If lim _{x rightarrow 0}left(frac{cos 4 x+a c | vedclass

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

Question

(C) Let $f(x) = \cos 4x + a \cos 2x + b$. For the limit to be finite as $x \rightarrow 0$,the numerator must approach $0$ as $x \rightarrow 0$.Using t

Vedclass · Accepted Answer

$-4, 3$

Vedclass · Answer

(C) Let $f(x) = \cos 4x + a \cos 2x + b$. For the limit to be finite as $x ightarrow 0$,the numerator must approach $0$ as $x ightarrow 0$.Using the Taylor series expansion: $\cos 	heta = 1 - \frac{	heta^2}{2!} + \frac{	heta^4}{4!} - \dots$$\cos 4x = 1 - \frac{(4x)^2}{2} + \frac{(4x)^4}{24} - \dots = 1 - 8x^2 + \frac{32}{3}x^4 - \dots$$a \cos 2x = a(1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} - \dots) = a - 2ax^2 + \frac{2a}{3}x^4 - \dots$Substituting these into the expression:$f(x) = (1 + a + b) - (8 + 2a)x^2 + (\frac{32}{3} + \frac{2a}{3})x^4 + \dots$For the limit to be finite,the coefficients of $x^0$ and $x^2$ must be zero.$1 + a + b = 0$ and $8 + 2a = 0$.From $8 + 2a = 0$,we get $a = -4$.Substituting $a = -4$ into $1 + a + b = 0$,we get $1 - 4 + b = 0$,which gives $b = 3$.Thus,the values are $a = -4$ and $b = 3$.

If $\lim _{x \rightarrow 0}\left(\frac{\cos 4 x+a \cos 2 x+b}{x^4}\right)$ is finite,then the values of $a, b$ are respectively :

Similar Questions

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$,then $\lim_{x \rightarrow \alpha} \frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2} = $

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

Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a > -1$. Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ respectively are

If $\lim_{x \to \infty} [\frac{x^2 + x + 1}{x + 1} - ax - b] = 3$, then $a - b = $

Let $f : R - \{0\} \rightarrow R$ be a function such that $f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$. If $\lim_{x \rightarrow 0} \left(\frac{1}{\alpha x} + f(x)\right) = \beta$,where $\alpha, \beta \in R$,then $\alpha + 2\beta$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module