Here \(\alpha+\beta=6\) and \(\alpha^2+\beta^2=26\). From \(36-2\alpha\beta=26\), \(\alpha\beta=5\), so the roots are (1) and (5).
Step 2
Why this answer is correct
The correct answer is A. (1) और (5) / (1) and (5). Here \(\alpha+\beta=6\) and \(\alpha^2+\beta^2=26\). From \(36-2\alpha\beta=26\), \(\alpha\beta=5\), so the roots are (1) and (5).
Step 3
Exam Tip
\(\alpha+\beta=6\) और \(\alpha^2+\beta^2=26\) है। \(36-2\alpha\beta=26\) से \(\alpha\beta=5\), इसलिए जड़ें (1) और (5) हैं।
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 \(0<\lambda<36\)।
Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+30=0\). Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=6\) हैं। नई जड़ें (5) और (6) हैं, इसलिए समीकरण \(x^2-11x+30=0\) है।
Here \(\alpha+\beta=5\) and \(\alpha^2+\beta^2=17\). From \(25-2\alpha\beta=17\), \(\alpha\beta=4\), so the roots are (1) and (4).
Step 2
Why this answer is correct
The correct answer is A. (1) और (4) / (1) and (4). Here \(\alpha+\beta=5\) and \(\alpha^2+\beta^2=17\). From \(25-2\alpha\beta=17\), \(\alpha\beta=4\), so the roots are (1) and (4).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha^2+\beta^2=17\) है। \(25-2\alpha\beta=17\) से \(\alpha\beta=4\), इसलिए जड़ें (1) और (4) हैं।
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 \(0<\lambda<25\)।
Here \(\alpha+\beta=4\) and \(\alpha^2+\beta^2=10\). From \(16-2\alpha\beta=10\), \(\alpha\beta=3\), so the roots are (1) and (3).
Step 2
Why this answer is correct
The correct answer is A. (1) और (3) / (1) and (3). Here \(\alpha+\beta=4\) and \(\alpha^2+\beta^2=10\). From \(16-2\alpha\beta=10\), \(\alpha\beta=3\), so the roots are (1) and (3).
Step 3
Exam Tip
\(\alpha+\beta=4\) और \(\alpha^2+\beta^2=10\) है। \(16-2\alpha\beta=10\) से \(\alpha\beta=3\), इसलिए जड़ें (1) और (3) हैं।
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 \(0<\lambda<1\)।
Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+4=0\). Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 3
Exam Tip
\(\alpha+\beta=3\) और \(\alpha\beta=-2\) है। इसलिए \(\alpha^2+\beta^2=13\) और \(\alpha^2\beta^2=4\), अतः समीकरण \(x^2-13x+4=0\) है।
Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-10x+3=0\). Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{10}{3}\) और \(\alpha\beta=1\) है। व्युत्क्रम जड़ों का योग \(\frac{10}{3}\) और गुणनफल (1) ही रहता है।
In the first equation, (D=100-92=8>0), and (8) is not a perfect square. So the roots are real, irrational and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+23=0\). In the first equation, (D=100-92=8>0), and (8) is not a perfect square. So the roots are real, irrational and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=100-92=8>0) है और (8) पूर्ण वर्ग नहीं है। इसलिए मूल वास्तविक, अपरिमेय और भिन्न हैं।
For (k=2), the discriminant is (16-8=8), positive but not a perfect square. Therefore the roots are real and irrational.
Step 2
Why this answer is correct
The correct answer is B. (k=2). For (k=2), the discriminant is (16-8=8), positive but not a perfect square. Therefore the roots are real and irrational.
Step 3
Exam Tip
(k=2) पर विविक्तकर (16-8=8), जो धनात्मक पर पूर्ण वर्ग नहीं है। इसलिए मूल वास्तविक और अपरिमेय होंगे।
A. कथन और कारण दोनों सही हैं/Both assertion and reason are correct
Step 1
Concept
Here (D=32-4(1)(7)=-19). Since (D<0), the assertion is correct.
Step 2
Why this answer is correct
The correct answer is A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct. Here (D=32-4(1)(7)=-19). Since (D<0), the assertion is correct.
