2 Problems, 1 Question (Part 1)

Calculus Level 2

Here are the instructions:

You must first find $\displaystyle a = \lim_{x \to -1} \frac {(2x-1)^2-(x-2)^2}{x+1}$ .
Then find $b$ as given in $\sqrt{a^2+4ab+4b^2} = 0$ .

Submit the value of $b$ .

The answer is 3.

1 solution

Chew-Seong Cheong
Feb 23, 2017

$\begin{aligned} a & = \lim_{x \to -1} \frac {(2x-1)^2 - (x-2)^2}{x+1} \\ & = \lim_{x \to -1} \frac {4x^2-4x+1-x^2+4x-4}{x+1} \\ & = \lim_{x \to -1} \frac {3x^2-3}{x+1} \\ & = \lim_{x \to -1} \frac {3(x^2-1)}{x+1} \\ & = \lim_{x \to -1} \frac {3(x-1)(x+1)}{x+1} \\ & = \lim_{x \to -1} 3(x-1) \\ & = -6 \end{aligned}$

$\begin{aligned} \sqrt{a^2+4ab+4b^2} & = 0 \\ \sqrt{36-24b+4b^2} & = 0 \\ \sqrt{(2b-6)^2} & = 0 \\ 2b-6 & = 0 \\ \implies b & = \boxed{3} \end{aligned}$

Sir dont you think this is calculus problem

Md Zuhair - 4 years, 3 months ago

Yes, it should be calculus. Do you want me to change it? In another problem that you post. What do you mean by $m(ABC) = 130^\circ$ ? It is $\angle ABC = 130^\circ$ . Use ^\circ instead of ^{o}. The braces {} are not required if there is only one character after the function. For example \frac 12 for $\frac 12$ .

Chew-Seong Cheong - 4 years, 3 months ago

Yes you can change it. But for 3 digit numbers what shall I do for fractions?

Md Zuhair - 4 years, 3 months ago

@Md Zuhair – Do you mean that $m(ABC)$ is $\angle ABC$ ? Just use \angle in the future. \frac 1{23} $\frac 1{23}$ , \frac {12}{34} $\frac {12}{34}$ . \sqrt 2 $\sqrt 2$ but \sqrt {12} $\sqrt{12}$ .

Chew-Seong Cheong - 4 years, 3 months ago

2 Problems, 1 Question (Part 1)

The answer is 3.

1 solution

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