- #1

Saitama

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## Homework Statement

(There are four questions based on the paragraph below. I was able to solve two of them but clueless about the other two.)

Given two curves ##y=f(x)## passing through the points (0,1) and the curve ##y=g(x)=\int_{-∞}^{x} f(t)dt## passing through the points (0,1/2). The tangents drawn to both the curves at the points with equal abscissas intersect on the x-axis. Then

Q.1) The value of f'(0) is:

(A)0

(B)1/2

(C)2

(D)1

**Attempt:**

Let the points on the curve at which the tangents are drawn be ##(x_1,y_1)## for ##y=f(x)## and ##(x_1,y_2)## for ##y=g(x)##.

The equation of tangent at ##(x_1,y_1)## is ##y-y_1=f'(x_1)(x-x_1)##. It intersects the x-axis at ##x=\frac{-y_1}{f'(x_1)}+x_1##.

The equation of tangent at ##(x_1,y_2)## is ##y-y_2=g'(x_1)(x-x_1)## and this intersects the x-axis at ##x=\frac{-y_2}{g'(x_1)}+x_1##.

According to the above paragraph, the tangents intersect at x-axis, the x intercept is equal. Hence ##\frac{y_1}{f'(x_1)}=\frac{y_2}{g'(x_1)}## which is equal to ##f(x_1) \cdot g'(x_1)=g(x_1) \cdot f'(x_1)## using ##y_1=f(x_1)## and ##y_2=g(x_1)##.

We have ##g'(x)=f(x)##. And for ##x_1=0##, i get ##f(0) \cdot f(0)=g(0) \cdot f'(0)##.

##f(0)=1## and ##g(0)=\frac{1}{2}##. Therefore, ##f'(0)=2##.

Q.2)##\lim_{x→0}\frac{f^2(x)-1}{x}## equals

(A)1

(B)2

(C)3

(D)4

**Attempt:**

Using L'Hôpital's rule,

[tex]\lim_{x→0}\frac{f^2(x)-1}{x}=\lim_{x→0} 2f(x) \cdot f'(x)[/tex]

Solving, I get the answer 4.

Q.3)The value of ##g'(\frac{1}{2})## is

(A)e

(B)e/2

(C)2e

(D)1

**Attempt:**

I can't obtain the value of g'(1/2) using any of the equations I have derived. I am stuck here.

Q.4)The area bounded by the x-axis, the tangent and normal to the curve y=f(x) at the point where it cuts the y-axis, is:

(A)3/4

(B)1

(C)5/4

(D)3/2

**Attempt:**

Haven't yet tried this. I would like to first discuss Q.3 and then move on to this one.

Thanks for taking your time to read this.

Any help is appreciated. Thanks!