1
KCET 2023
+1
-0

Let $$f: R \rightarrow R$$ be defined by $$f(x)=3 x^2-5$$ and $$g: R \rightarrow R$$ by $$g(x)=\frac{x}{x^2+1}$$, then $$g \circ f$$ is

A
$$\frac{3 x^2-5}{9 x^4-6 x^2+26}$$
B
$$\frac{3 x^2}{x^4+2 x^2-4}$$
C
$$\frac{3 x^2}{9 x^4+30 x^2-2}$$
D
$$\frac{3 x^2-5}{9 x^4-30 x^2+26}$$
2
KCET 2023
+1
-0

Let $$f(x)=\sin 2 x+\cos 2 x$$ and $$g(x)=x^2-1$$ then $$g(f(x))$$ is invertible in the domain

A
$$x \in\left[\frac{-\pi}{8}, \frac{\pi}{8}\right]$$
B
$$x \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$$
C
$$x \in\left[0, \frac{\pi}{4}\right]$$
D
$$x \in\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]$$
3
KCET 2023
+1
-0

If the function is $$f(x)=\frac{1}{x+2}$$, then the point of discontinuity of the composite function $$y=f(f(x))$$ is

A
$$\frac{5}{2}$$
B
$$\frac{2}{5}$$
C
$$\frac{1}{2}$$
D
$$\frac{-5}{2}$$
4
KCET 2022
+1
-0

The domain of the function $$f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}$$ is

A
$$[-2,0) \cap(0,1)$$
B
$$[-2,1)$$
C
$$[-2,0)$$
D
$$[-2,0) \cup(0,1)$$
EXAM MAP
Medical
NEET