Here \(\alpha+\beta=\frac{11}{3}\) and (\(\alpha-\beta\)2=\frac{25}{9}). Using (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta), we get \(\alpha\beta=\frac{8}{3}\), so (p=8).
Step 2
Why this answer is correct
The correct answer is A. (8). Here \(\alpha+\beta=\frac{11}{3}\) and (\(\alpha-\beta\)2=\frac{25}{9}). Using (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta), we get \(\alpha\beta=\frac{8}{3}\), so (p=8).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{11}{3}\) और (\(\alpha-\beta\)2=\frac{25}{9}) है। सूत्र (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) से \(\alpha\beta=\frac{8}{3}\), इसलिए (p=8)।
A. \(6\sqrt{2}\) और \(-6\sqrt{2}\)/\(6\sqrt{2}\) and \(-6\sqrt{2}\)
Step 1
Concept
Let the roots be (r) and (2r), then \(2r^2=4\) gives \(r=\pm\sqrt{2}\). Since \(3r=-\frac{m}{2}\), we get \(m=\pm6\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{2}\) और \(-6\sqrt{2}\) / \(6\sqrt{2}\) and \(-6\sqrt{2}\). Let the roots be (r) and (2r), then \(2r^2=4\) gives \(r=\pm\sqrt{2}\). Since \(3r=-\frac{m}{2}\), we get \(m=\pm6\sqrt{2}\).
Step 3
Exam Tip
जड़ें (r) और (2r) मानें, तब \(2r^2=4\) से \(r=\pm\sqrt{2}\) मिलता है। योग \(3r=-\frac{m}{2}\), इसलिए \(m=\pm6\sqrt{2}\)।
Putting (x=4) gives (16-4(a+4)+4a=0). Hence (4) is always one root and the other root is (a).
Step 2
Why this answer is correct
The correct answer is A. (4) हमेशा एक जड़ है / (4) is always a root. Putting (x=4) gives (16-4(a+4)+4a=0). Hence (4) is always one root and the other root is (a).
Step 3
Exam Tip
(x=4) रखने पर (16-4(a+4)+4a=0) मिलता है। अतः (4) हमेशा एक जड़ है और दूसरी जड़ (a) होती है।
We use (\frac{1}{\alpha-2}+\frac{1}{\beta-2}=\frac{\alpha-2+\beta-2}{\(\alpha\beta\)2}). Since \(\alpha^2+\beta^2=21\) and (\(\alpha\beta\)2=4), the value is \(\frac{21}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{21}{4}\). We use (\frac{1}{\alpha-2}+\frac{1}{\beta-2}=\frac{\alpha-2+\beta-2}{\(\alpha\beta\)2}). Since \(\alpha^2+\beta^2=21\) and (\(\alpha\beta\)2=4), the value is \(\frac{21}{4}\).
Step 3
Exam Tip
(\frac{1}{\alpha-2}+\frac{1}{\beta-2}=\frac{\alpha-2+\beta-2}{\(\alpha\beta\)2}) होता है। \(\alpha^2+\beta^2=21\) और (\(\alpha\beta\)2=4), इसलिए मान \(\frac{21}{4}\) 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 roots are (3) and (5). Substitution gives \(\frac{5}{1}+\frac{7}{3}=\frac{22}{3}\), so none of the options is correct; the correct value should be \(\frac{22}{3}\).
Step 2
Why this answer is correct
The correct answer is A. (8). The roots are (3) and (5). Substitution gives \(\frac{5}{1}+\frac{7}{3}=\frac{22}{3}\), so none of the options is correct; the correct value should be \(\frac{22}{3}\).
Step 3
Exam Tip
जड़ें (3) और (5) हैं। सीधे रखने पर \(\frac{5}{1}+\frac{7}{3}=\frac{22}{3}\) आता है, इसलिए विकल्पों में कोई सही नहीं; सही प्रश्न के लिए उत्तर \(\frac{22}{3}\) होना चाहिए।
One root can be \(\alpha=2\), which makes \(\alpha-2=0\). Therefore the expression is undefined; always check zero denominators first in exams.
Step 2
Why this answer is correct
The correct answer is A. अपरिभाषित / Undefined. One root can be \(\alpha=2\), which makes \(\alpha-2=0\). Therefore the expression is undefined; always check zero denominators first in exams.
Step 3
Exam Tip
जड़ों में से एक \(\alpha=2\) हो सकती है, जिससे \(\alpha-2=0\) बनता है। इसलिए व्यंजक अपरिभाषित है; परीक्षा में हर पहले शून्य हर देखें।
Here \(\alpha+\beta=\frac{5}{2}\) and \(\alpha\beta=\frac{3}{2}\). Using (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)), we get \(\frac{35}{8}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{35}{8}\). Here \(\alpha+\beta=\frac{5}{2}\) and \(\alpha\beta=\frac{3}{2}\). Using (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)), we get \(\frac{35}{8}\).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{5}{2}\) और \(\alpha\beta=\frac{3}{2}\) है। (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)) से \(\frac{35}{8}\) मिलता है।
Since \(\alpha^2=3\alpha+1\), the expression becomes (3\alpha+1+3\beta-1=3\(\alpha+\beta\)=9), so no option is correct. The correct value is (9).
