Concept-wise Practice

conjugate MCQ Questions for Class 10

conjugate se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.

Practice Questions

46 questions tagged with conjugate.

\(\dfrac{3}{2-\sqrt{3}}\) का हर परिमेय करने पर कौन सा रूप मिलेगा?

Which form is obtained by rationalising the denominator of \(\dfrac{3}{2-\sqrt{3}}\)?

Explanation opens after your attempt
Correct Answer

A. \(,6+3\sqrt{3},\)

Step 1

Concept

Multiplying by \(2+\sqrt{3}\) makes the denominator (4-3=1). In exams, multiply both numerator and denominator by the conjugate.

Step 2

Why this answer is correct

The correct answer is A. \(,6+3\sqrt{3},\). Multiplying by \(2+\sqrt{3}\) makes the denominator (4-3=1). In exams, multiply both numerator and denominator by the conjugate.

Step 3

Exam Tip

हर को \(2+\sqrt{3}\) से गुणा करने पर हर (4-3=1) हो जाता है। परीक्षा में conjugate से numerator और denominator दोनों को गुणा करें।

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किस विकल्प में अपरिमेय संख्या को परिमेय संख्या में बदलने के लिए सही संयुग्मी चुना गया है?

In which option is the correct conjugate chosen to rationalize an irrational denominator?

Explanation opens after your attempt
Correct Answer

A. \(\frac{1}{5+\sqrt{2}}\) के लिए \(5-\sqrt{2}\)For \(\frac{1}{5+\sqrt{2}}\) use \(5-\sqrt{2}\)

Step 1

Concept

The conjugate of \(5+\sqrt{2}\) is \(5-\sqrt{2}\). In exams changing the middle sign is the key idea of a conjugate.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{1}{5+\sqrt{2}}\) के लिए \(5-\sqrt{2}\) / For \(\frac{1}{5+\sqrt{2}}\) use \(5-\sqrt{2}\). The conjugate of \(5+\sqrt{2}\) is \(5-\sqrt{2}\). In exams changing the middle sign is the key idea of a conjugate.

Step 3

Exam Tip

\(5+\sqrt{2}\) का संयुग्मी \(5-\sqrt{2}\) है। परीक्षा में बीच का चिन्ह बदलना ही संयुग्मी बनाने की मुख्य बात है।

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यदि \(\frac{1}{\sqrt{7}+\sqrt{6}}\) को परिमेयकृत किया जाए, तो मान क्या होगा?

If \(\frac{1}{\sqrt{7}+\sqrt{6}}\) is rationalized, what is its value?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt{7}-\sqrt{6}\)

Step 1

Concept

The conjugate of the denominator is \(\sqrt{7}-\sqrt{6}\), and the denominator becomes (7-6=1). In exams the answer simplifies when the difference is (1).

Step 2

Why this answer is correct

The correct answer is A. \(\sqrt{7}-\sqrt{6}\). The conjugate of the denominator is \(\sqrt{7}-\sqrt{6}\), and the denominator becomes (7-6=1). In exams the answer simplifies when the difference is (1).

Step 3

Exam Tip

हर का संयुग्मी \(\sqrt{7}-\sqrt{6}\) है और हर (7-6=1) बनता है। परीक्षा में अंतर (1) होने पर उत्तर सरल हो जाता है।

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यदि \(x=\frac{1}{\sqrt{5}-2}\), तो (x) का सरल रूप क्या है?

If \(x=\frac{1}{\sqrt{5}-2}\), what is the simplified form of (x)?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt{5}+2\)

Step 1

Concept

\(\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\). Rationalise the denominator in exams.

Step 2

Why this answer is correct

The correct answer is A. \(\sqrt{5}+2\). \(\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\). Rationalise the denominator in exams.

Step 3

Exam Tip

\(\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\) है। परीक्षा में हर का परिमेयकरण करें।

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यदि \(x=3+\sqrt{8}\), तो \(\frac{1}{x}\) किसके बराबर है?

If \(x=3+\sqrt{8}\), what is \(\frac{1}{x}\) equal to?

Explanation opens after your attempt
Correct Answer

A. \(3-\sqrt{8}\)

Step 1

Concept

Because (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), the reciprocal is \(3-\sqrt{8}\). If the product is (1), the reciprocal is immediate.

