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15 results found for "lambda" in Class 10.

समीकरण (\(\lambda+1\)x-2-2\(\lambda-2\)x+\(\lambda+1\)=0) के वास्तविक और समान मूलों के लिए \(\lambda\) क्या होगा?

What will \(\lambda\) be for real and equal roots of (\(\lambda+1\)x-2-2\(\lambda-2\)x+\(\lambda+1\)=0)?

Explanation opens after your attempt
Correct Answer

A. \(\lambda=\frac{1}{4}\)

Step 1

Concept

Here (D=4\(\lambda-2\)2-4\(\lambda+1\)2=12\(1-2\lambda\)). From (D=0), \(\lambda=\frac{1}{2}\).

Step 2

Why this answer is correct

The correct answer is A. \(\lambda=\frac{1}{4}\). Here (D=4\(\lambda-2\)2-4\(\lambda+1\)2=12\(1-2\lambda\)). From (D=0), \(\lambda=\frac{1}{2}\).

Step 3

Exam Tip

यहाँ (D=4\(\lambda-2\)2-4\(\lambda+1\)2=12\(1-2\lambda\)) है। (D=0) से \(\lambda=\frac{1}{2}\) मिलता है।

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समीकरण (x-2+2\(\lambda-1\)x+\lambda-2+1=0) के वास्तविक मूलों के लिए \(\lambda\) पर सही शर्त क्या है?

What is the correct condition on \(\lambda\) for real roots of (x-2+2\(\lambda-1\)x+\lambda-2+1=0)?

Explanation opens after your attempt
Correct Answer

A. \(\lambda\le0\)

Step 1

Concept

Here (D=4\(\lambda-1\)2-4\(\lambda^2+1\)=-8\lambda). For real roots \(D\ge0\), so \(\lambda\le0\).

Step 2

Why this answer is correct

The correct answer is A. \(\lambda\le0\). Here (D=4\(\lambda-1\)2-4\(\lambda^2+1\)=-8\lambda). For real roots \(D\ge0\), so \(\lambda\le0\).

Step 3

Exam Tip

यहाँ (D=4\(\lambda-1\)2-4\(\lambda^2+1\)=-8\lambda) है। वास्तविक मूलों के लिए \(D\ge0\), इसलिए \(\lambda\le0\)।

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समीकरण \(x^2-2\lambda x+\lambda=0\) के समान मूलों के लिए \(\lambda\) के मान कौन से हैं?

What are the values of \(\lambda\) for equal roots of \(x^2-2\lambda x+\lambda=0\)?

Explanation opens after your attempt
Correct Answer

A. \(\lambda=0\) या \(\lambda=1\)\(\lambda=0\) or \(\lambda=1\)

Step 1

Concept

For equal roots (D=4\lambda\(\lambda-1\)=0). Therefore \(\lambda=0\) or \(\lambda=1\).

Step 2

Why this answer is correct

The correct answer is A. \(\lambda=0\) या \(\lambda=1\) / \(\lambda=0\) or \(\lambda=1\). For equal roots (D=4\lambda\(\lambda-1\)=0). Therefore \(\lambda=0\) or \(\lambda=1\).

Step 3

Exam Tip

समान मूलों के लिए (D=4\lambda\(\lambda-1\)=0) है। इसलिए \(\lambda=0\) या \(\lambda=1\)।

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यदि \(x^2-2\lambda x+\lambda=0\) के दो वास्तविक और असमान मूल हों, तो \(\lambda\) पर कौन सी शर्त सही है?

If \(x^2-2\lambda x+\lambda=0\) has two real and distinct roots, which condition on \(\lambda\) is correct?

Explanation opens after your attempt
Correct Answer

A. \(\lambda<0\) या \(\lambda>1\)\(\lambda<0\) or \(\lambda>1\)

Step 1

Concept

Here (D=4\lambda-2-4\lambda=4\lambda\(\lambda-1\)). For distinct real roots (D>0), so \(\lambda<0\) or \(\lambda>1\).

Step 2

Why this answer is correct

The correct answer is A. \(\lambda<0\) या \(\lambda>1\) / \(\lambda<0\) or \(\lambda>1\). Here (D=4\lambda-2-4\lambda=4\lambda\(\lambda-1\)). For distinct real roots (D>0), so \(\lambda<0\) or \(\lambda>1\).

Step 3

Exam Tip

यहाँ (D=4\lambda-2-4\lambda=4\lambda\(\lambda-1\)) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए \(\lambda<0\) या \(\lambda>1\)।

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\(2x^2+\lambda x+8=0\) की जड़ें समान हों, तो \(\lambda\) के मान क्या होंगे?

If \(2x^2+\lambda x+8=0\) has equal roots, what are the values of \(\lambda\)?

Explanation opens after your attempt
Correct Answer

A. (8) और (-8)(8) and (-8)

Step 1

Concept

For equal roots, put (D=0). From \(\lambda^2-64=0\), we get \(\lambda=\pm8\).

Step 2

Why this answer is correct

The correct answer is A. (8) और (-8) / (8) and (-8). For equal roots, put (D=0). From \(\lambda^2-64=0\), we get \(\lambda=\pm8\).

Step 3

Exam Tip

समान जड़ों के लिए (D=0) रखें। \(\lambda^2-64=0\) से \(\lambda=\pm8\) मिलता है।

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किसी समांतर श्रेणी के पहले दो पद \(\lambda-4\) और \(3\lambda+2\) हैं। सामान्य अंतर क्या है?

The first two terms of an AP are \(\lambda-4\) and \(3\lambda+2\). What is the common difference?

Explanation opens after your attempt
Correct Answer

A. \(2\lambda+6\)

Step 1

Concept

The common difference is second term minus first term, so (3\lambda+2-\(\lambda-4\)=2\lambda+6). In exams, handle the minus sign before brackets carefully.

Step 2

Why this answer is correct

The correct answer is A. \(2\lambda+6\). The common difference is second term minus first term, so (3\lambda+2-\(\lambda-4\)=2\lambda+6). In exams, handle the minus sign before brackets carefully.

Step 3

Exam Tip

सामान्य अंतर दूसरा पद घटा पहला पद है, इसलिए (3\lambda+2-\(\lambda-4\)=2\lambda+6)। परीक्षा में कोष्ठक हटाते समय ऋण चिन्ह संभालें।

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युग्म \(3x+2y=\lambda\) और (18x+12y=48) के अनंत हलों के लिए \(\lambda\) क्या होगा?

For infinitely many solutions of \(3x+2y=\lambda\) and (18x+12y=48), what is \(\lambda\)?

Explanation opens after your attempt
Correct Answer

B. \(\lambda=8\)

Step 1

Concept

The coefficient ratio is \(\frac{1}{6}\). For infinitely many solutions, \(\frac{\lambda}{48}=\frac{1}{6}\), so \(\lambda=8\).

Step 2

Why this answer is correct

The correct answer is B. \(\lambda=8\). The coefficient ratio is \(\frac{1}{6}\). For infinitely many solutions, \(\frac{\lambda}{48}=\frac{1}{6}\), so \(\lambda=8\).

Step 3

Exam Tip

गुणांक अनुपात \(\frac{1}{6}\) है। अनंत हलों के लिए \(\frac{\lambda}{48}=\frac{1}{6}\) इसलिए \(\lambda=8\)।

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यदि \(x^2+2x+\lambda=0\) के वास्तविक मूल नहीं हैं, तो \(\lambda\) के लिए सही शर्त क्या है?

If \(x^2+2x+\lambda=0\) has no real roots, what is the correct condition for \(\lambda\)?

Explanation opens after your attempt
Correct Answer

A. \(\lambda>1\)

Step 1

Concept

For no real roots, (D<0), so \(4-4\lambda<0\) gives \(\lambda>1\). In exams, keep strict inequality separate from equality.

Step 2

Why this answer is correct

The correct answer is A. \(\lambda>1\). For no real roots, (D<0), so \(4-4\lambda<0\) gives \(\lambda>1\). In exams, keep strict inequality separate from equality.

Step 3

Exam Tip

वास्तविक मूल न होने के लिए (D<0), अतः \(4-4\lambda<0\) से \(\lambda>1\)। परीक्षा में strict inequality को बराबरी से अलग रखें।

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यदि \(7x^2-6x+\lambda=0\) की जड़ें वास्तविक नहीं हैं, तो \(\lambda\) पर सही शर्त क्या है?

If the roots of \(7x^2-6x+\lambda=0\) are not real, what is the correct condition on \(\lambda\)?

Explanation opens after your attempt
Correct Answer

A. \(\lambda>\frac{9}{7}\)

Step 1

Concept

For non-real roots, (D<0) is required. From \(36-28\lambda<0\), we get \(\lambda>\frac{9}{7}\).

Step 2

Why this answer is correct

The correct answer is A. \(\lambda>\frac{9}{7}\). For non-real roots, (D<0) is required. From \(36-28\lambda<0\), we get \(\lambda>\frac{9}{7}\).

Step 3

Exam Tip

वास्तविक नहीं होने के लिए (D<0) चाहिए। \(36-28\lambda<0\) से \(\lambda>\frac{9}{7}\) मिलता है।

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यदि \(5x^2-4x+\lambda=0\) की जड़ें वास्तविक नहीं हैं, तो \(\lambda\) पर सही शर्त क्या है?

