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

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(B) Let $p(x) = ax^4 + bx^3 + cx^2 + dx + e$.Since $\lim_{x \rightarrow 0} (1 + \frac{p(x)}{x^2}) = 2$,we have $\lim_{x \rightarrow 0} \frac{p(x)}{x^2

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(B) Let $p(x) = ax^4 + bx^3 + cx^2 + dx + e$.Since $\lim_{x \rightarrow 0} (1 + \frac{p(x)}{x^2}) = 2$,we have $\lim_{x \rightarrow 0} \frac{p(x)}{x^2} = 1$. This implies $e = 0$ and $d = 0$,and $c = 1$.Thus,$p(x) = ax^4 + bx^3 + x^2$.The derivative is $p'(x) = 4ax^3 + 3bx^2 + 2x$.Given extrema at $x=1$ and $x=2$,$p'(1) = 0$ and $p'(2) = 0$.$p'(1) = 4a + 3b + 2 = 0 \implies 4a + 3b = -2$ (Equation $1$).$p'(2) = 4a(8) + 3b(4) + 2(2) = 32a + 12b + 4 = 0 \implies 8a + 3b = -1$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(8a - 4a) = -1 - (-2) \implies 4a = 1 \implies a = \frac{1}{4}$.Substituting $a = \frac{1}{4}$ into Equation $1$: $4(\frac{1}{4}) + 3b = -2 \implies 1 + 3b = -2 \implies 3b = -3 \implies b = -1$.So,$p(x) = \frac{1}{4}x^4 - x^3 + x^2$.Calculating $p(2)$: $p(2) = \frac{1}{4}(16) - (8) + (4) = 4 - 8 + 4 = 0$.

Let $p(x)$ be a polynomial of degree $4$ having extrema at $x=1$ and $x=2$,and $\lim_{x \rightarrow 0} \left(1+\frac{p(x)}{x^2}\right) = 2$. Then the value of $p(2)$ is

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