Calculus problems - 4 questions

[Saitama]

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,
$$\lim_{x→0}\frac{f^2(x)-1}{x}=\lim_{x→0} 2f(x) \cdot f'(x)$$
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!

 

[Mark44]

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,
$$\lim_{x→0}\frac{f^2(x)-1}{x}=\lim_{x→0} 2f(x) \cdot f'(x)$$
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.
​

Use the Fundamental Theorem of Calculus (2nd part) to find g'(x).

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!

 

[Saitama]

Use the Fundamental Theorem of Calculus (2nd part) to find g'(x).

I already found g'(x) in the attempt of Q.1 and is equal to f(x). I do not know the value of f(1/2).

 

[Mark44]

I'm having a hard time understanding what the initial paragraph is trying to say.

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.

"Given two curves " - presumably this refers to y = f(x) and y = g(x), but the lack of any punctuation makes this unclear to me.

"...passing through the points (0, 1) ..." This is a single point.
"...passing through the points (0, 1/2) ... " This is also a single point.
"The tangents drawn to both the curves at the points with equal abscissas..." - Does this refer to all pairs of points on the two curves where one point is above or below the corresponding point on the other curve?

The wording is very confusing to me - is there a drawing that shows the relationship better than this description?

 

[Saitama]

The wording is very confusing to me - is there a drawing that shows the relationship better than this description?

No, there is no drawing with this paragraph.

 

[Saitama]

Anyone?

 

[Dick]

Ooo. Ooo. I've got it. "The tangents drawn to both the curves at the points with equal abscissas intersect on the x-axis." They mean that for any value of x that the tangents for f and g there will intersect the x-axis. Not just the two given points. So you've got f(x)*f(x)=g(x)*f'(x) holds for all x. Try to turn that into a differential equation involving only f and solve it.

 

[Saitama]

Ooo. Ooo. I've got it. "The tangents drawn to both the curves at the points with equal abscissas intersect on the x-axis." They mean that for any value of x that the tangents for f and g there will intersect the x-axis. Not just the two given points. So you've got f(x)*f(x)=g(x)*f'(x) holds for all x. Try to turn that into a differential equation involving only f and solve it.

How will I express g(x) in terms of f(x)?

Even if I differentiate f(x)*f(x)=g(x)*f'(x), g(x) still remains.

 

[Dick]

How will I express g(x) in terms of f(x)?

Even if I differentiate f(x)*f(x)=g(x)*f'(x), g(x) still remains.

Solve for g(x) first. Then differentiate!

 

[Saitama]

Solve for g(x) first. Then differentiate!

$$g'(x)=\frac{2(f'(x))^2 \cdot f(x)-f^2(x) \cdot f''(x)}{(f'(x))^2}$$

I am still clueless. I can't obtain value of g'(1/2) from this equation.

 

[Dick]

$$g'(x)=\frac{2(f'(x))^2 \cdot f(x)-f^2(x) \cdot f''(x)}{(f'(x))^2}$$

I am still clueless. I can't obtain value of g'(1/2) from this equation.

No you can't. But you can certainly simplify that relation. You want to get a differential equation to solve for f(x).

 

[Saitama]

No you can't. But you can certainly simplify that relation. You want to get a differential equation to solve for f(x).

Wouldn't that give me a differential equation in f(x) of second order?

 

[Dick]

Wouldn't that give me a differential equation in f(x) of second order?

Yes it will. But it will be easy to solve. And you have two boundary values, you know f(0) and f'(0).

 

[Saitama]

Yes it will. But it will be easy to solve. And you have two boundary values, you know f(0) and f'(0).

I don't know how to solve differential equations of second order and it is not in our course too.

Simplifying, i get
$$f'(x)=\sqrt{f(x) \cdot f''(x)}$$
At x=0,
$$f''(0)=4$$

 

[Dick]

I don't know how to solve differential equations of second order and it is not in our course too.

Simplifying, i get
$$f'(x)=\sqrt{f(x) \cdot f''(x)}$$
At x=0,
$$f''(0)=4$$

This is really no harder to solve than a first order equation. Write it as f'(x)/f(x)=f''(x)/f'(x). Remember (log(f))'=f'/f?

 

[Saitama]

Thanks a lot Dick!
The f''(x) term made me feel that now its impossible for me to solve this. Your hint was really helpful.

I got f(x)=e^(2x). I have solved the q.3 & 4. Answer to Q.3 is e and to Q.4 is 5/4.

Thanks once again.