Step 2

Why this answer is correct

The correct answer is A. \(3-\sqrt{8}\). Because (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), the reciprocal is \(3-\sqrt{8}\). If the product is (1), the reciprocal is immediate.

Step 3

Exam Tip

क्योंकि (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), इसलिए व्युत्क्रम \(3-\sqrt{8}\) है। परीक्षा में गुणनफल (1) होने पर व्युत्क्रम तुरंत मिल जाता है।

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यदि (p(x)=x-2-2x-2) का एक शून्यक \(1+\sqrt{3}\) है, तो दूसरा शून्यक क्या है?

If one zero of (p(x)=x-2-2x-2) is \(1+\sqrt{3}\), what is the other zero?

Explanation opens after your attempt
Correct Answer

A. \(1-\sqrt{3}\)

Step 1

Concept

The sum of zeroes is (2), so the other zero is (2-\(1+\sqrt{3}\)=1-\sqrt{3}). With rational coefficients, the conjugate also appears.

Step 2

Why this answer is correct

The correct answer is A. \(1-\sqrt{3}\). The sum of zeroes is (2), so the other zero is (2-\(1+\sqrt{3}\)=1-\sqrt{3}). With rational coefficients, the conjugate also appears.

Step 3

Exam Tip

शून्यकों का योग (2) है, इसलिए दूसरा शून्यक (2-\(1+\sqrt{3}\)=1-\sqrt{3}) है। परिमेय गुणांकों में संयुग्मी भी मिलता है।

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किस बहुपद के शून्यक \(1+\sqrt{10}\) और \(1-\sqrt{10}\) हैं?

Which polynomial has zeroes \(1+\sqrt{10}\) and \(1-\sqrt{10}\)?

Explanation opens after your attempt
Correct Answer

A. \(x^2-2x-9\)

Step 1

Concept

The sum is (2) and the product is (1-10=-9). So the polynomial is \(x^2-2x-9\).

Step 2

Why this answer is correct

The correct answer is A. \(x^2-2x-9\). The sum is (2) and the product is (1-10=-9). So the polynomial is \(x^2-2x-9\).

Step 3

Exam Tip

योग (2) और गुणनफल (1-10=-9) है। इसलिए बहुपद \(x^2-2x-9\) है।

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यदि (p(x)=x-2-6x+n) के शून्यक \(3+\sqrt{m}\) और \(3-\sqrt{m}\) हैं तथा (n=5), तो (m) क्या है?

If zeroes of (p(x)=x-2-6x+n) are \(3+\sqrt{m}\) and \(3-\sqrt{m}\), and (n=5), what is (m)?

Explanation opens after your attempt
Correct Answer

A. (4)

Step 1

Concept

The product is (9-m), and it equals (n=5). Thus (9-m=5), so (m=4).

Step 2

Why this answer is correct

The correct answer is A. (4). The product is (9-m), and it equals (n=5). Thus (9-m=5), so (m=4).

Step 3

Exam Tip

गुणनफल (9-m) है और वह (n=5) के बराबर है। अतः (9-m=5), इसलिए (m=4)।

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यदि \(\alpha=\sqrt{3}+1\) और \(\beta=\sqrt{3}-1\), तो \(\alpha\beta\) क्या है?

If \(\alpha=\sqrt{3}+1\) and \(\beta=\sqrt{3}-1\), what is \(\alpha\beta\)?

Explanation opens after your attempt
Correct Answer

A. (2)

Step 1

Concept

(\alpha\beta=\(\sqrt{3}+1\)\(\sqrt{3}-1\)=3-1=2). The irrational part cancels in conjugate multiplication.

Step 2

Why this answer is correct

The correct answer is A. (2). (\alpha\beta=\(\sqrt{3}+1\)\(\sqrt{3}-1\)=3-1=2). The irrational part cancels in conjugate multiplication.

Step 3

Exam Tip

(\alpha\beta=\(\sqrt{3}+1\)\(\sqrt{3}-1\)=3-1=2) है। संयुग्मी गुणनफल से अपरिमेय भाग कट जाता है।

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किस मान के लिए (p(x)=x-2-10x+k) के शून्यक \(5+\sqrt{2}\) और \(5-\sqrt{2}\) होंगे?

For which value of (k) will (p(x)=x-2-10x+k) have zeroes \(5+\sqrt{2}\) and \(5-\sqrt{2}\)?

Explanation opens after your attempt
Correct Answer

A. (23)

Step 1

Concept

The product is (\(5+\sqrt{2}\)\(5-\sqrt{2}\)=25-2=23). So (k=23).

Step 2

Why this answer is correct

The correct answer is A. (23). The product is (\(5+\sqrt{2}\)\(5-\sqrt{2}\)=25-2=23). So (k=23).

Step 3

Exam Tip

गुणनफल (\(5+\sqrt{2}\)\(5-\sqrt{2}\)=25-2=23) है। इसलिए (k=23) होगा।

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कौन सा विकल्प \(\frac{1}{5-\sqrt{6}}\) का परिमेय हर वाला रूप है?

Which option is the rationalized form of \(\frac{1}{5-\sqrt{6}}\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{5+\sqrt{6}}{19}\)

Step 1

Concept

The conjugate of the denominator is \(5+\sqrt{6}\), and the denominator becomes (25-6=19). Hence the first option is correct.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{5+\sqrt{6}}{19}\). The conjugate of the denominator is \(5+\sqrt{6}\), and the denominator becomes (25-6=19). Hence the first option is correct.

Step 3

Exam Tip

हर का संयुग्मी \(5+\sqrt{6}\) है और हर (25-6=19) बनता है। इसलिए पहला विकल्प सही है।

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यदि \(a=\sqrt{7}+\sqrt{2}\) और \(b=\sqrt{7}-\sqrt{2}\) हैं तो (ab) का मान क्या है?

If \(a=\sqrt{7}+\sqrt{2}\) and \(b=\sqrt{7}-\sqrt{2}\), what is the value of (ab)?

Explanation opens after your attempt
Correct Answer

A. (5)

Step 1

Concept

In conjugate multiplication, (ab=7-2=5). The difference of squares formula gives the answer quickly.

Step 2

Why this answer is correct

The correct answer is A. (5). In conjugate multiplication, (ab=7-2=5). The difference of squares formula gives the answer quickly.

Step 3

Exam Tip

संयुग्मी गुणन में (ab=7-2=5) होता है। अंतर वर्ग सूत्र जल्दी उत्तर देता है।

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कौन सा विकल्प (\left\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right\)) का सरल रूप है?

Which option is the simplified form of (\left\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right\))?

Explanation opens after your attempt
Correct Answer

A. \(5+2\sqrt{6}\)

Step 1

Concept

Multiplying by the conjugate makes the denominator (1). The numerator is (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}).

Step 2

Why this answer is correct

The correct answer is A. \(5+2\sqrt{6}\). Multiplying by the conjugate makes the denominator (1). The numerator is (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}).

Step 3

Exam Tip

हर के संयुग्मी से गुणा करने पर हर (1) बनता है। अंश (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}) है।

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यदि \(x=2+\sqrt{3}\) है तो कौन सा समीकरण सत्य है?

If \(x=2+\sqrt{3}\), which equation is true?

Explanation opens after your attempt
Correct Answer

A. \(x^2-4x+1=0\)

Step 1

Concept

The conjugate is \(2-\sqrt{3}\), with sum (4) and product (1). Hence the equation is \(x^2-4x+1=0\).

Step 2

Why this answer is correct

The correct answer is A. \(x^2-4x+1=0\). The conjugate is \(2-\sqrt{3}\), with sum (4) and product (1). Hence the equation is \(x^2-4x+1=0\).

Step 3

Exam Tip

(x) का संयुग्मी \(2-\sqrt{3}\) है और योग (4), गुणनफल (1) है। इसलिए समीकरण \(x^2-4x+1=0\) है।

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यदि \(m=\frac{1}{\sqrt{5}+\sqrt{2}}\) है तो (m) का सरल रूप क्या है?

If \(m=\frac{1}{\sqrt{5}+\sqrt{2}}\), what is the simplified form of (m)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{\sqrt{5}-\sqrt{2}}{3}\)

Step 1

Concept

Multiplying by the conjugate makes the denominator (5-2=3). So the rationalized form is \(\frac{\sqrt{5}-\sqrt{2}}{3}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{\sqrt{5}-\sqrt{2}}{3}\). Multiplying by the conjugate makes the denominator (5-2=3). So the rationalized form is \(\frac{\sqrt{5}-\sqrt{2}}{3}\).

Step 3

Exam Tip

संयुग्मी से गुणा करने पर हर (5-2=3) हो जाता है। इसलिए परिमेय हर वाला रूप \(\frac{\sqrt{5}-\sqrt{2}}{3}\) है।

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कौन सा विकल्प (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)) का मान है?

Which option is the value of (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\))?

Explanation opens after your attempt
Correct Answer

A. (1)

Step 1

Concept

By difference of squares the value is (13-12=1). Product of conjugate pairs often gives a rational number.

Step 2

Why this answer is correct

The correct answer is A. (1). By difference of squares the value is (13-12=1). Product of conjugate pairs often gives a rational number.

Step 3

Exam Tip

अंतर वर्ग सूत्र से मान (13-12=1) है। संयुग्मी जोड़े का गुणन अक्सर परिमेय देता है।

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यदि \(r=\sqrt{2}+\sqrt{3}\) है तो \(r+\frac{1}{r}\) का सरल रूप क्या है?

If \(r=\sqrt{2}+\sqrt{3}\), what is the simplified form of \(r+\frac{1}{r}\)?

Explanation opens after your attempt
Correct Answer

A. \(2\sqrt{3}\)

Step 1

Concept

\(\frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\). Adding gives \(2\sqrt{3}\).

Step 2

Why this answer is correct

The correct answer is A. \(2\sqrt{3}\). \(\frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\). Adding gives \(2\sqrt{3}\).

Step 3

Exam Tip

\(\frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\) होता है। जोड़ने पर \(2\sqrt{3}\) मिलता है।

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कौन सा विकल्प (\(5+\sqrt{6}\)\(5-\sqrt{6}\)+\sqrt{24}) की प्रकृति सही बताता है?

Which option correctly describes the nature of (\(5+\sqrt{6}\)\(5-\sqrt{6}\)+\sqrt{24})?

Explanation opens after your attempt
Correct Answer

A. अपरिमेय संख्याIrrational number

Step 1

Concept

The first product is (25-6=19) and \(\sqrt{24}=2\sqrt{6}\) is irrational. A rational plus an irrational is irrational.

Step 2

Why this answer is correct

The correct answer is A. अपरिमेय संख्या / Irrational number. The first product is (25-6=19) and \(\sqrt{24}=2\sqrt{6}\) is irrational. A rational plus an irrational is irrational.

Step 3

Exam Tip

पहला गुणनफल (25-6=19) है और \(\sqrt{24}=2\sqrt{6}\) अपरिमेय है। परिमेय और अपरिमेय का योग अपरिमेय होता है।

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यदि \(p=2+\sqrt{3}\) है तो \(\frac{1}{p}\) किसके बराबर है?

If \(p=2+\sqrt{3}\), what is \(\frac{1}{p}\) equal to?

Explanation opens after your attempt
Correct Answer

A. \(2-\sqrt{3}\)

Step 1

Concept

Since (\(2+\sqrt{3}\)\(2-\sqrt{3}\)=1), the reciprocal is \(2-\sqrt{3}\). Recognizing conjugates is a fast method.

Step 2

Why this answer is correct

The correct answer is A. \(2-\sqrt{3}\). Since (\(2+\sqrt{3}\)\(2-\sqrt{3}\)=1), the reciprocal is \(2-\sqrt{3}\). Recognizing conjugates is a fast method.

Step 3

Exam Tip

क्योंकि (\(2+\sqrt{3}\)\(2-\sqrt{3}\)=1), इसलिए व्युत्क्रम \(2-\sqrt{3}\) है। संयुग्मी को पहचानना तेज तरीका है।

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कौन सा विकल्प \(\frac{1}{3+\sqrt{5}}\) का परिमेय हर वाला रूप है?

Which option is the rationalized form of \(\frac{1}{3+\sqrt{5}}\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{3-\sqrt{5}}{4}\)

Step 1

Concept

The conjugate of the denominator is \(3-\sqrt{5}\) and the denominator becomes (9-5=4). Multiply by the conjugate to rationalize.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{3-\sqrt{5}}{4}\). The conjugate of the denominator is \(3-\sqrt{5}\) and the denominator becomes (9-5=4). Multiply by the conjugate to rationalize.

Step 3

Exam Tip

हर का संयुग्मी \(3-\sqrt{5}\) है और हर (9-5=4) बनता है। परिमेयकरण में संयुग्मी से गुणा करें।

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यदि \(a=3+\sqrt{7}\) और \(b=3-\sqrt{7}\) हैं तो \(a^2+b^2\) का मान क्या है?

If \(a=3+\sqrt{7}\) and \(b=3-\sqrt{7}\), what is the value of \(a^2+b^2\)?

Explanation opens after your attempt
Correct Answer

A. (32)

Step 1

Concept

On adding the two squares the radical terms cancel and the result is (32). Identify cancelling terms in conjugates.

Step 2

Why this answer is correct

The correct answer is A. (32). On adding the two squares the radical terms cancel and the result is (32). Identify cancelling terms in conjugates.

Step 3

Exam Tip

दोनों वर्ग जोड़ने पर जड़ वाले पद कट जाते हैं और (32) मिलता है। संयुग्मी संख्याओं में कटने वाले पद पहचानें।

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यदि \(p=7+\sqrt{11}\) और \(q=7-\sqrt{11}\) हैं तो (pq) का मान क्या है?

If \(p=7+\sqrt{11}\) and \(q=7-\sqrt{11}\), what is the value of (pq)?

Explanation opens after your attempt
Correct Answer

A. (38)

Step 1

Concept

Conjugate multiplication gives (pq=72-\(\sqrt{11}\)2=49-11=38). Use \(a^2-b^2\) in such questions.

Step 2

Why this answer is correct

The correct answer is A. (38). Conjugate multiplication gives (pq=72-\(\sqrt{11}\)2=49-11=38). Use \(a^2-b^2\) in such questions.

Step 3

Exam Tip

संयुग्मी गुणन से (pq=72-\(\sqrt{11}\)2=49-11=38) मिलता है। ऐसे प्रश्नों में \(a^2-b^2\) लगाएँ।

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कौन सा विकल्प (\(\sqrt{15}+\sqrt{6}\)\(\sqrt{15}-\sqrt{6}\)) का मान है?

Which option is the value of (\(\sqrt{15}+\sqrt{6}\)\(\sqrt{15}-\sqrt{6}\))?

Explanation opens after your attempt
Correct Answer

A. (9)

Step 1

Concept

This is the difference of squares formula and the value is (15-6=9). In conjugate multiplication the irrational part cancels.

Step 2

Why this answer is correct

The correct answer is A. (9). This is the difference of squares formula and the value is (15-6=9). In conjugate multiplication the irrational part cancels.

Step 3

Exam Tip

यह अंतर वर्ग सूत्र है और मान (15-6=9) मिलता है। संयुग्मी गुणन में अपरिमेय भाग हट जाता है।

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कौन सा विकल्प (\(\sqrt{10}+\sqrt{5}\)\(\sqrt{10}-\sqrt{5}\)) का मान है?

Which option is the value of (\(\sqrt{10}+\sqrt{5}\)\(\sqrt{10}-\sqrt{5}\))?

Explanation opens after your attempt
Correct Answer

A. (5)

Step 1

Concept

This is the difference of squares formula. The value is (10-5=5).

Step 2

Why this answer is correct

The correct answer is A. (5). This is the difference of squares formula. The value is (10-5=5).

Step 3

Exam Tip

यह अंतर वर्ग सूत्र है। मान (10-5=5) होगा।

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कौन सा विकल्प \(3+\sqrt{8}\) और \(3-\sqrt{8}\) के योग और गुणनफल को सही बताता है?

Which option correctly gives the sum and product of \(3+\sqrt{8}\) and \(3-\sqrt{8}\)?

Explanation opens after your attempt
Correct Answer

A. योग (6), गुणनफल (1)Sum (6), product (1)

Step 1

Concept

The radical terms cancel in the sum and the product is (9-8=1). This method is quick for conjugates.

Step 2

Why this answer is correct

The correct answer is A. योग (6), गुणनफल (1) / Sum (6), product (1). The radical terms cancel in the sum and the product is (9-8=1). This method is quick for conjugates.

