फलन (f(x)=\frac{1}{x-2-9}) के ग्राफ के लंबवत आसमापी कौन-से हैं?

What are the vertical asymptotes of the graph of (f(x)=\frac{1}{x-2-9})?

Explanation opens after your attempt
Correct Answer

A. (x=-3) और (x=3)(x=-3) and (x=3)

Step 1

Concept

The denominator (x-2-9=(x-3)(x+3)) is zero at \(x=\pm3\). In exams, find vertical asymptotes from the zeroes of the denominator.

Step 2

Why this answer is correct

The correct answer is A. (x=-3) और (x=3) / (x=-3) and (x=3). The denominator (x-2-9=(x-3)(x+3)) is zero at \(x=\pm3\). In exams, find vertical asymptotes from the zeroes of the denominator.

Step 3

Exam Tip

हर (x-2-9=(x-3)(x+3)) शून्य होने पर \(x=\pm3\) मिलता है। परीक्षा में हर के शून्यों से लंबवत आसमापी खोजें।

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Mathematics Answer, Explanation and Revision Hints

फलन (f(x)=\frac{1}{x-2-9}) के ग्राफ के लंबवत आसमापी कौन-से हैं? / What are the vertical asymptotes of the graph of (f(x)=\frac{1}{x-2-9})?

Correct Answer: A. (x=-3) और (x=3) / (x=-3) and (x=3). Explanation: हर (x-2-9=(x-3)(x+3)) शून्य होने पर \(x=\pm3\) मिलता है। परीक्षा में हर के शून्यों से लंबवत आसमापी खोजें। / The denominator (x-2-9=(x-3)(x+3)) is zero at \(x=\pm3\). In exams, find vertical asymptotes from the zeroes of the denominator.

Which concept should I revise for this Mathematics MCQ?

The denominator (x-2-9=(x-3)(x+3)) is zero at \(x=\pm3\). In exams, find vertical asymptotes from the zeroes of the denominator.

What exam hint can help solve this Mathematics question?

हर (x-2-9=(x-3)(x+3)) शून्य होने पर \(x=\pm3\) मिलता है। परीक्षा में हर के शून्यों से लंबवत आसमापी खोजें।