Step 3
Exam Tip
यहाँ (D=32-4(1)(7)=-19) है। (D<0) होने से कथन सही है।
A. \(k\neq0\) और \(k^2\le36\)/\(k\neq0\) and \(k^2\le36\)
Step 1
Concept
The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(144-4k^2\ge0\), hence \(k^2\le36\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq0\) और \(k^2\le36\) / \(k\neq0\) and \(k^2\le36\). The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(144-4k^2\ge0\), hence \(k^2\le36\).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{k}{k}=1\) है, इसलिए \(k\neq0\) चाहिए। वास्तविक जड़ों के लिए \(144-4k^2\ge0\), अतः \(k^2\le36\)।
C. \(k\neq0\) और \(k^2\le25\)/\(k\neq0\) and \(k^2\le25\)
Step 1
Concept
The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(100-4k^2\ge0\), hence \(k^2\le25\).
Step 2
Why this answer is correct
The correct answer is C. \(k\neq0\) और \(k^2\le25\) / \(k\neq0\) and \(k^2\le25\). The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(100-4k^2\ge0\), hence \(k^2\le25\).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{k}{k}=1\) है, इसलिए \(k\neq0\) चाहिए। वास्तविक जड़ों के लिए \(100-4k^2\ge0\), अतः \(k^2\le25\)।
In the given equation, the sum of roots is (2r+5) and the product is (r-2+5r+6=(r+2)(r+3)). Hence the roots are (r+2) and (r+3), so the positive difference is (1).
Step 2
Why this answer is correct
The correct answer is A. (1). In the given equation, the sum of roots is (2r+5) and the product is (r-2+5r+6=(r+2)(r+3)). Hence the roots are (r+2) and (r+3), so the positive difference is (1).
Step 3
Exam Tip
दिए गए समीकरण में जड़ों का योग (2r+5) और गुणनफल (r-2+5r+6=(r+2)(r+3)) है। इसलिए जड़ें (r+2) और (r+3) हैं, अतः धनात्मक अंतर (1) है।
A. \(k\neq0\) और \(k^2\le16\)/\(k\neq0\) and \(k^2\le16\)
Step 1
Concept
For reciprocal roots, \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(64-4k^2\ge0\), hence \(k^2\le16\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq0\) और \(k^2\le16\) / \(k\neq0\) and \(k^2\le16\). For reciprocal roots, \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(64-4k^2\ge0\), hence \(k^2\le16\).
Step 3
Exam Tip
व्युत्क्रम जड़ों के लिए \(\frac{k}{k}=1\) है, इसलिए \(k\neq0\) चाहिए। वास्तविक जड़ों के लिए \(64-4k^2\ge0\), अतः \(k^2\le16\)।
The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+27=0\). The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 3
Exam Tip
पुराने योग (4) और गुणनफल (3) हैं। नए योग (12) और गुणनफल (27) होंगे इसलिए \(x^2-12x+27=0\) है।
The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x+8=0\). The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 3
Exam Tip
पुराने योग (3) और गुणनफल (2) हैं। नए योग (6) और गुणनफल (8) होंगे इसलिए \(x^2-6x+8=0\) है।
A. मूल एक दूसरे के विपरीत हैं/The roots are opposites of each other
Step 1
Concept
If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 2
Why this answer is correct
The correct answer is A. मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other. If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 3
Exam Tip
यदि \(\alpha+\beta=0\) है तो \(\beta=-\alpha\) होता है। इसलिए मूल विपरीत हो सकते हैं।
The sum of new roots is (4\alpha+4\beta=4\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 2
Why this answer is correct
The correct answer is A. (12). The sum of new roots is (4\alpha+4\beta=4\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 3
Exam Tip
नए मूलों का योग (4\alpha+4\beta=4\(\alpha+\beta\)=12) है। गुणक लगे मूलों में योग भी उसी गुणक से गुणा होता है।
The sum of new roots is (3\alpha+3\beta=3\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.
Step 2
Why this answer is correct
The correct answer is A. (12). The sum of new roots is (3\alpha+3\beta=3\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.
Step 3
Exam Tip
नए मूलों का योग (3\alpha+3\beta=3\(\alpha+\beta\)=12) है। गुणक लगे मूलों में योग पर भी वही गुणक लगता है।
The sum of new roots is (2\alpha+2\beta=2\(\alpha+\beta\)=10). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 2
Why this answer is correct
The correct answer is A. (10). The sum of new roots is (2\alpha+2\beta=2\(\alpha+\beta\)=10). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 3
Exam Tip
नए मूलों का योग (2\alpha+2\beta=2\(\alpha+\beta\)=10) है। गुणक लगे मूलों में योग पर भी वही गुणक लगता है।
A positive product means both roots have the same sign. A negative sum means both roots are negative.
