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Question: If $f(x) = \int \frac{ln x}{\sqrt{x}} dx$ and $f(1) = -4$ then, which of the following option(s) is/...

If f(x)=∫lnxxdxf(x) = \int \frac{ln x}{\sqrt{x}} dx and f(1)=−4f(1) = -4 then, which of the following option(s) is/are correct?

A

f(e)=−2ef(e) = -2\sqrt{e}

B

f(1e)=−6ef(\frac{1}{e}) = -\frac{6}{\sqrt{e}}

C

f(2)=−22f(2) = -2\sqrt{2}

D

f(e2)=0f(e^2) = 0

Answer

A, B, D

Explanation

Solution

To solve this problem, we need to find the function f(x)f(x) by integrating ln⁡xx\frac{\ln x}{\sqrt{x}} and use the given condition f(1)=−4f(1) = -4 to determine the constant of integration.

1. Integrate f(x)=∫ln⁡xxdxf(x) = \int \frac{\ln x}{\sqrt{x}} dx

We use integration by parts, which states ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.

Let u=ln⁡xu = \ln x and dv=1xdx=x−1/2dxdv = \frac{1}{\sqrt{x}} dx = x^{-1/2} dx.
Then, we find dudu and vv:
du=1xdxdu = \frac{1}{x} dx
v=∫x−1/2dx=x−1/2+1−1/2+1=x1/21/2=2xv = \int x^{-1/2} dx = \frac{x^{-1/2+1}}{-1/2+1} = \frac{x^{1/2}}{1/2} = 2\sqrt{x}.

Now, apply the integration by parts formula:
f(x)=(ln⁡x)(2x)−∫(2x)(1x)dxf(x) = (\ln x)(2\sqrt{x}) - \int (2\sqrt{x}) \left(\frac{1}{x}\right) dx
f(x)=2xln⁡x−∫2x1/2x−1dxf(x) = 2\sqrt{x} \ln x - \int 2x^{1/2} x^{-1} dx
f(x)=2xln⁡x−∫2x−1/2dxf(x) = 2\sqrt{x} \ln x - \int 2x^{-1/2} dx
f(x)=2xln⁡x−2(x1/21/2)+Cf(x) = 2\sqrt{x} \ln x - 2 \left(\frac{x^{1/2}}{1/2}\right) + C
f(x)=2xln⁡x−4x+Cf(x) = 2\sqrt{x} \ln x - 4\sqrt{x} + C

2. Use the condition f(1)=−4f(1) = -4 to find CC

Substitute x=1x=1 into the expression for f(x)f(x):
f(1)=21ln⁡1−41+Cf(1) = 2\sqrt{1} \ln 1 - 4\sqrt{1} + C
Since ln⁡1=0\ln 1 = 0 and 1=1\sqrt{1} = 1:
f(1)=2(1)(0)−4(1)+Cf(1) = 2(1)(0) - 4(1) + C
f(1)=0−4+Cf(1) = 0 - 4 + C
f(1)=−4+Cf(1) = -4 + C

Given f(1)=−4f(1) = -4:
−4=−4+C-4 = -4 + C
C=0C = 0

So, the function is f(x)=2xln⁡x−4xf(x) = 2\sqrt{x} \ln x - 4\sqrt{x}, which can be factored as f(x)=2x(ln⁡x−2)f(x) = 2\sqrt{x}(\ln x - 2).

3. Check the given options

A. f(e)=−2ef(e) = -2\sqrt{e}
Substitute x=ex=e into f(x)f(x):
f(e)=2e(ln⁡e−2)f(e) = 2\sqrt{e}(\ln e - 2)
Since ln⁡e=1\ln e = 1:
f(e)=2e(1−2)f(e) = 2\sqrt{e}(1 - 2)
f(e)=2e(−1)f(e) = 2\sqrt{e}(-1)
f(e)=−2ef(e) = -2\sqrt{e}
Option A is correct.

B. f(1e)=−6ef(\frac{1}{e}) = -\frac{6}{\sqrt{e}}
Substitute x=1ex=\frac{1}{e} into f(x)f(x):
f(1e)=21e(ln⁡1e−2)f\left(\frac{1}{e}\right) = 2\sqrt{\frac{1}{e}}\left(\ln \frac{1}{e} - 2\right)
f(1e)=2e(ln⁡e−1−2)f\left(\frac{1}{e}\right) = \frac{2}{\sqrt{e}}(\ln e^{-1} - 2)
Since ln⁡e−1=−1ln⁡e=−1\ln e^{-1} = -1 \ln e = -1:
f(1e)=2e(−1−2)f\left(\frac{1}{e}\right) = \frac{2}{\sqrt{e}}(-1 - 2)
f(1e)=2e(−3)f\left(\frac{1}{e}\right) = \frac{2}{\sqrt{e}}(-3)
f(1e)=−6ef\left(\frac{1}{e}\right) = -\frac{6}{\sqrt{e}}
Option B is correct.

C. f(2)=−22f(2) = -2\sqrt{2}
Substitute x=2x=2 into f(x)f(x):
f(2)=22(ln⁡2−2)f(2) = 2\sqrt{2}(\ln 2 - 2)
For this to be equal to −22-2\sqrt{2}, we would need ln⁡2−2=−1\ln 2 - 2 = -1, which implies ln⁡2=1\ln 2 = 1. This means 2=e2 = e, which is false (since e≈2.718e \approx 2.718).
Therefore, Option C is incorrect.

D. f(e2)=0f(e^2) = 0
Substitute x=e2x=e^2 into f(x)f(x):
f(e2)=2e2(ln⁡e2−2)f(e^2) = 2\sqrt{e^2}(\ln e^2 - 2)
f(e2)=2e(2ln⁡e−2)f(e^2) = 2e(2\ln e - 2)
Since ln⁡e=1\ln e = 1:
f(e2)=2e(2(1)−2)f(e^2) = 2e(2(1) - 2)
f(e2)=2e(2−2)f(e^2) = 2e(2 - 2)
f(e2)=2e(0)f(e^2) = 2e(0)
f(e2)=0f(e^2) = 0
Option D is correct.

The correct options are A, B, and D.