Question

Let $M = \begin{bmatrix} x & 2x & 3x \\ f(x) & g(x) & h(x) \\ 0 & 1 & 1 \end{bmatrix}$ be a singular matrix. If $f(x) = ln(e^x + 1)$ and $g(x) = ln(e^x - 1)$, then the value of h'(ln 3) is

• A. $\frac{9}{8}$
• B. $\frac{9}{4}$
• C. 3
• D. 6

Answer 1

Answer: (B)

$\frac{9}{4}$

Answer 2

The given matrix is $M = \begin{bmatrix} x & 2x & 3x \\ f(x) & g(x) & h(x) \\ 0 & 1 & 1 \end{bmatrix}$. The matrix M is singular, which means its determinant is zero, i.e., $\det(M) = 0$. The determinant of M is calculated as:

$\det(M) = x \begin{vmatrix} g(x) & h(x) \\ 1 & 1 \end{vmatrix} - 2x \begin{vmatrix} f(x) & h(x) \\ 0 & 1 \end{vmatrix} + 3x \begin{vmatrix} f(x) & g(x) \\ 0 & 1 \end{vmatrix}$

$\det(M) = x(g(x) \cdot 1 - h(x) \cdot 1) - 2x(f(x) \cdot 1 - h(x) \cdot 0) + 3x(f(x) \cdot 1 - g(x) \cdot 0)$

$\det(M) = x(g(x) - h(x)) - 2x f(x) + 3x f(x)$

$\det(M) = x(g(x) - h(x) + f(x))$

Since $\det(M) = 0$, we have $x(g(x) - h(x) + f(x)) = 0$.

The functions are $f(x) = \ln(e^x + 1)$ and $g(x) = \ln(e^x - 1)$. The function $g(x)$ is defined for $e^x - 1 > 0$, which means $e^x > 1$, or $x > 0$. The relation holds for $x > 0$.

For $x > 0$, we can divide by $x$:

$g(x) - h(x) + f(x) = 0$

$h(x) = f(x) + g(x)$

Substitute the expressions for $f(x)$ and $g(x)$:

$h(x) = \ln(e^x + 1) + \ln(e^x - 1)$

Using the logarithm property $\ln A + \ln B = \ln(AB)$:

$h(x) = \ln((e^x + 1)(e^x - 1))$

Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$:

$h(x) = \ln((e^x)^2 - 1^2) = \ln(e^{2x} - 1)$

This expression for $h(x)$ is valid for $x > 0$.

Next, we need to find the derivative of $h(x)$, denoted by $h'(x)$.

$h(x) = \ln(e^{2x} - 1)$

Using the chain rule, $\frac{d}{dx} \ln(u) = \frac{1}{u} \frac{du}{dx}$. Let $u = e^{2x} - 1$.

$\frac{du}{dx} = \frac{d}{dx}(e^{2x} - 1) = e^{2x} \cdot \frac{d}{dx}(2x) - 0 = e^{2x} \cdot 2 = 2e^{2x}$.

So, $h'(x) = \frac{1}{e^{2x} - 1} \cdot 2e^{2x} = \frac{2e^{2x}}{e^{2x} - 1}$.

Finally, we need to evaluate $h'(\ln 3)$. Substitute $x = \ln 3$ into the expression for $h'(x)$. Note that $\ln 3 > 0$, so it is in the domain of $h'(x)$.

$h'(\ln 3) = \frac{2e^{2(\ln 3)}}{e^{2(\ln 3)} - 1}$

Using the logarithm and exponent properties $a \ln b = \ln b^a$ and $e^{\ln c} = c$:

$2 \ln 3 = \ln 3^2 = \ln 9$.

So, $e^{2(\ln 3)} = e^{\ln 9} = 9$.

Substitute this value into the expression for $h'(\ln 3)$:

$h'(\ln 3) = \frac{2 \cdot 9}{9 - 1} = \frac{18}{8} = \frac{9}{4}$.

The value of $h'(\ln 3)$ is $\frac{9}{4}$.