Step 2
Why this answer is correct
The correct answer is B. दोनों ऋणात्मक / Both negative. A positive product means both roots have the same sign. A negative sum means both roots are negative.
Step 3
Exam Tip
गुणनफल धनात्मक होने पर दोनों मूलों का चिन्ह समान होता है। योग ऋणात्मक होने से दोनों मूल ऋणात्मक होंगे।
A. वास्तविक, अपरिमेय और भिन्न/Real, irrational and distinct
Step 1
Concept
Here (D=\(2\sqrt{5}\)2-4(1)(1)=16>0). The roots are \(\sqrt{5}\pm2\), so they are irrational and distinct.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक, अपरिमेय और भिन्न / Real, irrational and distinct. Here (D=\(2\sqrt{5}\)2-4(1)(1)=16>0). The roots are \(\sqrt{5}\pm2\), so they are irrational and distinct.
Step 3
Exam Tip
यहाँ (D=\(2\sqrt{5}\)2-4(1)(1)=16>0) है। मूल \(\sqrt{5}\pm2\) होंगे, इसलिए वे अपरिमेय और भिन्न हैं।
A. वास्तविक, अपरिमेय और भिन्न/Real, irrational and distinct
Step 1
Concept
Here (D=(-5)2-4(1)(3)=13), and (13) is not a perfect square. So the roots are real, irrational and distinct.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक, अपरिमेय और भिन्न / Real, irrational and distinct. Here (D=(-5)2-4(1)(3)=13), and (13) is not a perfect square. So the roots are real, irrational and distinct.
Step 3
Exam Tip
यहाँ (D=(-5)2-4(1)(3)=13) है और (13) पूर्ण वर्ग नहीं है। इसलिए मूल वास्तविक, अपरिमेय और भिन्न हैं।
A. कथन गलत है, कारण सही है/Assertion is wrong, reason is correct
Step 1
Concept
Here (D=(-6)2-4(1)(11)=-8). Since (D<0), real roots do not exist, so the assertion is wrong.
Step 2
Why this answer is correct
The correct answer is A. कथन गलत है, कारण सही है / Assertion is wrong, reason is correct. Here (D=(-6)2-4(1)(11)=-8). Since (D<0), real roots do not exist, so the assertion is wrong.
Step 3
Exam Tip
यहाँ (D=(-6)2-4(1)(11)=-8) है। (D<0) होने पर वास्तविक मूल नहीं होते, इसलिए कथन गलत है।
(\(\sqrt{6}+\sqrt{2}\)\(\sqrt{6}-\sqrt{2}\)=6-2=4), which is rational.
Step 3
Exam Tip
Identifying conjugates helps remove radicals quickly. चरण 1: पहला विकल्प संयुग्मी पदों का गुणनफल है। चरण 2: (\(\sqrt{6}+\sqrt{2}\)\(\sqrt{6}-\sqrt{2}\)=6-2=4), जो परिमेय है। चरण 3: संयुग्मी पद पहचानने से वर्गमूल जल्दी हट जाते हैं।
(4) is rational and \(\sqrt{13}\) is irrational, so the sum is irrational. In exams identify square roots of perfect squares first.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{13}\). (4) is rational and \(\sqrt{13}\) is irrational, so the sum is irrational. In exams identify square roots of perfect squares first.
Step 3
Exam Tip
(4) परिमेय है और \(\sqrt{13}\) अपरिमेय है, इसलिए योग अपरिमेय है। परीक्षा में पूर्ण वर्ग के वर्गमूल को पहले पहचानें।
\(\sqrt{2}\times\sqrt{5}=\sqrt{10}\), which is irrational. Multiplying equal roots can often give a rational number.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\times\sqrt{5}\). \(\sqrt{2}\times\sqrt{5}=\sqrt{10}\), which is irrational. Multiplying equal roots can often give a rational number.
Step 3
Exam Tip
\(\sqrt{2}\times\sqrt{5}=\sqrt{10}\) है जो अपरिमेय है। समान जड़ों का गुणन अक्सर परिमेय दे सकता है।
A. जब (a) कोई भी परिमेय संख्या हो/When (a) is any rational number
Step 1
Concept
Adding a rational number to an irrational number gives an irrational result. This simple property often appears in MCQs.
Step 2
Why this answer is correct
The correct answer is A. जब (a) कोई भी परिमेय संख्या हो / When (a) is any rational number. Adding a rational number to an irrational number gives an irrational result. This simple property often appears in MCQs.
