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Anvisha [2.4K]
2 years ago
11

Please help, AND READ DIRECTIONS, get done as soon as possible.

Mathematics
1 answer:
katen-ka-za [31]2 years ago
7 0

Answer:

IT would be 13a to 17b

Step-by-step explanation:

I had this.

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A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient
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<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

y=(x-a) \sqrt{(x-b)}  --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=(b+1-a) \sqrt{(b+1-b)}

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

y^{\prime}=u^{\prime} y+y^{\prime} u

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{u}^{\prime}=1

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y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}

By putting value of point A and putting value of eq 2 we get

y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}

y^{\prime}=\frac{d y}{d x}=1

Hence proved that the gradient of the curve at A is 1.

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