JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 2)
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 :
$${{11} \over 3}$$
$${{7} \over 3}$$
$${{13} \over 3}$$
$${{14} \over 3}$$
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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