Shown below is a solved anti-differentiation problem to obtain : such that Taking anti-derivative…
Shown below is a solved anti-differentiation problem to obtain $f(x)$:
$\frac{d}{dx}f(x) = \frac{1}{x(\log x)^2}$ such that $f(e) = -1$
Taking anti-derivative:
$f(x) = \int \frac{1}{x(\log x)^2} dx + C$
Step 1: $\Rightarrow f(x) = \int \frac{d(\log x)}{(\log x)^2} + C$
Step 2: $\Rightarrow f(x) = -\frac{1}{\log x} + C$
Given $f(e) = -1 \Rightarrow f(e) = -1 + C \Rightarrow C = 0$
Step 3: $f(x) = -\frac{1}{\log x}$
In which step is there an error (if any) in the solution?
- aStep 1
- bStep 2
- cStep 3
- dNo error
Answer
Answer
AIWritten by AI - it can contain mistakes.
Correct option: d
(d) No error
All steps in the solution are mathematically correct:
Step 1: Expressing $\frac{1}{x}dx$ as $d(\log x)$ is valid.
Step 2: Integrating $\int (\log x)^{-2} d(\log x) = -(\log x)^{-1} + C$ and applying $f(e) = -1 \Rightarrow -1 + C = -1 \Rightarrow C = 0$ is correct.
Step 3: Stating $f(x) = -\frac{1}{\log x}$ is correct.
From ISC 2027 Specimen Mathematics Paper 1, question 1(ix).