Step 3
Exam Tip
परिमेय में अपरिमेय जोड़ने पर परिणाम अपरिमेय रहता है। यह आसान गुण अक्सर MCQ में आता है।
\(\sqrt{5}\) is irrational and \(2-\sqrt{5}\) is also irrational.
Step 2
Why this answer is correct
Their sum is (2) which is rational.
Step 3
Exam Tip
There is no single always rule for the sum of two irrational numbers. चरण 1: \(\sqrt{5}\) अपरिमेय है और \(2-\sqrt{5}\) भी अपरिमेय है। चरण 2: उनका योग (2) है जो परिमेय है। चरण 3: दो अपरिमेय संख्याओं के योग के लिए एक ही नियम हर बार लागू नहीं होता।
Its reciprocal \(\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\) is also irrational.
Step 3
Exam Tip
Do not assume the reciprocal of a non-zero irrational surd is rational. चरण 1: \(\sqrt{12}=2\sqrt{3}\) अपरिमेय है। चरण 2: इसका व्युत्क्रम \(\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\) भी अपरिमेय है। चरण 3: अशून्य अपरिमेय मूल के व्युत्क्रम को परिमेय मानने की गलती न करें।
The sum of two irrational numbers can be rational.
Step 2
Why this answer is correct
For example, (\sqrt{2}+\(-\sqrt{2}\)=0). Therefore, saying (a+b) is always irrational is false.
Step 3
Exam Tip
Be careful with universal statements about two irrational numbers. चरण 1: दो अपरिमेय संख्याओं का योग कभी परिमेय भी हो सकता है। चरण 2: उदाहरण (\sqrt{2}+\(-\sqrt{2}\)=0) है। इसलिए (a+b) हमेशा अपरिमेय कहना गलत है। चरण 3: दो अपरिमेय संख्याओं पर हमेशा वाले नियम बहुत सावधानी से लगाएँ।
A rational number minus an irrational number is irrational.
Step 2
Why this answer is correct
If (r-s) were rational, then (s=r-(r-s)) would be rational, which is impossible.
Step 3
Exam Tip
Use the same reasoning for subtraction as for addition. चरण 1: परिमेय संख्या में से अपरिमेय संख्या घटाने पर परिणाम अपरिमेय रहता है। चरण 2: यदि (r-s) परिमेय हो, तो (s=r-(r-s)) परिमेय हो जाएगा, जो असंभव है। चरण 3: घटाव में भी वही सोच रखें जो योग में रखते हैं।
A. दो वास्तविक परिमेय और असमान ((D=1))/Two real rational and distinct ((D=1))
Step 1
Concept
Here (D=(-11)2-4(2)(15)=1). A positive perfect-square (D) gives rational distinct roots.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक परिमेय और असमान ((D=1)) / Two real rational and distinct ((D=1)). Here (D=(-11)2-4(2)(15)=1). A positive perfect-square (D) gives rational distinct roots.
Step 3
Exam Tip
यहाँ (D=(-11)2-4(2)(15)=1) है। धनात्मक पूर्ण वर्ग (D) परिमेय असमान मूल देता है।
Here \(\alpha+\beta=2a+1\) and \(\alpha\beta=a^2+a-6\). Since (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=25), the positive difference is (5).
Step 2
Why this answer is correct
The correct answer is A. (5). Here \(\alpha+\beta=2a+1\) and \(\alpha\beta=a^2+a-6\). Since (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=25), the positive difference is (5).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=2a+1\) और \(\alpha\beta=a^2+a-6\) है। (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=25), इसलिए धनात्मक अंतर (5) है।
A. हर वास्तविक (a) के लिए वास्तविक नहीं/Not real for every real (a)
Step 1
Concept
The discriminant is (D=4-4\(a^2+3\)=-4a-2-8). It is negative for every real (a), so the roots are not real.
Step 2
Why this answer is correct
The correct answer is A. हर वास्तविक (a) के लिए वास्तविक नहीं / Not real for every real (a). The discriminant is (D=4-4\(a^2+3\)=-4a-2-8). It is negative for every real (a), so the roots are not real.
Step 3
Exam Tip
विविक्तकर (D=4-4\(a^2+3\)=-4a-2-8) है। यह हर वास्तविक (a) के लिए ऋणात्मक है, इसलिए जड़ें वास्तविक नहीं हैं।
Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Since (\(\alpha-\beta\)2=25-9=16), the positive difference is (4).
