Consider the polynomial f(x)=1+2x+3x^2+4x^3. | vedclass

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(C, A, B) $1.$ $f(x) = 4x^3 + 3x^2 + 2x + 1$. Since $f(-1) = -4+3-2+1 = -2$ and $f(-1/2) = 4(-1/8) + 3(1/4) + 2(-1/2) + 1 = -0.5 + 0.75 - 1 + 1 = 0.25

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$(C, A, B)$

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(C, A, B) $1.$ $f(x) = 4x^3 + 3x^2 + 2x + 1$. Since $f(-1) = -4+3-2+1 = -2$ and $f(-1/2) = 4(-1/8) + 3(1/4) + 2(-1/2) + 1 = -0.5 + 0.75 - 1 + 1 = 0.25$. Since $f(-1) 0$,the root $s$ lies in $(-1, -1/2)$. Checking the options,$s$ lies in $(-3/4, -1/2)$ is a subset of this,so $(C)$ is correct.$2.$ $t = |s|$. Since $s \in (-3/4, -1/2)$,$t \in (1/2, 3/4)$. The area $A = \int_0^t (4x^3+3x^2+2x+1) dx = [x^4+x^3+x^2+x]_0^t = t^4+t^3+t^2+t$. For $t=1/2$,$A = 1/16 + 1/8 + 1/4 + 1/2 = 15/16 = 0.9375$. For $t=3/4$,$A = 81/256 + 27/64 + 9/16 + 3/4 = (81+108+144+192)/256 = 525/256 \approx 2.05$. The interval $(21/64, 11/16)$ is $(0.328, 0.6875)$,which is incorrect. Re-evaluating: $f(x)$ is increasing. The area is $A(t)$. For $t \in (0.5, 0.75)$,$A(0.5) = 0.9375$ and $A(0.75) \approx 2.05$. Option $(A)$ is $(0.75, 3)$,which contains the range $(0.9375, 2.05)$. Thus $(A)$ is correct.$3.$ $f'(x) = 12x^2 + 6x + 2$. $f''(x) = 24x + 6$. $f''(x) = 0$ at $x = -1/4$. For $x > -1/4$,$f''(x) > 0$ (increasing). For $x < -1/4$,$f''(x) < 0$ (decreasing). Thus $(B)$ is correct.

List $I$	List $II$
$A. 3x^4 - 2x^3 - 6x^2 + 6x + 1$	$(I)$ has minimum value at $x = 4$
$B. x + \frac{1}{x}, \forall x < 0$	$(II)$ has maximum value at $x = -1$
$C. x^4(7 - x)^3$	$(III)$ has maximum value at $x = 4$
$D. x^4 + (8 - x)^4$	$(IV)$ is decreasing in $[2, \infty)$
	$(V)$ is increasing in $[2, \infty)$

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