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reciprocal-expression MCQ Questions for Class 10

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4 questions tagged with reciprocal-expression.

Question 1/4 Expert Mathematics Polynomials Polynomials in one variable Class 10 Level 48

यदि (p(x)=x-2-12x+35), तो \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}\) क्या है, जहाँ \(\alpha,\beta\) शून्यक हैं?

If (p(x)=x-2-12x+35), what is \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}\), where \(\alpha,\beta\) are zeroes?

Explanation opens after your attempt
Correct Answer

D. \(\frac{11}{24}\)

Step 1

Concept

\(\alpha+\beta=12\) and \(\alpha\beta=35\). \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}=\frac{\alpha+\beta-2}{\alpha\beta-\alpha-\beta+1}=\frac{10}{24}=\frac{5}{12}\).

Step 2

Why this answer is correct

The correct answer is D. \(\frac{11}{24}\). \(\alpha+\beta=12\) and \(\alpha\beta=35\). \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}=\frac{\alpha+\beta-2}{\alpha\beta-\alpha-\beta+1}=\frac{10}{24}=\frac{5}{12}\).

Step 3

Exam Tip

\(\alpha+\beta=12\) और \(\alpha\beta=35\) हैं। \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}=\frac{\alpha+\beta-2}{\alpha\beta-\alpha-\beta+1}=\frac{10}{24}=\frac{5}{12}\)।

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Question 2/4 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 45

यदि \(x+\frac{1}{x}=7\), तो \(x^{2}+\frac{1}{x^{2}}\) का मान क्या है?

If \(x+\frac{1}{x}=7\), what is the value of \(x^{2}+\frac{1}{x^{2}}\)?

Explanation opens after your attempt
Correct Answer

B. (47)

Step 1

Concept

We use (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2). Thus \(49=x^{2}+\frac{1}{x^{2}}+2\), so the value is (47).

Step 2

Why this answer is correct

The correct answer is B. (47). We use (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2). Thus \(49=x^{2}+\frac{1}{x^{2}}+2\), so the value is (47).

Step 3

Exam Tip

(\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2) होता है। इसलिए \(49=x^{2}+\frac{1}{x^{2}}+2\) और मान (47) है।

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Question 3/4 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

यदि \(x-\frac{1}{x}=6\), तो \(x^{2}+\frac{1}{x^{2}}\) का मान क्या है?

If \(x-\frac{1}{x}=6\), what is the value of \(x^{2}+\frac{1}{x^{2}}\)?

Explanation opens after your attempt
Correct Answer

C. (38)

Step 1

Concept

We use (\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2). Thus \(36=x^{2}+\frac{1}{x^{2}}-2\), so the value is (38).

Step 2

Why this answer is correct

The correct answer is C. (38). We use (\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2). Thus \(36=x^{2}+\frac{1}{x^{2}}-2\), so the value is (38).

Step 3

Exam Tip

(\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2) होता है। इसलिए \(36=x^{2}+\frac{1}{x^{2}}-2\) और मान (38) है।

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Question 4/4 Expert Mathematics Polynomials Irrational numbers and real numbers Class 10 Level 25

यदि (p(x)=x-2-4x-1) है, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या है, जहाँ \(\alpha\) और \(\beta\) इसके शून्यक हैं?

If (p(x)=x-2-4x-1), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\), where \(\alpha\) and \(\beta\) are its zeroes?

Explanation opens after your attempt
Correct Answer

A. (-4)

Step 1

Concept

\(\alpha+\beta=4\) and \(\alpha\beta=-1\), so \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4\). Find sum and product first.

Step 2

Why this answer is correct

The correct answer is A. (-4). \(\alpha+\beta=4\) and \(\alpha\beta=-1\), so \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4\). Find sum and product first.

Step 3

Exam Tip

\(\alpha+\beta=4\) और \(\alpha\beta=-1\), इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4\)। पहले योग और गुणनफल निकालें।

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