Step 2
Why this answer is correct
The correct answer is A. (4). Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Since (\(\alpha-\beta\)2=25-9=16), the positive difference is (4).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=\frac{9}{4}\) है। (\(\alpha-\beta\)2=25-9=16), इसलिए धनात्मक अंतर (4) है।
Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Thus (\(\alpha-\beta\)2=25-9=16), so the positive difference is (4); option (A) should be correct.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{7}{2}\). Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Thus (\(\alpha-\beta\)2=25-9=16), so the positive difference is (4); option (A) should be correct.
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=\frac{9}{4}\) है। (\(\alpha-\beta\)2=25-9=16), इसलिए धनात्मक अंतर (4) है, अतः विकल्प (A) सही होना चाहिए।
The prime pairs with sum (14) are ((3,11)) and ((7,7)). Thus (m=33) or (m=49), and the sum is (82), so none of the options is correct.
Step 2
Why this answer is correct
The correct answer is D. (94). The prime pairs with sum (14) are ((3,11)) and ((7,7)). Thus (m=33) or (m=49), and the sum is (82), so none of the options is correct.
Step 3
Exam Tip
योग (14) वाली अभाज्य जोड़ियाँ ((3,11)) और ((7,7)) हैं। इसलिए (m=33) या (m=49), और योग (82) है, अतः विकल्पों में कोई सही नहीं है।
The sum of roots is \(\frac{1}{4+\sqrt{3}}+\frac{1}{4-\sqrt{3}}=\frac{8}{13}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{8}{13}\).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{8}{13}\). The sum of roots is \(\frac{1}{4+\sqrt{3}}+\frac{1}{4-\sqrt{3}}=\frac{8}{13}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{8}{13}\).
Step 3
Exam Tip
जड़ों का योग \(\frac{1}{4+\sqrt{3}}+\frac{1}{4-\sqrt{3}}=\frac{8}{13}\) है। \(x^2+ax+b=0\) में योग (-a) होता है, इसलिए \(a=-\frac{8}{13}\)।
The product of the two roots is (1), so the product condition is satisfied. For real roots, the discriminant \(s^2-4\ge0\), so \(s^2\ge4\).
Step 2
Why this answer is correct
The correct answer is A. \(s^2\ge4\). The product of the two roots is (1), so the product condition is satisfied. For real roots, the discriminant \(s^2-4\ge0\), so \(s^2\ge4\).
Step 3
Exam Tip
दोनों जड़ों का गुणनफल (1) है, इसलिए समीकरण का गुणनफल सही है। वास्तविक जड़ों के लिए विविक्तकर \(s^2-4\ge0\), इसलिए \(s^2\ge4\)।
The sum of these roots is \(\frac{5t+3}{4}\), and the product is (\frac{t(t+3)}{4}). These match \(-\frac{b}{a}\) and \(\frac{c}{a}\) of the given equation.
Step 2
Why this answer is correct
The correct answer is A. हर (t) के लिए / For every (t). The sum of these roots is \(\frac{5t+3}{4}\), and the product is (\frac{t(t+3)}{4}). These match \(-\frac{b}{a}\) and \(\frac{c}{a}\) of the given equation.
Step 3
Exam Tip
इन जड़ों का योग \(\frac{5t+3}{4}\) और गुणनफल (\frac{t(t+3)}{4}) है। ये दिए गए समीकरण के \(-\frac{b}{a}\) और \(\frac{c}{a}\) से मेल खाते हैं।
The sum (6) is positive and (c>0) is needed for both positive roots. For real roots, \(36-4c\ge0\), so \(0<c\le9\).
Step 2
Why this answer is correct
The correct answer is A. \(0<c\le9\). The sum (6) is positive and (c>0) is needed for both positive roots. For real roots, \(36-4c\ge0\), so \(0<c\le9\).
Step 3
Exam Tip
योग (6) धनात्मक है और दोनों धनात्मक जड़ों के लिए (c>0) चाहिए। वास्तविक जड़ों के लिए \(36-4c\ge0\), इसलिए \(0<c\le9\)।
For real roots, \(D\ge0\) is required. Here (D=36(a-1)2-36\(a^2-4a-5\)=72a+216), so the exact condition is \(a\ge-3\), not \(a\ge-\frac{7}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(a\ge-\frac{7}{2}\). For real roots, \(D\ge0\) is required. Here (D=36(a-1)2-36\(a^2-4a-5\)=72a+216), so the exact condition is \(a\ge-3\), not \(a\ge-\frac{7}{2}\).
