f: R rightarrow R and g:[0, infty) rightar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $ f(x) = x^2 $ and $ g(x) = \sqrt{x} $.Check option $ (A) $: $ f \circ g(-4) = f(g(-4)) $. Since the domain of $ g $ is $ [0, \infty) $,$ g(

Vedclass · Accepted Answer

$ f \circ g(-4) = 4 $

Vedclass · Answer

(A) Given $ f(x) = x^2 $ and $ g(x) = \sqrt{x} $.Check option $ (A) $: $ f \circ g(-4) = f(g(-4)) $. Since the domain of $ g $ is $ [0, \infty) $,$ g(-4) $ is not defined. Thus,$ f \circ g(-4) $ is not defined. This statement is not true.Check option $ (B) $: $ g \circ f(-2) = g(f(-2)) = g((-2)^2) = g(4) = \sqrt{4} = 2 $. This is true.Check option $ (C) $: $ g \circ f(4) = g(f(4)) = g(4^2) = g(16) = \sqrt{16} = 4 $. This is true.Check option $ (D) $: $ f \circ g(2) = f(g(2)) = f(\sqrt{2}) = (\sqrt{2})^2 = 2 $. This is true.Therefore,the statement that is not true is $ (A) $.

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

Similar Questions

Water is used to cool radiators of engines in cars because:

Let $\bar{a} = 3 \hat{i} + 2 \hat{j} + x \hat{k}$ and $\bar{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\bar{a} \times \bar{b}| = r$ is possible,if

The oxidation number of $Pt$ in $[Pt(C_2H_4)Cl_3]^-$ is

$A$ body of mass $m_1 = 4 \ kg$ moves at $5 \hat{i} \ m/s$ and another body of mass $m_2 = 2 \ kg$ moves at $10 \hat{i} \ m/s$. The kinetic energy of the centre of mass is

$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module