1
KCET 2023
MCQ (Single Correct Answer)
+1
-0

$$f: R \rightarrow R$$ and $$g:[0, \infty) \rightarrow R$$ defined by $$f(x)=x^2$$ and $$g(x)=\sqrt{x}$$. Which one of the following is not true?

A
$$(f \circ g)(-4)=4$$
B
$$(f \circ g)(2)=2$$
C
$$(g \circ f)(-2)=2$$
D
$$(g \circ f)(4)=4$$
2
KCET 2023
MCQ (Single Correct Answer)
+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
$$\frac{3 x^2-5}{9 x^4-30 x^2+26}$$
3
KCET 2023
MCQ (Single Correct Answer)
+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
$$x \in\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]$$
4
KCET 2023
MCQ (Single Correct Answer)
+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
$$\frac{-5}{2}$$
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