Let f and g be twice differentiable even functions on (–2, 2) such that
\(ƒ(\frac{1}{4})=0, ƒ(\frac{1}{2})=0, ƒ(1) =1\) and \(g(\frac{3}{4}) = 0 , g(1)=2\)
.Then, the minimum number of solutions of f(x)g′′(x) + f′(x)g′(x) = 0 in (–2, 2) is equal to_____.
Let f and g be twice differentiable even functions on (–2, 2) such that
\(ƒ(\frac{1}{4})=0, ƒ(\frac{1}{2})=0, ƒ(1) =1\) and \(g(\frac{3}{4}) = 0 , g(1)=2\)
.Then, the minimum number of solutions of f(x)g′′(x) + f′(x)g′(x) = 0 in (–2, 2) is equal to_____.
Correct Answer: 4
Solution and Explanation
The correct answer is 4
Suppose h(x) = f(x).g′(x)
As f(x) is even
\(ƒ(\frac{1}{2}) = (\frac{1}{4}) = 0\)
\(⇒ ƒ(-\frac{1}{2}) = ƒ(-\frac{1}{4}) = 0\)
and g(x) is even ⇒ g′(x) is odd
and g(1) = 2 ensures one root of g′(x) is 0.
So , h(x) = f(x).g′(x) has minimum fives zeroes
Therefore h′(x) = f′(x).g′(x) + f(x).g′(x)=0,
has minimum 4 zeroes
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Concepts Used:
Derivatives
Derivatives are defined as a function's changing rate of change with relation to an independent variable. When there is a changing quantity and the rate of change is not constant, the derivative is utilised. The derivative is used to calculate the sensitivity of one variable (the dependent variable) to another one (independent variable). Derivatives relate to the instant rate of change of one quantity with relation to another. It is beneficial to explore the nature of a quantity on a moment-to-moment basis.
Few formulae for calculating derivatives of some basic functions are as follows:
