Logarithmic Differentiation

Step 1: Understanding the Concept:
This problem involves differentiating a function defined implicitly and manipulating the resulting differential equation to match a given form.
Step 2: Detailed Explanation:
Given .
Let , then .
.
So, .
Differentiating with respect to :

So, .
Squaring both sides: .
Differentiating again with respect to :

Dividing by (assuming is not zero):

Comparing with :
and .
.
Step 3: Final Answer:
The value of is 17.

The correct answer is 14.
$f(x)=x^2+g^{\prime }(1)x+g^{\prime \prime }(2)$
$f^{\prime }(x)=2x+g^{\prime }(1)$
$f^{\prime \prime }(x)=2$
$g(x)=f(1)x^2+x\left[2x+g^{\prime }(1)\right]+2$
$g^{\prime }(x)=2f(1)x+4x+g^{\prime }(1)$
$g^{\prime \prime }(x)=2f(1)+4$
$g^{\prime \prime }(x)=0$
$2f(1)+4=0$
$f(1)=-2$
$-2=1+g^{\prime }(1)=g^{\prime }(1)=-3$
So $f^{\prime }(x)=2x-3$
$f(x)=x^2-3x+c$
$c=0$
$f(x)=x^2-3x$
$g(x)=-3x+2$
$f(4)-g(4)=14$

The correct answer is 1.
$\frac{dy}{dx}+y=k$
$y\cdot e^x=k\cdot e^x+c$
$f(0)=e^{-2}$
$\Rightarrow c=e^{-2}-k$
$\therefore y=k+\left(e^{-2}-k\right)e^{-x}$
$\text{now}k=\underset{0}{\overset{2}{\int }}\left(k+\left(e^{-2}-k\right)e^{-x}\right)dx$
$\Rightarrow k=e^{-2}-1$
$\therefore y=\left(e^{-2}-1\right)+e^{-x}$
$f(2)=2e^{-2}-1,f(0)=e^{-2}$
$2f(0)-f(2)=1$

The correct answer is (B) : 4
${\log }_{\cos x}\cot x+4{\log }_{\sin x}\tan x=1$
$\Rightarrow \frac{\ln \cos x-\ln \sin x}{\ln \cos x}+4\frac{\ln \sin x-\ln \cos x}{\ln \sin x}=1$
$\Rightarrow (\ln \sin x{)}^2-4(\ln \sin x)(\ln \cos x)+4(\ln \cos x{)}^2=1$
$\Rightarrow \ln \sin x=2\ln \cos x$
$\Rightarrow {\sin }^{2}x+\sin x-1=0\Rightarrow \sin x=\frac{-1+\sqrt{5}}{2}$
$\therefore \alpha +\beta =4$

The correct answer is 27
$f(x)+\underset{0}{\overset{x}{\int }}f(t)\sqrt{1-{\left({\log }_{e}f(t)\right)}^2}dt=e$
$\Rightarrow f(0)=e$
$f^{\prime }(x)+f(x)\sqrt{1-(\ln f(x){)}^2}=0$
$f(x)=y$
$\frac{dy}{dx}=-y\sqrt{1-(\ln y{)}^2}$
$\int \frac{dy}{y\sqrt{1-(\ln y{)}^2}}=-\int dx$
Put $\ln y=t$
$\int \frac{dt}{\sqrt{1-t^2}}=-x+C$
${\sin }^{-1}t=-x+C\Rightarrow {\sin }^{-1}(\ln y)=-x+C$
${\sin }^{-1}(\ln f(x))=-x+C$
$f(0)=e$
$\Rightarrow \frac{\pi }{2}=C$
$\Rightarrow {\sin }^{-1}(\ln f(x))=-x+\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}\left(\ln f\left(\frac{\pi }{6}\right)\right)=\frac{-\pi }{6}+\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}\left(\ln f\left(\frac{\pi }{6}\right)\right)=\frac{\pi }{3}$
$\Rightarrow \ln f\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2},$ we need ${\left(6\times \frac{\sqrt{3}}{2}\right)}^2=27$

We are given the function: Now, differentiate to find : Next, evaluate and : - For : - For : Now, using the given condition , substitute the values of and to solve for . Simplifying, we find: Thus, .