JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 22)
Let a be an integer such that all the real roots of the polynomial
2x5 + 5x4 + 10x3 + 10x2 + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to ___________.
2x5 + 5x4 + 10x3 + 10x2 + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to ___________.
Answer
2
Explanation
Let, $$f(x) = 2{x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 10x + 10$$
$$ \Rightarrow f'(x) = 10({x^4} + 2{x^3} + 3{x^2} + 2x + 1)$$
$$ = 10\left( {{x^2} + {1 \over {{x^2}}} + 2\left( {x + {1 \over x}} \right) + 3} \right)$$
$$ = 10\left( {{{\left( {x + {1 \over x}} \right)}^2} + 2\left( {x + {1 \over x}} \right) + 1} \right)$$
$$ = 10{\left( {\left( {x + {1 \over x}} \right) + 1} \right)^2} > 0;\forall x \in R$$
$$ \therefore $$ f(x) is strictly increasing function. Since, it is an odd degree polynomial it will have exactly one real root.
Now, by observation.
$$f( - 1) = 3 > 0$$
$$f( - 2) = - 64 + 80 - 80 + 40 - 20 + 10$$
$$ = - 34 < 0$$
$$ \Rightarrow f(x)$$ has at least one root in $$( - 2, - 1) \equiv (a,a + 1)$$
$$ \Rightarrow a = - 2$$
$$ \Rightarrow $$ |a| = - 2
$$ \Rightarrow f'(x) = 10({x^4} + 2{x^3} + 3{x^2} + 2x + 1)$$
$$ = 10\left( {{x^2} + {1 \over {{x^2}}} + 2\left( {x + {1 \over x}} \right) + 3} \right)$$
$$ = 10\left( {{{\left( {x + {1 \over x}} \right)}^2} + 2\left( {x + {1 \over x}} \right) + 1} \right)$$
$$ = 10{\left( {\left( {x + {1 \over x}} \right) + 1} \right)^2} > 0;\forall x \in R$$
$$ \therefore $$ f(x) is strictly increasing function. Since, it is an odd degree polynomial it will have exactly one real root.
Now, by observation.
$$f( - 1) = 3 > 0$$
$$f( - 2) = - 64 + 80 - 80 + 40 - 20 + 10$$
$$ = - 34 < 0$$
$$ \Rightarrow f(x)$$ has at least one root in $$( - 2, - 1) \equiv (a,a + 1)$$
$$ \Rightarrow a = - 2$$
$$ \Rightarrow $$ |a| = - 2
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