Question

If \[ A= \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \] where \[ a=7^x,\quad b=7^{7^x},\quad c=7^{7^{7^x}} \] then the value of \[ \int |A| \, dx \] is:

• A. \((\frac{7^{7^x}}{(\log 7)^3} + k) \)
• B. \((\frac{7^{7^{7^x}}}{\log 7} + k) \)
• C. \((\frac{7^{7^{7^x}}}{(\log 7)^3} + k) \)
• D. \((7^{7^{7^x}} (\log 7)^3 + k)\)

Hint:

In nested exponentials, the integral of the product of all terms is simply the highest tower divided by $(\ln a)^n$.

Answer 1

The Correct Option is C

Step 1: Find determinant (|A|) 
For a diagonal matrix, \((|A| = a \cdot b \cdot c = 7^x \cdot 7^{7^x} \cdot 7^{7^{7^x}}). \)
Step 2: Recognize the derivative pattern 
\(Let (u = 7^{7^{7^x}}). \)
\((\frac{du}{dx} = 7^{7^{7^x}} \cdot \ln 7 \cdot \frac{d}{dx}(7^{7^x}) = 7^{7^{7^x}} \cdot \ln 7 \cdot 7^{7^x} \cdot \ln 7 \cdot 7^x \cdot \ln 7). \)\( (\frac{du}{dx} = (7^{7^{7^x}} \cdot 7^{7^x} \cdot 7^x) \cdot (\ln 7)^3). .\)

Step 3: Integrate 
\((\int (7^{7^{7^x}} \cdot 7^{7^x} \cdot 7^x) , dx = \frac{u}{(\ln 7)^3} + k = \frac{7^{7^{7^x}}}{(\ln 7)^3} + k).\)
Final Answer: (C)