![\bf \begin{cases} f(x)=\cfrac{2}{x}\\[1em] g(x)=x^2+9 \end{cases}~\hspace{5em}f(~~g(x)~~)=\cfrac{2}{g(x)}\implies f(~~g(x)~~)=\cfrac{2}{x^2+9}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20f%28x%29%3D%5Ccfrac%7B2%7D%7Bx%7D%5C%5C%5B1em%5D%20g%28x%29%3Dx%5E2%2B9%20%5Cend%7Bcases%7D~%5Chspace%7B5em%7Df%28~~g%28x%29~~%29%3D%5Ccfrac%7B2%7D%7Bg%28x%29%7D%5Cimplies%20f%28~~g%28x%29~~%29%3D%5Ccfrac%7B2%7D%7Bx%5E2%2B9%7D)
that's one combination for f(x) and g(x), off many combinations.
0 is the real number for which there is no multiplicative inverse.
<u>Solution:</u>
Need to find a real number not having multiplicative inverse. Let’s first understand what is multiplicative inverse.
<em>Multiplicative inverse of a real number</em> is that number which when multiplied to the real number gives result as 1.
That is product of real number and multiplicative inverse is always 1.
Can you imagine any real number which whatever be the case, when multiplied by any number will never give result as 1.
Think about 0.
Whenever any number multiplied by 0, result is always zero. So there will never be a case when zero multiplied by any number gives result as 1.
Hence 0 is that real number for which there is no multiplicative inverse.
the answer to the question is c
