1
VITEEE 2024
MCQ (Single Correct Answer)
+1
-0

The sum of the infinite terms of the series $\cot ^{-1}\left(1^2+\frac{3}{4}\right)+\cot ^{-1}\left(2^2+\frac{3}{4}\right)$ $+\cot ^{-1}\left(3^2+\frac{3}{4}\right)+\ldots$ is equal to

A
$\tan ^{-1}(1)$
B
$\tan ^{-1}(2)$
C
$\tan ^{-1}(3)$
D
$\tan ^{-1}(4)$
2
VITEEE 2023
MCQ (Single Correct Answer)
+1
-0

If $$\left(\sin ^{-1} x\right)^2-\left(\cos ^{-1} x\right)^2=a \pi^2$$, then the range of $$a$$ is

A
$$\left[\frac{-3}{4}, \frac{1}{4}\right]$$
B
$$\left[\frac{-3}{4}, \frac{3}{4}\right]$$
C
$$[-1,1]$$
D
$$\left[-1, \frac{3}{4}\right]$$
3
VITEEE 2023
MCQ (Single Correct Answer)
+1
-0

The equation $$3 \cos ^{-1} x-\pi x-\pi / 2=0$$ has

A
one negative solution
B
one positive solution
C
no solution
D
more than one solution
4
VITEEE 2023
MCQ (Single Correct Answer)
+1
-0

If $$\cot ^{-1}(y)=\cot ^{-1}(x)+\cot ^{-1}\left(\frac{x^2-1}{2 x}\right)$$, then the value of $$y$$ is

A
$$\frac{3 x-x^3}{3 x^2-1}$$
B
$$\frac{x^3-3 x}{3 x^2-1}$$
C
$$\frac{x^3-3 x}{1+3 x^2}$$
D
$$\frac{3 x+x^3}{1+3 x^2}$$
VITEEE Subjects
EXAM MAP