If the roots of \(5x^2-4x+\lambda=0\) are not real, what is the correct condition on \(\lambda\)?

Explanation opens after your attempt
Correct Answer

C. \(\lambda>\frac{4}{5}\)

Step 1

Concept

For non-real roots, (D<0) is required. From \(16-20\lambda<0\), we get \(\lambda>\frac{4}{5}\).

Step 2

Why this answer is correct

The correct answer is C. \(\lambda>\frac{4}{5}\). For non-real roots, (D<0) is required. From \(16-20\lambda<0\), we get \(\lambda>\frac{4}{5}\).

Step 3

Exam Tip

वास्तविक नहीं होने के लिए (D<0) चाहिए। \(16-20\lambda<0\) से \(\lambda>\frac{4}{5}\) मिलता है।

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यदि \(3x^2+2x+\lambda=0\) की जड़ें वास्तविक नहीं हैं, तो \(\lambda\) पर सही शर्त क्या है?

If the roots of \(3x^2+2x+\lambda=0\) are not real, what is the correct condition on \(\lambda\)?

Explanation opens after your attempt
Correct Answer

A. \(\lambda>\frac{1}{3}\)

Step 1

Concept

For non-real roots, (D<0) is required. From \(4-12\lambda<0\), we get \(\lambda>\frac{1}{3}\).

Step 2

Why this answer is correct

The correct answer is A. \(\lambda>\frac{1}{3}\). For non-real roots, (D<0) is required. From \(4-12\lambda<0\), we get \(\lambda>\frac{1}{3}\).

Step 3

Exam Tip

वास्तविक नहीं होने के लिए (D<0) चाहिए। \(4-12\lambda<0\) से \(\lambda>\frac{1}{3}\) मिलता है।

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\(x^2+12x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?

For \(x^2+12x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?

Explanation opens after your attempt
Correct Answer

A. \(0<\lambda<36\)

Step 1

Concept

For both roots to be negative, the sum (-12) and product \(\lambda>0\) are needed. For real distinct roots, \(144-4\lambda>0\), so \(0<\lambda<36\).

Step 2

Why this answer is correct

The correct answer is A. \(0<\lambda<36\). For both roots to be negative, the sum (-12) and product \(\lambda>0\) are needed. For real distinct roots, \(144-4\lambda>0\), so \(0<\lambda<36\).

Step 3

Exam Tip

दोनों ऋणात्मक जड़ों के लिए योग (-12) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(144-4\lambda>0\), इसलिए \(0<\lambda<36\)।

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\(x^2+10x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?

For \(x^2+10x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?

Explanation opens after your attempt
Correct Answer

B. \(0<\lambda<25\)

Step 1

Concept

For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).

Step 2

Why this answer is correct

The correct answer is B. \(0<\lambda<25\). For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).

Step 3

Exam Tip

दोनों ऋणात्मक जड़ों के लिए योग (-10) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(100-4\lambda>0\), इसलिए \(0<\lambda<25\)।

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\(x^2+2x+\lambda=0\) की जड़ें वास्तविक और भिन्न हों तथा दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?

For \(x^2+2x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?

Explanation opens after your attempt
Correct Answer

A. \(0<\lambda<1\)

Step 1

Concept

For both roots to be negative, the sum (-2) and product \(\lambda>0\) are needed. For real distinct roots, \(4-4\lambda>0\), hence \(0<\lambda<1\).

Step 2

Why this answer is correct

The correct answer is A. \(0<\lambda<1\). For both roots to be negative, the sum (-2) and product \(\lambda>0\) are needed. For real distinct roots, \(4-4\lambda>0\), hence \(0<\lambda<1\).

Step 3

Exam Tip

दोनों ऋणात्मक जड़ों के लिए योग (-2) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(4-4\lambda>0\), इसलिए \(0<\lambda<1\)।

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समीकरण \(2x^2+3x+\lambda=0\) के वास्तविक और असमान मूलों के लिए कौन सी शर्त सही है?

For \(2x^2+3x+\lambda=0\) to have real and distinct roots, which condition is correct?

Explanation opens after your attempt
Correct Answer

A. \(\lambda<\frac{9}{8}\)

Step 1

Concept

(D=32-4(2)\lambda=9-8\lambda). From (D>0), we get \(\lambda<\frac{9}{8}\).

Step 2

Why this answer is correct

The correct answer is A. \(\lambda<\frac{9}{8}\). (D=32-4(2)\lambda=9-8\lambda). From (D>0), we get \(\lambda<\frac{9}{8}\).

Step 3

Exam Tip

(D=32-4(2)\lambda=9-8\lambda) है। (D>0) से \(\lambda<\frac{9}{8}\) मिलता है।

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