2. If x = sin^-1 (sin 10) and y = cos^-1 (cos 10), then y - x is equal to:   (2019)
(1) 0
(2) 10
(3) 7 π
(4) π
Ans. (4)
Solution.

 
 

Q.3.     (2019)
(1) 21/19
(2) 19/21
(3) 22/23
(4) 23/22
Ans. (1)
Solution.






Q.4. All x-satisfying the inequality (cot^-1 x)² - 7(cot^-1x)+ 10 > 0, lie in the interval:    (2019)
(1) (-∞, cot 5) ∪ (cot 4, cot 2)
(2) (cot 2, ∞)
(3) (-∞, cot 5) ∪ (cot 2, ∞)
(4) (cot 5, cot 4)
Ans. (2)
Solution.


⇒  ...(1)
But cot^-1 x lies in (0, π)
Now, from equation (1)
cot^-1x ∈  (0, 2)
Now, it is clear from the graph
x ∈ (cot 2, ∞)

Q.5. Considering only the principal values of inverse functions, the set     (2019)
(1) contains two elements
(2) contains more than two elements
(3)  is a singleton
(4) is an empty set
Ans. (3)
Solution.



Therefore, A is a singleton set.

Q.6.then α - β is equal to:    (2019)




Ans. (4)
Solution.

Q.7.where -1 ≤ x ≤ 1, -2 ≤ y ≤ 2,  then for all x, y, 4x² -4xy cosα +y² is equal to:    (2019)
(1) 4 sin²α
(2) 2 sin²α
(3) 4 sin²α - 2x²y²
(d) 4 cos² α + 2x²y²
Ans. (1)
Solution.




Q.8. The value of  is equal to:    (2019)




Ans. (2)
Solution.



Q.9. If for x∈  the derivative of tan^-1 is √x × g(x) , then g(x) equals     (2017)
(1)
(2)
(3)
(4)
Ans. (2)
Solution.


⇒
⇒


Q.10. The value of  is equal to:     (2017)




Ans. (1)
Solution.
x² = cos2θ ; θ =
=
=

=

Q.11. A value of x satisfying the equation sin[cot^-1(1+x)]=cos[tan^-1x], is:    (2017)
(1)
(2) 0
(3) -1
(4) 1/2
Ans. (1)
Solution.
Given that sin[cot^-1 (x+1)] = cos (tan^-1x) ......(i)
We know that,

Here, A = x + 1







or 1 + 2x = 0
or   

Q.12. Consider 
A normal to y = f(x) at x = π/6 also passes through the point:    (2016)
(1) (0, 0)
(2) (0, 2π/3)
(3) (π/6, 0)
(4) (π/4, 0)
Ans. (2)