Let P(x) be a polynomial of degree 3 having a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $P(x) = ax^3 + bx^2 + cx + d$. Then $P'(x) = 3ax^2 + 2bx + c$. Since $P(x)$ has an extreme value at $x=1$, we have $P'(1) = 0$, which implies

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let $P(x) = ax^3 + bx^2 + cx + d$. Then $P'(x) = 3ax^2 + 2bx + c$. Since $P(x)$ has an extreme value at $x=1$, we have $P'(1) = 0$, which implies $3a + 2b + c = 0$ ... $(i)$. Given $\lim_{x \rightarrow 0} \left(\frac{P(x)+4}{x^2} + 2\right) = 6$, we have $\lim_{x \rightarrow 0} \frac{P(x)+4}{x^2} = 4$. Substituting $P(x)$, we get $\lim_{x \rightarrow 0} \frac{ax^3 + bx^2 + cx + d + 4}{x^2} = 4$. For the limit to be finite, the coefficients of $x^{-2}$ and $x^{-1}$ must be zero. Thus, $d+4 = 0 \Rightarrow d = -4$ and $c = 0$. Then $\lim_{x \rightarrow 0} (ax + b) = 4 \Rightarrow b = 4$. Substituting $b=4$ and $c=0$ into equation $(i)$, we get $3a + 2(4) + 0 = 0 \Rightarrow 3a = -8 \Rightarrow a = -\frac{8}{3}$. So, $P(x) = -\frac{8}{3}x^3 + 4x^2 - 4$. The derivative is $P'(x) = -8x^2 + 8x$. Evaluating at $x = \frac{1}{2}$, we get $P'(\frac{1}{2}) = -8(\frac{1}{4}) + 8(\frac{1}{2}) = -2 + 4 = 2$.

Let $P(x)$ be a polynomial of degree $3$ having an extreme value at $x=1$. If $\lim _{x \rightarrow 0}\left(\frac{P(x)+4}{x^2}+2\right)=6$, then $\left(\frac{d P}{d x}\right)_{x=\frac{1}{2}}=$

Similar Questions

Find the point where $f(x) = 2\sin x + \cos 2x$ has a maximum value in the interval $0 \leq x \leq 2\pi$.

The number of points in $(-\infty, \infty)$ for which $x^2 - x \sin x - \cos x = 0$ is

If $ab = 2a + 3b$ where $a > 0$ and $b > 0$,then the minimum value of $ab$ is:

If $x = -1$ and $x = 2$ are extreme points of $f(x) = \alpha \log |x| + \beta x^2 + x$,then find the values of $(\alpha, \beta)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module