If alpha = lim_{x rightarrow pi/4} frac{tan^{ | vedclass

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

Question

(D) First,calculate $\alpha = \lim_{x ightarrow \pi/4} \frac{	an^{3} x - 	an x}{\cos(x + \pi/4)}$. This is a $\frac{0}{0}$ form.Applying $L'H\hat{

Vedclass · Accepted Answer

$(1, 3)$

Vedclass · Answer

(D) First,calculate $\alpha = \lim_{x ightarrow \pi/4} \frac{	an^{3} x - 	an x}{\cos(x + \pi/4)}$. This is a $\frac{0}{0}$ form.Applying $L'H\hat{o}pital's$ rule:$\alpha = \lim_{x ightarrow \pi/4} \frac{3 	an^{2} x \sec^{2} x - \sec^{2} x}{-\sin(x + \pi/4)} = \frac{3(1)^{2}(2) - 2}{-\sin(\pi/2)} = \frac{6 - 2}{-1} = -4$.Next,calculate $\beta = \lim_{x ightarrow 0} (\cos x)^{\cot x}$. This is a $1^{\infty}$ form.$\beta = e^{\lim_{x ightarrow 0} \cot x (\cos x - 1)} = e^{\lim_{x ightarrow 0} \frac{\cos x - 1}{	an x}} = e^{\lim_{x ightarrow 0} \frac{-\sin x}{\sec^{2} x}} = e^{0} = 1$.Since $\alpha = -4$ and $\beta = 1$ are roots of $ax^{2} + bx - 4 = 0$,the sum of roots $\alpha + \beta = -b/a \Rightarrow -4 + 1 = -3 = -b/a \Rightarrow b = 3a$.The product of roots $\alpha \beta = -4/a \Rightarrow (-4)(1) = -4/a \Rightarrow a = 1$.Substituting $a = 1$ into $b = 3a$,we get $b = 3$.Thus,the ordered pair $(a, b)$ is $(1, 3)$.

If $\alpha = \lim_{x \rightarrow \pi/4} \frac{\tan^{3} x - \tan x}{\cos(x + \pi/4)}$ and $\beta = \lim_{x \rightarrow 0} (\cos x)^{\cot x}$ are the roots of the equation $ax^{2} + bx - 4 = 0$,then the ordered pair $(a, b)$ is:

Similar Questions

Let $f(x) = \lim_{n \rightarrow \infty} \sum_{r=0}^n \left( \frac{2\tan(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})} \right)$. Then $\lim_{x \rightarrow 0} \frac{e^x - e^{f(x)}}{x - f(x)}$ is equal to . . . . . . .

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\int_{\frac{\pi }{2}}^x t \,dt}}{{\sin (2x - \pi )}}$ is

$\lim _{x \rightarrow 0} \frac{\cos 2x - \cos 3x}{\cos 4x - \cos 5x} = $

If $f(x)$ is a differentiable function of $x$,then $\mathop {Limit}\limits_{h \to 0} \frac{f(x + 3h) - f(x - 2h)}{h} = $

If $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$,then $\lim_{\alpha \rightarrow 0} \frac{f(1-\alpha) - f(1)}{\alpha^3 + 3\alpha} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module