Let f(x) be a quadratic polynomial such that f(–2) + f(3) = 0. If one of the roots of f(x) = 0 is –1, then the sum of the roots of f(x) = 0 is equal to:
\(\frac{11}{3}\)
\(\frac{7}{3}\)
\(\frac{13}{3}\)
\(\frac{14}{3}\)
The Correct Option is A
Solution and Explanation
The correct answer is (A) : \(\frac{11}{3}\)
∵ x = –1 be the roots of f(x) = 0
∴ let f(x) = A(x + 1)(x – b) …(i)
Now, f(–2) + f(3) = 0
⇒ A[–1(–2 – b) + 4(3 – b)] = 0
b=14/3
∴ Second root of f(x) = 0 will be \(\frac{14}{3}\).
∴ Sum of roots
\(=\frac{14}{3}−1=\frac{11}{3}\)
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions