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 :
JEE · Math · previous-year question
- A.$${{11} \over 3}$$correct
- B.$${{7} \over 3}$$
- C.$${{13} \over 3}$$
- D.$${{14} \over 3}$$
Answer
A. $${{11} \over 3}$$
Explanation
$$\because$$ x = $$-$$1 be the roots of f(x) = 0 $$\therefore$$ Let $$f(x) = A(x + 1)(x - 1)$$ ...... (i) Now, $$f( - 2) + f(3) = 0$$ $$ \Rightarrow A[ - 1( - 2 - b) + 4(3 - b)] = 0$$ $$b = {{14} \over 3}$$ $$\therefore$$ Second root of f(x) = 0 will be $${{14} \over 3}$$ $$\therefore$$ Sum of roots $$ = {{14} \over 3} - 1 = {{11} \over 3}$$
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