Q)The number of solutions of the equation a^f(x) + g(x) = 0,a>0,g(x) is not equal to zero and has minimum value of ½,is

a)one b)two c) zero d)infinitely many

PLZ solve it sir

To determine the number of solutions for the equation \( a f(x) + g(x) = 0 \) under the given conditions, we need to analyze the functions involved and their properties. Let's break it down step by step.

Understanding the Functions

We have two functions here: \( f(x) \) and \( g(x) \). The parameter \( a \) is a positive constant, and it is given that \( g(x) \) is not equal to zero and has a minimum value of \( \frac{1}{2} \). This means that for all \( x \), \( g(x) \geq \frac{1}{2} \).

Rearranging the Equation

The equation can be rearranged to express \( g(x) \) in terms of \( f(x) \):

\( g(x) = -a f(x) \)

Since \( a > 0 \), the sign of \( g(x) \) will depend on the sign of \( f(x) \). If \( f(x) \) is positive, \( g(x) \) will be negative, which contradicts the condition that \( g(x) \) is always at least \( \frac{1}{2} \). Therefore, \( f(x) \) must be negative or zero for \( g(x) \) to remain non-negative.

Analyzing the Minimum Value of g(x)

Given that the minimum value of \( g(x) \) is \( \frac{1}{2} \), we can infer that:

• For \( g(x) \) to equal \( 0 \), \( a f(x) \) would have to equal \( -g(x) \). However, since \( g(x) \) cannot be less than \( \frac{1}{2} \), \( a f(x) \) can never reach \( -\frac{1}{2} \) or lower.
• This means that \( a f(x) + g(x) = 0 \) can never be satisfied because \( g(x) \) is always at least \( \frac{1}{2} \), and thus \( a f(x) \) would need to be at least \( -\frac{1}{2} \), which is not possible if \( g(x) \) is always positive.

Conclusion on the Number of Solutions

Since \( g(x) \) cannot equal \( -a f(x) \) for any \( x \) due to the constraints on \( g(x) \), we conclude that there are no values of \( x \) that satisfy the equation \( a f(x) + g(x) = 0 \). Therefore, the number of solutions to the equation is:

c) zero