If $\mathrm{A}=\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 &…
- $\frac{7^{7^x}}{(\log 7)^3}+k, \quad$ where $k$ is constant of integration
- $\frac{7^{7^{7^x}}}{\log 7}+\mathrm{k}, \quad$ where k is constant of integration
- $\frac{7^{7^{7^x}}}{(\log 7)^3}+\mathrm{k} \quad, \quad$ where k is constant of integration
- $7^{7^{7^x}}(\log 7)^3+\mathrm{k}, \quad$ where k is constant of integration
Solution
Determinant Evaluation: The determinant of the diagonal matrix $A$ is the product of its diagonal elements: $|A| = a \cdot b \cdot c = 7^x \cdot 7^{7^x} \cdot 7^{7^{7^x}}$.
Integral Setup: The required integral is $\int |A| dx = \int 7^x \cdot 7^{7^x} \cdot 7^{7^{7^x}} dx$.
Substitution: Let $u = 7^{7^{7^x}}$.
Differentiating using the chain rule: $\frac{du}{dx} = 7^{7^{7^x}} \ln 7 \cdot \frac{d}{dx}(7^{7^x}) = 7^{7^{7^x}} \ln 7 \cdot (7^{7^x} \ln 7 \cdot 7^x \ln 7) = 7^{7^{7^x}} \cdot 7^{7^x} \cdot 7^x \cdot (\ln 7)^3$.
This gives $dx = \frac{du}{u \cdot 7^{7^x} \cdot 7^x \cdot (\ln 7)^3}$.
Integration: Substituting into the integral yields $\int u \cdot \frac{du}{u \cdot (\ln 7)^3} = \frac{1}{(\ln 7)^3} \int du = \frac{u}{(\ln 7)^3} + k$.
Replacing $u$ gives the result $\frac{7^{7^{7^x}}}{(\ln 7)^3} + k$.
Answer Selection: This matches option C, interpreting $\log 7$ as $\ln 7$.
Asked in: MHT CET 2025 (05 May Shift 2)