Question

If \(y =\)\(e^{\sqrt{x\sqrt{x\sqrt{x\ldots}}}}\) \(x>1,\) then \(\left. \frac{{d^2y}}{{dx^2}} \right|_{x = \log_e 3}\) is

• A. 3
• B. 0
• C. 5
• D. 1

Answer 1

The Correct Option is A

Let \[ y = e^{\sqrt{x\sqrt{x\sqrt{x\ldots}}}} = e^{\sqrt{x\cdot y}} \quad \text{(since the nested root tends to a limit)} \] Take natural logarithm: \[ \ln y = \sqrt{x \cdot y} \] Differentiate both sides with respect to \(x\): LHS: \[ \frac{1}{y} \cdot \frac{dy}{dx} \] RHS: \[ \frac{1}{2\sqrt{xy}} \cdot \left( y + x \cdot \frac{dy}{dx} \right) \] So, \[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{y + x \cdot \frac{dy}{dx}}{2\sqrt{xy}} \] Multiply both sides by \(2\sqrt{xy}\): \[ \frac{2\sqrt{xy}}{y} \cdot \frac{dy}{dx} = y + x \cdot \frac{dy}{dx} \] Now, simplify: \[ \frac{2\sqrt{x} \cdot \sqrt{y}}{y} \cdot \frac{dy}{dx} = y + x \cdot \frac{dy}{dx} \Rightarrow \frac{2\sqrt{x}}{\sqrt{y}} \cdot \frac{dy}{dx} = y + x \cdot \frac{dy}{dx} \] Rewriting: \[ \frac{dy}{dx} \left( \frac{2\sqrt{x}}{\sqrt{y}} - x \right) = y \Rightarrow \frac{dy}{dx} = \frac{y}{\frac{2\sqrt{x}}{\sqrt{y}} - x} \] Now evaluate at \(x = \ln 3\). First, compute \(y\): \[ \ln y = \sqrt{x \cdot y} \Rightarrow \ln y = \sqrt{\ln 3 \cdot y} \] Try \(y = 3\): \[ \ln 3 = \sqrt{\ln 3 \cdot 3} \Rightarrow \ln 3 = \sqrt{3\ln 3} \Rightarrow (\ln 3)^2 = 3 \ln 3 \Rightarrow \ln 3 = 3 \Rightarrow \text{False} \] Try \(y = e^{\sqrt{x y}}\) again with \(x = \ln 3\) Guess: Let’s take \(y = 3\), try if it satisfies \[ \ln y = \ln 3 \Rightarrow \sqrt{x y} = \ln 3 \Rightarrow \sqrt{\ln 3 \cdot 3} = \ln 3 \Rightarrow \ln 3 = \sqrt{3 \ln 3} \Rightarrow (\ln 3)^2 = 3 \ln 3 \Rightarrow \ln 3 = 3 \Rightarrow \text{Contradiction} \] Let’s take \(y = 1\) Then: \[ \ln y = 0, \quad \sqrt{x \cdot y} = \sqrt{x} \Rightarrow \ln y = \sqrt{x} \Rightarrow 0 = \sqrt{x} \Rightarrow x = 0 \quad \text{Not valid} \] Let’s define: \[ z = \sqrt{x \cdot z} \Rightarrow z^2 = xz \Rightarrow z(z - x) = 0 \Rightarrow z = 0 \text{ or } z = x \] We discard 0, so: \[ z = x \Rightarrow y = e^{x} \] So, \[ y = e^x, \quad \frac{dy}{dx} = e^x, \quad \frac{d^2y}{dx^2} = e^x \] Now at \(x = \log_e 3\): \[ \frac{d^2y}{dx^2} = e^{\log_e 3} = 3 \] Hence, \(\left. \frac{d^2y}{dx^2} \right|_{x = \log_e 3} = 3\) 

Answer 2

Let \(z = \sqrt{x\sqrt{x\sqrt{x\ldots}}}\). Then, \(z = \sqrt{xz}\), so \(z^2 = xz\). Since x > 1, then z cannot be zero, so \(z = x\).

Thus, \(y = e^z = e^x\).

Now, we find the first and second derivatives of y with respect to x:

\(\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x\)

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(e^x) = e^x\)

We want to evaluate \(\frac{d^2y}{dx^2}\) at \(x = \log_e 3\). So we have:

\(\left. \frac{d^2y}{dx^2} \right|_{x = \log_e 3} = e^{\log_e 3}\)

Since \(e^{\log_e a} = a\),

\(\left. \frac{d^2y}{dx^2} \right|_{x = \log_e 3} = 3\)

Answer: 3