Step 3

Exam Tip

योग में जड़ वाले पद कटते हैं और गुणनफल (9-8=1) है। संयुग्मी संख्याओं में यह तरीका तेज है।

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कौन सा विकल्प \(2+\sqrt{5}\) और \(2-\sqrt{5}\) के गुणनफल का मान है?

Which option is the value of the product of \(2+\sqrt{5}\) and \(2-\sqrt{5}\)?

Explanation opens after your attempt
Correct Answer

A. (-1)

Step 1

Concept

(\(2+\sqrt{5}\)\(2-\sqrt{5}\)=4-5=-1). In conjugate multiplication the irrational part cancels.

Step 2

Why this answer is correct

The correct answer is A. (-1). (\(2+\sqrt{5}\)\(2-\sqrt{5}\)=4-5=-1). In conjugate multiplication the irrational part cancels.

Step 3

Exam Tip

(\(2+\sqrt{5}\)\(2-\sqrt{5}\)=4-5=-1) है। संयुग्मी गुणन में अपरिमेय भाग हट जाता है।

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कौन सा विकल्प (\(\sqrt{5}+\sqrt{2}\)\(\sqrt{5}-\sqrt{2}\)) का मान है?

Which option is the value of (\(\sqrt{5}+\sqrt{2}\)\(\sqrt{5}-\sqrt{2}\))?

Explanation opens after your attempt
Correct Answer

A. (3)

Step 1

Concept

This is ((a+b)(a-b)=a-2-b-2). The value is (5-2=3).

Step 2

Why this answer is correct

The correct answer is A. (3). This is ((a+b)(a-b)=a-2-b-2). The value is (5-2=3).

Step 3

Exam Tip

यह ((a+b)(a-b)=a-2-b-2) है। मान (5-2=3) मिलेगा।

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कौन सा विकल्प (\(\sqrt{13}-\sqrt{3}\)\(\sqrt{13}+\sqrt{3}\)) का मान है?

Which option is the value of (\(\sqrt{13}-\sqrt{3}\)\(\sqrt{13}+\sqrt{3}\))?

Explanation opens after your attempt
Correct Answer

A. (10)

Step 1

Concept

This is ((a-b)(a+b)=a-2-b-2). The value is (13-3=10).

Step 2

Why this answer is correct

The correct answer is A. (10). This is ((a-b)(a+b)=a-2-b-2). The value is (13-3=10).

Step 3

Exam Tip

यह ((a-b)(a+b)=a-2-b-2) है। मान (13-3=10) होगा।

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कौन सा विकल्प (\(4+\sqrt{7}\)\(4-\sqrt{7}\)) का मान है?

Which option is the value of (\(4+\sqrt{7}\)\(4-\sqrt{7}\))?

Explanation opens after your attempt
Correct Answer

A. (9)

Step 1

Concept

Conjugate multiplication gives (42-\(\sqrt{7}\)2=16-7=9). Use the difference of squares formula in such questions.

Step 2

Why this answer is correct

The correct answer is A. (9). Conjugate multiplication gives (42-\(\sqrt{7}\)2=16-7=9). Use the difference of squares formula in such questions.

Step 3

Exam Tip

संयुग्मी गुणन से (42-\(\sqrt{7}\)2=16-7=9) मिलता है। ऐसे प्रश्नों में अंतर वर्ग सूत्र लगाएँ।

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कौन सा विकल्प \(2+\sqrt{3}\) और \(2-\sqrt{3}\) के योग और गुणनफल को सही बताता है?

Which option correctly gives the sum and product of \(2+\sqrt{3}\) and \(2-\sqrt{3}\)?

Explanation opens after your attempt
Correct Answer

A. योग (4), गुणनफल (1)Sum (4), product (1)

Step 1

Concept

The radical terms cancel in the sum and the product is (4-3=1). This method is quick for conjugates.

Step 2

Why this answer is correct

The correct answer is A. योग (4), गुणनफल (1) / Sum (4), product (1). The radical terms cancel in the sum and the product is (4-3=1). This method is quick for conjugates.

Step 3

Exam Tip

योग में जड़ वाले पद कटते हैं और गुणनफल (4-3=1) है। संयुग्मी संख्याओं में यह तरीका तेज है।

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