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=36(a-1)2-36\(a^2-4a-5\)=72a+216), इसलिए \(a\ge-\frac{7}{2}\) नहीं बल्कि \(a\ge-3\) होगा।
For real roots, \(D\ge0\) is required. Here (D=64(a-2)2-64\(a^2-6a\)=128(a+2)), so \(a\ge-2\); hence none of these options is exact.
Step 2
Why this answer is correct
The correct answer is A. \(a\ge1\). For real roots, \(D\ge0\) is required. Here (D=64(a-2)2-64\(a^2-6a\)=128(a+2)), so \(a\ge-2\); hence none of these options is exact.
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=64(a-2)2-64\(a^2-6a\)=128(a+2)), इसलिए \(a\ge-2\) होगा, अतः विकल्पों में सही शर्त नहीं है।
A. \(7\sqrt{3}\) और \(-7\sqrt{3}\)/\(7\sqrt{3}\) and \(-7\sqrt{3}\)
Step 1
Concept
Let the roots be (2r) and (5r). From \(10r^2=\frac{10}{3}\), \(r=\pm\frac{1}{\sqrt{3}}\), so \(7r=-\frac{m}{3}\) gives \(m=\pm7\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(7\sqrt{3}\) और \(-7\sqrt{3}\) / \(7\sqrt{3}\) and \(-7\sqrt{3}\). Let the roots be (2r) and (5r). From \(10r^2=\frac{10}{3}\), \(r=\pm\frac{1}{\sqrt{3}}\), so \(7r=-\frac{m}{3}\) gives \(m=\pm7\sqrt{3}\).
Step 3
Exam Tip
जड़ें (2r) और (5r) मानें। \(10r^2=\frac{10}{3}\) से \(r=\pm\frac{1}{\sqrt{3}}\), इसलिए \(7r=-\frac{m}{3}\) से \(m=\pm7\sqrt{3}\)।
Here \(\alpha+\beta=\frac{16}{5}\) and \(\alpha\beta=\frac{p}{5}\). Using (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta), we get (p=11).
Step 2
Why this answer is correct
The correct answer is C. (11). Here \(\alpha+\beta=\frac{16}{5}\) and \(\alpha\beta=\frac{p}{5}\). Using (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta), we get (p=11).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{16}{5}\) और \(\alpha\beta=\frac{p}{5}\) है। (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) से (p=11) मिलता है।
A. \(m\ge0\) और \(m\neq1\)/\(m\ge0\) and \(m\neq1\)
Step 1
Concept
The product of roots is \(\frac{m-1}{m-1}=1\), so \(m\neq1\) is needed. For real roots, \(D=16m\ge0\), hence \(m\ge0\) and \(m\neq1\).
Step 2
Why this answer is correct
The correct answer is A. \(m\ge0\) और \(m\neq1\) / \(m\ge0\) and \(m\neq1\). The product of roots is \(\frac{m-1}{m-1}=1\), so \(m\neq1\) is needed. For real roots, \(D=16m\ge0\), hence \(m\ge0\) and \(m\neq1\).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{m-1}{m-1}=1\) है, इसलिए \(m\neq1\) चाहिए। वास्तविक जड़ों के लिए \(D=16m\ge0\), अतः \(m\ge0\) और \(m\neq1\)।
C. हर (a) के लिए वास्तविक नहीं/Not real for every (a)
Step 1
Concept
The discriminant is (D=4-4\(a^2+2\)=-4\(a^2+1\)). It is negative for every real (a), so the roots are not real.
Step 2
Why this answer is correct
The correct answer is C. हर (a) के लिए वास्तविक नहीं / Not real for every (a). The discriminant is (D=4-4\(a^2+2\)=-4\(a^2+1\)). It is negative for every real (a), so the roots are not real.
Step 3
Exam Tip
विविक्तकर (D=4-4\(a^2+2\)=-4\(a^2+1\)) है। यह हर वास्तविक (a) के लिए ऋणात्मक है, इसलिए जड़ें वास्तविक नहीं हैं।
Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). With \(\alpha+\beta=\frac{13}{3}\) and \(\alpha\beta=\frac{4}{3}\), the positive difference is \(\frac{11}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{11}{3}\). Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). With \(\alpha+\beta=\frac{13}{3}\) and \(\alpha\beta=\frac{4}{3}\), the positive difference is \(\frac{11}{3}\).
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) लगाएँ। \(\alpha+\beta=\frac{13}{3}\) और \(\alpha\beta=\frac{4}{3}\), इसलिए धनात्मक अंतर \(\frac{11}{3}\) है।
