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Question: Let f'(x) > 0 ∀ x ∈ R$^+$ where f: R$^+$ → R and f(x) + $\frac{1}{x}$ = f$^{-1}$($\frac{1}{f(x)}$) a...

Let f'(x) > 0 ∀ x ∈ R+^+ where f: R+^+ → R and f(x) + 1x\frac{1}{x} = f−1^{-1}(1f(x)\frac{1}{f(x)}) and f−1^{-1}(1f(x)\frac{1}{f(x)}) > 0. If sin−1^{-1}(f(2)) is −πλ\frac{-\pi}{\lambda} then 'λ\lambda' equals

Answer

10

Explanation

Solution

The given equation is f(x)+1x=f−1(1f(x))f(x) + \frac{1}{x} = f^{-1}\left(\frac{1}{f(x)}\right).
We are also given that f′(x)>0f'(x) > 0 for all x∈R+x \in R^+. This implies that f(x)f(x) is strictly increasing, so it is a one-to-one function and its inverse f−1(x)f^{-1}(x) exists.

To simplify the functional equation, apply ff to both sides:
f(f(x)+1x)=f(f−1(1f(x)))f\left(f(x) + \frac{1}{x}\right) = f\left(f^{-1}\left(\frac{1}{f(x)}\right)\right)
f(f(x)+1x)=1f(x)f\left(f(x) + \frac{1}{x}\right) = \frac{1}{f(x)}

Let's assume a solution of the form f(x)=cxf(x) = \frac{c}{x} for some constant cc.
First, let's check the condition f′(x)>0f'(x) > 0:
f′(x)=ddx(cx)=−cx2f'(x) = \frac{d}{dx}\left(\frac{c}{x}\right) = -\frac{c}{x^2}.
For f′(x)>0f'(x) > 0 for x∈R+x \in R^+, we must have −c>0-c > 0, which means c<0c < 0.

Now, substitute f(x)=cxf(x) = \frac{c}{x} into the functional equation f(f(x)+1x)=1f(x)f\left(f(x) + \frac{1}{x}\right) = \frac{1}{f(x)}:
Left Hand Side (LHS):
f(f(x)+1x)=f(cx+1x)=f(c+1x)f\left(f(x) + \frac{1}{x}\right) = f\left(\frac{c}{x} + \frac{1}{x}\right) = f\left(\frac{c+1}{x}\right)
Using f(x)=cxf(x) = \frac{c}{x}, we get:
f(c+1x)=cc+1x=cxc+1f\left(\frac{c+1}{x}\right) = \frac{c}{\frac{c+1}{x}} = \frac{cx}{c+1}

Right Hand Side (RHS):
1f(x)=1cx=xc\frac{1}{f(x)} = \frac{1}{\frac{c}{x}} = \frac{x}{c}

Equating LHS and RHS:
cxc+1=xc\frac{cx}{c+1} = \frac{x}{c}
Since x∈R+x \in R^+, we can divide by xx:
cc+1=1c\frac{c}{c+1} = \frac{1}{c}
Cross-multiply:
c2=c+1c^2 = c+1
c2−c−1=0c^2 - c - 1 = 0

This is a quadratic equation for cc. Using the quadratic formula:
c=−(−1)±(−1)2−4(1)(−1)2(1)c = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}
c=1±1+42c = \frac{1 \pm \sqrt{1+4}}{2}
c=1±52c = \frac{1 \pm \sqrt{5}}{2}

We established earlier that c<0c < 0. Therefore, we must choose the negative root:
c=1−52c = \frac{1 - \sqrt{5}}{2}

So the function is f(x)=1−52xf(x) = \frac{1 - \sqrt{5}}{2x}.

Now, let's check the condition f−1(1f(x))>0f^{-1}\left(\frac{1}{f(x)}\right) > 0.
For f(x)=cxf(x) = \frac{c}{x}, its inverse f−1(y)f^{-1}(y) is found by letting y=cx  ⟹  x=cyy = \frac{c}{x} \implies x = \frac{c}{y}. So f−1(y)=cyf^{-1}(y) = \frac{c}{y}.
Then f−1(1f(x))=f−1(xc)=cxc=c2xf^{-1}\left(\frac{1}{f(x)}\right) = f^{-1}\left(\frac{x}{c}\right) = \frac{c}{\frac{x}{c}} = \frac{c^2}{x}.
We need c2x>0\frac{c^2}{x} > 0. Since x∈R+x \in R^+, we need c2>0c^2 > 0.
Our value c=1−52c = \frac{1 - \sqrt{5}}{2} is not zero, so c2=(1−52)2=1+5−254=6−254=3−52c^2 = \left(\frac{1 - \sqrt{5}}{2}\right)^2 = \frac{1 + 5 - 2\sqrt{5}}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}.
Since 3>53 > \sqrt{5} (as 9>59 > 5), 3−5>03-\sqrt{5} > 0. Thus c2>0c^2 > 0, and the condition is satisfied.

Now we need to find sin⁡−1(f(2))\sin^{-1}(f(2)).
Substitute x=2x=2 into f(x)f(x):
f(2)=1−52(2)=1−54f(2) = \frac{1 - \sqrt{5}}{2(2)} = \frac{1 - \sqrt{5}}{4}

We need to calculate sin⁡−1(1−54)\sin^{-1}\left(\frac{1 - \sqrt{5}}{4}\right).
Recall the known trigonometric value sin⁡(π10)=sin⁡(18∘)=5−14\sin\left(\frac{\pi}{10}\right) = \sin(18^\circ) = \frac{\sqrt{5}-1}{4}.
Therefore, sin⁡−1(5−14)=π10\sin^{-1}\left(\frac{\sqrt{5}-1}{4}\right) = \frac{\pi}{10}.
Since 1−54=−(5−14)\frac{1 - \sqrt{5}}{4} = -\left(\frac{\sqrt{5}-1}{4}\right), and sin⁡−1(−y)=−sin⁡−1(y)\sin^{-1}(-y) = -\sin^{-1}(y):
sin⁡−1(1−54)=−sin⁡−1(5−14)=−π10\sin^{-1}\left(\frac{1 - \sqrt{5}}{4}\right) = -\sin^{-1}\left(\frac{\sqrt{5}-1}{4}\right) = -\frac{\pi}{10}.

We are given that sin⁡−1(f(2))=−πλ\sin^{-1}(f(2)) = \frac{-\pi}{\lambda}.
So, −π10=−πλ-\frac{\pi}{10} = \frac{-\pi}{\lambda}.
Comparing both sides, we find λ=10\lambda = 10.