JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 22)
Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x = $$-$$1 and x = 1. If $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {{x^3}}} = 1$$, then $$5.f(2)$$ is equal to _________.
Answer
144
Explanation
$$f(x) = {x^6} + a{x^5} + b{x^4} + {x^3}$$
$$\therefore$$ $$f'(x) = 6{x^5} + 5a{x^4} + 4b{x^3} + 3{x^2}$$
Roots 1 & $$-$$1
$$ \therefore $$ $$6 + 5z + 4b + 3 = 0$$ & $$ - 6 + 5a - 4b + 3 = 0$$ solving
$$a = - {3 \over 5}$$
$$b = - {3 \over 2}$$
$$ \therefore $$ $$f(x) = {x^6} - {3 \over 5}{x^5} - {3 \over 2}{x^4} + {x^3}$$
$$ \therefore $$ $$5.f(2) = 5\left[ {64 - {{96} \over 5} - 24 + 8} \right] = 144$$
$$\therefore$$ $$f'(x) = 6{x^5} + 5a{x^4} + 4b{x^3} + 3{x^2}$$
Roots 1 & $$-$$1
$$ \therefore $$ $$6 + 5z + 4b + 3 = 0$$ & $$ - 6 + 5a - 4b + 3 = 0$$ solving
$$a = - {3 \over 5}$$
$$b = - {3 \over 2}$$
$$ \therefore $$ $$f(x) = {x^6} - {3 \over 5}{x^5} - {3 \over 2}{x^4} + {x^3}$$
$$ \therefore $$ $$5.f(2) = 5\left[ {64 - {{96} \over 5} - 24 + 8} \right] = 144$$
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