Relations and Functions Test - 85

Given here,

\(f(x)=x^{4}-\frac{1}{x^{4}}\) ......(i)

Replace \(\mathrm{x}\) by \(\frac{1}{\mathrm{x}}\), we get,

\(f\left(\frac{1}{x}\right)=\frac{1}{x^{4}}-x^{4}\) ......(ii)

So, from (i) and (ii),

\(\mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{x}^{4}-\frac{1}{\mathrm{x}^{4}}+\frac{1}{x^{4}}-\mathrm{x}^{4}\)

\(=0\)

Given:

\(\mathrm{f}(\mathrm{x})=\mathrm{e}^{2 \mathrm{x}}\) and \(\mathrm{g}(\mathrm{x})=\ln \mathrm{x}\)

\(\operatorname{gof}(x)=g(f(x))\)

\(\operatorname{gof}(x)=g\left(e^{2 x}\right)\)

\(\operatorname{gof}(x)=\ln e^{2 x}\)

\(\operatorname{gof}(x)=2 x \ln e\)

\(\operatorname{gof}(x)=2 x \quad \{\because \ln e=1\}\)

Given:

\(a * b=\sqrt{a^{2}+b^{2}}\)

\(\therefore(3 * 4) * 5\)

\(=\sqrt{(3 * 4)^{2}+5^{2}}\)

\(=\sqrt{\left(\sqrt{3^{2}+4^{2}}\right)^{2}+5^{2}}\)

\(=\sqrt{3^{2}+4^{2}+5^{2}}\)

\(=\sqrt{9+16+25}\)

\(=\sqrt{50}\)

\(=5 \sqrt{2}\)

Given here,

\(R=\{(a, b): a, b \in Z\) and \((a+b)\) is even \(\}\)

We know that:

R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Therefore,

1. Since, \(a+a=2 a\), which is even, so \(R\) is reflexive.

2. If \(a+b\) is even, then \(b+a\) will also be even. So, \(R\) is symmetric.

3. Let, \(a=3, b=5\), and \(c=7\), then:

\(a+b=3+5=8\) (which is even),

b \(+c=5+7=12\) (which is again even)

and, \(a+c=3+7=10\) (which is also even)

So, R is transitive.

Therefore, \(R\) is an equivalence relation on \(Z\).

Given here,

\(f(x)=\frac{x+1}{x-1}\)

We know that:

\(\operatorname{fog}(x)=f[g(x)]\)

\(f\{f(x)\}=\frac{f(x)+1}{f(x)-1}\)

\(=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}\)

\(=\frac{2 x}{2}\)

\(=x\)

Given that:

\({ }^{*}\) is a binary operation on \(Z\) such that:

\(a^{*} b=a+b+1 \forall a, b \in Z\)

Let e be the identity element of \(Z\) with respect to *.

As we know that if e is an identity element of a non-empty set S with respect to a binary operation * then a * e = e * a = a ∀ a ∈ S.

Let a ∈ Z and because e is the identity element of Z with respect to given operation *.

i.e., a * e = a = e * a ∀ a ∈ Z.

According to the definition of *, we have:

⇒ a * e = a + e + 1 = a ∀ a ∈ Z

⇒ e = - 1

Therefore, - 1 is the identity element of Z with respect to given operation *.

If the function is onto and one-to-one, then it is called as Bijection.

If a function is said to be one-to-one, then it is called as Injective.

If a function is said to be onto, then it is called as Surjective.

If f is a bijection from set A to set B, then the function g is called as Inverse function.

Given, 

\(f(x)=3 x, g(x)=3 x^{2}+9\) 

Then, 

\((f+g)(2)=f(2)+g(2)\)

\(=(3 \times 2)+\left(3 \times 2^{2}+9\right)\)

\(=6+21\) \(=27\)

\(\therefore\) The value of \((f+g)(2)\) is 27.

Given that: 

\(R=\left\{(a, b): a^{2} \geq b\right\}\)

As we know:

A relation R on a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

We know that: 

\(\mathrm{a}^{2} \geq \mathrm{a}\)

Therefore, \((\mathrm{a}, \mathrm{a}) \in \mathrm{R}\), for all \(\mathrm{a} \in \mathrm{Z}\).

Thus, relation \(\mathrm{R}\) is reflexive.

Let \((\mathrm{a}, \mathrm{b}) \in \mathrm{R}\)

\(\Rightarrow a^{2} \geq b\) but \(b^{2} \ngeq a\) for all \(a, b \in Z\).

So, if \((a, b) \in R\), then it does not implies that \((b, a)\) also belongs to \(\mathrm{R}\).

Thus, relation \(\mathrm{R}\) is not symmetric.

Now, let \((a, b) \in R\) and \((b, c) \in R\).

\(\Rightarrow a^{2} \geq b\) and \(b^{2} \geq c\)

This does not implies that \(\mathrm{a}^{2} \geq \mathrm{c}\), therefore \((\mathrm{a}, \mathrm{c})\) does not belong to \(\mathrm{R}\) for all \(\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{Z}\).

Thus, relation \(\mathrm{R}\) is not transitive.

Hence \(R\) is not a equivalence relation, but is a reflexive and symmetric relation.

Given: 

\(y=5^{\log x}\)

Here, we have to find the inverse of the given function.

By applying log to base 5 on both sides of \(y=5^{\log x}\), we get:

Let \(f: A \rightarrow B\) be a bijective function.

Step - I: Put \(f(x)=y\)

Step - II: Solve the equation \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) to obtain \(\mathrm{x}\) in terms of \(\mathrm{y}\). 

Interchange \(x\) and \(y\) to obtain the inverse of the given function \(f\)

\(\Rightarrow \log _{5} y=\log _{5}\left(5^{\log x}\right)\)

\(\Rightarrow \log _{5} y=\log x\)

\(\Rightarrow \frac{\log y}{\log 5}=\log x\)

\(\Rightarrow \log \left[y^{\frac{1}{\log 5} 5}\right]=\log x\)

\(\Rightarrow x=y^{\frac{1}{\log 5}}\)