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Question: Let \(f:\mathbb{R}\to \mathbb{R}\) be any function defining \(g:\mathbb{R}\to \mathbb{R}\) by \(g\le...

Let f:RRf:\mathbb{R}\to \mathbb{R} be any function defining g:RRg:\mathbb{R}\to \mathbb{R} by g(x)=f(x)g\left( x \right)=\left| f\left( x \right) \right| for all xRx\in \mathbb{R} then g is
A. onto if f is into
B. one-one if f is one-one
C. continuous if f is continuous
D. differentiable if f is differentiable

Explanation

Solution

The domain of the given function g(x)=f(x)g\left( x \right)=\left| f\left( x \right) \right| is real as f:RRf:\mathbb{R}\to \mathbb{R}. We try to form the modulus function and find the continuity of the curve. We assume the function y=f(x)y=f\left( x \right) and put the values in the main function g. We show that g(x)=f(x)g\left( x \right)=\left| f\left( x \right) \right| is continuous as long as f:RRf:\mathbb{R}\to \mathbb{R} and the domain of g is in real values.

Complete step-by-step answer:
In the function g(x)=f(x)g\left( x \right)=\left| f\left( x \right) \right|, the main function is the modulus function.
Let’s check the continuity of y\left| y \right| for y:RRy:\mathbb{R}\to \mathbb{R} where y=f(x)y=f\left( x \right).
yR\forall y\in \mathbb{R}, we can put the graph of the modulus function as

The graph of modulus is continuous itself. So, the condition for y\left| y \right| to be continuous yRy\in \mathbb{R}.
If there exists some y where yRy\notin \mathbb{R}, then y\left| y \right| is discontinuous.
Now it’s given that f:RRf:\mathbb{R}\to \mathbb{R}. So, the range of f is only real. So, all values of y=f(x)y=f\left( x \right) are accounted for in the real value range.
So, g(x)=f(x)g\left( x \right)=\left| f\left( x \right) \right| is also continuous as g:RRg:\mathbb{R}\to \mathbb{R}.
The correct option is C.

So, the correct answer is “Option C”.

Note: The relation between continuity and differentiability is that if a function is differentiable then the function is definitely continuous. But the opposite is not always true. So, if we show that y=f(x)y=f\left( x \right) is continuous, its differentiability is only known when we have the exact function. Now if we assume that y=f(x)=x2y=f\left( x \right)={{x}^{2}}, then we can see that the function f is not one-one but the function g is one-one. So, all the other options were not considered. In case of the modulus function whatever be the function of the outer always remains continuous.