Question

The number of continuous and derivable function(s) $f\left( x \right)$ such that $f\left( 1 \right) = - 1$, $f\left( 4 \right) = 7$ and \[f\left( x \right) > 3\] for all $x \in R$ is/are
A) 0
B) 1
C) 2
D) Infinite

Answer 1

Hint:
Write the given conditions for the function. Interpret the meaning of each given condition. Combine the given conditions. Since, \[f\left( x \right) > 3\] the range of $f\left( x \right)$ is greater than 3 and $f\left( 1 \right) = - 1$ which is a contradicting statement.

Complete step by step solution:
A function is a relation that relates each element of a set to exactly one element of the other set or same set.
A function should give an output for each value of $x \in R$.
Also, for each input, there should be only one output of $f\left( x \right)$
We have to find the number of functions that satisfies the given conditions, $f\left( 1 \right) = - 1$ ,$f\left( 4 \right) = 7$ and \[f\left( x \right) > 3\].
In a function, if $f\left( 1 \right) = - 1$, when 1 is substituted for \[x\], then the value of $f\left( x \right)$ is $ - 1$.
Similarly, in a function, if $f\left( 1 \right) = - 1$, when 4 is substituted for \[x\], then the value of $f\left( x \right)$ is 7.
Also, we are given that, \[f\left( x \right) > 3\], then the value of $f\left( x \right)$ is greater than 3. But, this condition contradicts the statement $f\left( 1 \right) = - 1$ as $ - 1 < 3$.
Thus, there is no function such that $f\left( 1 \right) = - 1$,$f\left( 4 \right) = 7$ and \[f\left( x \right) > 3\] for all $x \in R$.

Hence, option A is correct.

Note:
A function is a relation that relates each element of a set to exactly one element of the other set or same set. The elements that we can put for $x$ in $f\left( x \right)$ are the domain of the function and the values we get for $f\left( x \right)$ are the range of $f\left( x \right)$.