Let f(x) be a polynomial of degree 4 having e | vedclass

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(D) Since $f(x)$ is a polynomial of degree $4$,let $f(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E$.Given $\lim_{x 	o 0} \left( \frac{f(x)}{x^2} + 1 ight) = 3$

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$\frac{9}{2}$

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(D) Since $f(x)$ is a polynomial of degree $4$,let $f(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E$.Given $\lim_{x 	o 0} \left( \frac{f(x)}{x^2} + 1 ight) = 3$,we have $\lim_{x 	o 0} \left( \frac{Ax^4 + Bx^3 + Cx^2 + Dx + E}{x^2} + 1 ight) = 3$.This implies $\lim_{x 	o 0} \left( Ax^2 + Bx + C + \frac{D}{x} + \frac{E}{x^2} + 1 ight) = 3$.For the limit to exist and be finite,we must have $D = 0$ and $E = 0$. Thus,$C + 1 = 3$,which gives $C = 2$.Now,$f(x) = Ax^4 + Bx^3 + 2x^2$. The derivative is $f'(x) = 4Ax^3 + 3Bx^2 + 4x$.Since $f(x)$ has extreme values at $x = 1$ and $x = 2$,$f'(1) = 0$ and $f'(2) = 0$.$f'(1) = 4A + 3B + 4 = 0 \implies 4A + 3B = -4$ (Equation $1$).$f'(2) = 4A(8) + 3B(4) + 4(2) = 32A + 12B + 8 = 0 \implies 8A + 3B = -2$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(8A + 3B) - (4A + 3B) = -2 - (-4) \implies 4A = 2 \implies A = \frac{1}{2}$.Substituting $A = \frac{1}{2}$ into Equation $1$: $4(\frac{1}{2}) + 3B = -4 \implies 2 + 3B = -4 \implies 3B = -6 \implies B = -2$.Thus,$f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$.Finally,$f(-1) = \frac{1}{2}(-1)^4 - 2(-1)^3 + 2(-1)^2 = \frac{1}{2} + 2 + 2 = \frac{9}{2}$.

Let $f(x)$ be a polynomial of degree $4$ having extreme values at $x = 1$ and $x = 2$. If $\mathop {\lim }\limits_{x \to 0} \left( {\frac{{f(x)}}{{{x^2}}} + 1} \right) = 3$,then $f(-1)$ is equal to

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