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QJEEM21Maa6eb: JEE Main 2021 Mathematics - Class 12 JEEM21Maa6eb Solution: Let f(x) be a polynomial of degree 6 in x, in which the…

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 lim⁡x→0f(x)x3=1\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {{x^3}}} = 1 , then 5.f(2)5.f(2) is equal to _________.

Step-by-Step Verified Solution

f(x)=x6+ax5+bx4+x3f(x) = {x^6} + a{x^5} + b{x^4} + {x^3}

∴\therefore f′(x)=6x5+5ax4+4bx3+3x2f'(x) = 6{x^5} + 5a{x^4} + 4b{x^3} + 3{x^2}

Roots 1 & −- 1

∴\therefore 6+5z+4b+3=06 + 5z + 4b + 3 = 0 & −6+5a−4b+3=0- 6 + 5a - 4b + 3 = 0 solving

a=−35a = - {3 \over 5}

b=−32b = - {3 \over 2}

∴\therefore f(x)=x6−35x5−32x4+x3f(x) = {x^6} - {3 \over 5}{x^5} - {3 \over 2}{x^4} + {x^3}

∴\therefore 5.f(2)=5[64−965−24+8]=1445.f(2) = 5\left[ {64 - {{96} \over 5} - 24 + 8} \right] = 144

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