1
KCET 2026
MCQ (Single Correct Answer)
+1
-0
The second order derivative of $\cos^{-1}(4x^3 - 3x)$ with respect to $\cos^{-1}(2x^2 - 1)$, where $\dfrac{1}{2} < x < 1$ is
A
$0$
B
$-\dfrac{1}{\sqrt{1 - x^2}}$
C
$\dfrac{3}{2}$
D
$-\dfrac{3}{2}$
2
KCET 2026
MCQ (Single Correct Answer)
+1
-0
$\tan^{-1}\left(\dfrac{1}{1 + 1 \cdot 2}\right) + \tan^{-1}\left(\dfrac{1}{1 + 2 \cdot 3}\right) + \ldots + \tan^{-1}\left(\dfrac{1}{1 + n(n+1)}\right) = $
A
$\tan^{-1}\left(\dfrac{n}{n+2}\right)$
B
$\tan^{-1}\left(\dfrac{n+1}{n}\right)$
C
$\tan^{-1}\left(\dfrac{n}{n+1}\right)$
D
$\tan^{-1}\left(\dfrac{n+2}{n}\right)$
3
KCET 2026
MCQ (Single Correct Answer)
+1
-0
If $\sin^{-1} x + \sin^{-1} y = \dfrac{\pi}{2}$, then $x^2$ is equal to
A
$1 - y^2$
B
$\sqrt{1 - y^2}$
C
$-\sqrt{1 - y^2}$
D
$1 + y^2$
4
KCET 2026
MCQ (Single Correct Answer)
+1
-0
$f(x) = (x-1)^2$ for $x \geq 1$, $g(x)$ is a function whose graph is the reflection of the graph of $f(x)$ in the line $y = x$, then $g(x)$ is
A
$-\sqrt{x} + 1$
B
$\sqrt{x} - 1$
C
$\sqrt{x} + 1$
D
$\sqrt{x - 1}$