JEE Advance - Mathematics (2010 - Paper 2 Offline - No. 16)
Consider the polynomial
$$f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}.$$
Let $$s$$ be the sum of all distinct real roots of $$f(x)$$ and let $$t = \left| s \right|.$$
The real numbers lies in the interval
$$\left( { - {1 \over 4},0} \right)$$
$$\left( { - 11, - {3 \over 4}} \right)$$
$$\left( { - {3 \over 4}, - {1 \over 2}} \right)$$
$$\left( {0,{1 \over 4}} \right)$$
Explanation
Given, $$f(x) = 4{x^3} + 3{x^2} + 2x + 1$$
$$f'(x) = 2(6{x^2} + 3x + 1)$$
$$D = 9 - 24 < 0$$
Hence, f(x) = 0 has only one real root.
$$f\left( { - {1 \over 2}} \right) = 1 - 1 + {3 \over 4} - {4 \over 8} > 0$$
$$f\left( { - {3 \over 4}} \right) = 1 - {6 \over 4} + {{27} \over {16}} - {{108} \over {64}}$$
$$ = {{64 - 96 + 108 - 108} \over {64}} < 0$$
f(x) changes its sign in $$\left( { - {3 \over 4},{{ - 1} \over 2}} \right)$$,
hence f(x) = 0 has a root in $$\left( { - {3 \over 4},{{ - 1} \over 2}} \right)$$.
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