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3 years ago

What is the solution of x-1/1-x < 0?

Mathematics

1 answer:

kykrilka [37]3 years ago

8 0

Option C is correct option.

Step-by-step explanation:

We need to find the solution of $\frac{x-1}{1-x}$

The given inequality becomes undefined when x = 1 because the denomiator is: 1-x and if x = 1 then it becomes 1-1 = 0 and anything divided by zero is undefined.

So, all values other than 1 are included in the solution of the given inequality.

So, solution is: x<1 or x>1 but x≠1

So, Option C is correct because all numbers are included in number line except 1. An unfilled circle on 1 shows that it is not included in the solution. Rest of numbers are included.

Option C is correct option.

Keywords: Solving inequalities

Learn more about Solving inequalities at:

#learnwithBrainly

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There exist a similar question where b = 68 instead of 6. First, determine the measure of the third angle, angle C,
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1 year ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

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