If lim_{x to 1} frac{sin(3x^2 - 4x + 1) - x^2 | vedclass

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

Question

(C) Since the limit exists and is finite, the denominator must be $0$ at $x = 1$. Thus, $2(1)^3 - 7(1)^2 + a(1) + b = 0 \implies a + b = 5$.Using $L$'

Vedclass · Accepted Answer

$x^2 - 5x - 24 = 0$

Vedclass · Answer

(C) Since the limit exists and is finite, the denominator must be $0$ at $x = 1$. Thus, $2(1)^3 - 7(1)^2 + a(1) + b = 0 \implies a + b = 5$.Using $L$'$H$ôpital's rule, $\lim_{x 	o 1} \frac{\cos(3x^2 - 4x + 1)(6x - 4) - 2x}{6x^2 - 14x + a} = -2$.Substituting $x = 1$: $\frac{\cos(0)(2) - 2}{6 - 14 + a} = -2 \implies \frac{2 - 2}{a - 8} = -2$. This implies the numerator derivative is $0$, so we apply $L$'$H$ôpital's rule again.Derivative of numerator: $-\sin(3x^2 - 4x + 1)(6x - 4)^2 + \cos(3x^2 - 4x + 1)(6) - 2$.At $x = 1$: $-0 + 1(6) - 2 = 4$.Derivative of denominator: $12x - 14$.At $x = 1$: $12 - 14 = -2$.Limit $= \frac{4}{-2} = -2$. This confirms the limit value is independent of $a$ and $b$ as long as $a+b=5$ and the denominator is $0$ at $x=1$. However, for the limit to be defined, the denominator must have a factor $(x-1)$. Since $2x^3 - 7x^2 + ax + b = (x-1)(2x^2 - 5x + (a-5))$, for the limit to exist, $2x^2 - 5x + (a-5)$ must be divisible by $(x-1)$, so $2 - 5 + a - 5 = 0 \implies a = 8$. Then $b = 5 - 8 = -3$.Roots are $8$ and $-3$. The equation is $(x - 8)(x + 3) = x^2 - 5x - 24 = 0$.

If $\lim_{x \to 1} \frac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2$, then the quadratic equation having roots $a$ and $b$ is

Similar Questions

If $\lim _{x \rightarrow 2} \frac{3 x^2-a x+5 b}{x-2}=17$,then $a b=$

If $\lim _{x \rightarrow 5} \frac{x^{k}-5^{k}}{x-5}=500$,then the value of $k$,where $k \in N$ is

If $\lim_{x \to 1} \frac{x^3 + ax^2 + bx + c}{x^2 - 2x + 1} = 2026$, then the value of $a - c$ is...

Define $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$. If $\lim_{x \rightarrow 0} f(x)$ exists,then $p =$

If $\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1$,where $\lambda, \mu \in \mathbb{R}$,then $\lambda+\mu$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module