Step 2
Why this answer is correct
The correct answer is A. (8). Since \(\alpha^2=3\alpha+1\), the expression becomes (3\alpha+1+3\beta-1=3\(\alpha+\beta\)=9), so no option is correct. The correct value is (9).
Step 3
Exam Tip
क्योंकि \(\alpha^2=3\alpha+1\), व्यंजक (3\alpha+1+3\beta-1=3\(\alpha+\beta\)=9) बनता है, इसलिए कोई विकल्प सही नहीं है। सही मान (9) होगा।
A. जड़ों का अंतर हमेशा (6) है/The difference of roots is always (6)
Step 1
Concept
The equation can be written as ((x-p)2-9=0). The roots are (p+3) and (p-3), so the difference is (6).
Step 2
Why this answer is correct
The correct answer is A. जड़ों का अंतर हमेशा (6) है / The difference of roots is always (6). The equation can be written as ((x-p)2-9=0). The roots are (p+3) and (p-3), so the difference is (6).
Step 3
Exam Tip
समीकरण को ((x-p)2-9=0) लिखा जा सकता है। जड़ें (p+3) और (p-3) हैं, इसलिए अंतर (6) है।
The sum (-4) is already negative and the product must be positive, so (c>0). For real roots, \(16-4c\ge0\), hence \(0<c\le4\).
Step 2
Why this answer is correct
The correct answer is A. \(0<c\le4\). The sum (-4) is already negative and the product must be positive, so (c>0). For real roots, \(16-4c\ge0\), hence \(0<c\le4\).
Step 3
Exam Tip
योग (-4) पहले से ऋणात्मक है और गुणनफल धनात्मक चाहिए, इसलिए (c>0)। वास्तविक जड़ों के लिए \(16-4c\ge0\), अतः \(0<c\le4\)।
We use (\(\alpha+3\)\(\beta+3\)=\alpha\beta+3\(\alpha+\beta\)+9). Since \(\alpha+\beta=2\) and \(\alpha\beta=-8\), the value is (7).
Step 2
Why this answer is correct
The correct answer is A. (7). We use (\(\alpha+3\)\(\beta+3\)=\alpha\beta+3\(\alpha+\beta\)+9). Since \(\alpha+\beta=2\) and \(\alpha\beta=-8\), the value is (7).
Step 3
Exam Tip
(\(\alpha+3\)\(\beta+3\)=\alpha\beta+3\(\alpha+\beta\)+9) है। \(\alpha+\beta=2\) और \(\alpha\beta=-8\), इसलिए मान (7) है।
The sum is \(\frac{3t+1}{2}\) and the product is (\frac{t(t+1)}{2}). These match \(-\frac{b}{a}\) and \(\frac{c}{a}\) for every (t).
Step 2
Why this answer is correct
The correct answer is A. हर (t) के लिए / For every (t). The sum is \(\frac{3t+1}{2}\) and the product is (\frac{t(t+1)}{2}). These match \(-\frac{b}{a}\) and \(\frac{c}{a}\) for every (t).
Step 3
Exam Tip
इन जड़ों का योग \(\frac{3t+1}{2}\) और गुणनफल (\frac{t(t+1)}{2}) है। ये दिए गए समीकरण के \(-\frac{b}{a}\) और \(\frac{c}{a}\) से हर (t) पर मेल खाते हैं।
Putting (x=3) gives (9-3(m-2)+m-6=0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 2
Why this answer is correct
The correct answer is A. (1). Putting (x=3) gives (9-3(m-2)+m-6=0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 3
Exam Tip
(x=3) रखने पर (9-3(m-2)+m-6=0), इसलिए \(m=\frac{9}{2}\)। गुणनफल \(m-6=-\frac{3}{2}\) है, अतः दूसरी जड़ \(-\frac{1}{2}\) होगी, इसलिए कोई विकल्प सही नहीं है।
A. जड़ें (a) और (a+1) हैं/The roots are (a) and (a+1)
Step 1
Concept
The sum of roots is (2a+1) and the product is (a(a+1)). These match the pair (a) and (a+1).
Step 2
Why this answer is correct
The correct answer is A. जड़ें (a) और (a+1) हैं / The roots are (a) and (a+1). The sum of roots is (2a+1) and the product is (a(a+1)). These match the pair (a) and (a+1).
Step 3
Exam Tip
जड़ों का योग (2a+1) और गुणनफल (a(a+1)) है। ये (a) और (a+1) की योग-गुणनफल जोड़ी से मेल खाते हैं।
We use \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Here \(\alpha^2+\beta^2=45\) and \(\alpha\beta=18\), so the value is \(\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{2}\). We use \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Here \(\alpha^2+\beta^2=45\) and \(\alpha\beta=18\), so the value is \(\frac{5}{2}\).
Step 3
Exam Tip
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\) है। यहाँ \(\alpha^2+\beta^2=45\) और \(\alpha\beta=18\), इसलिए मान \(\frac{5}{2}\) है।
A. ऐसा कोई वास्तविक \(\theta\) नहीं/No such real \(\theta\)
Step 1
Concept
We would need \(\sin \theta+\cos \theta=4\), but its maximum is \(\sqrt{2}\). Therefore no real \(\theta\) is possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा कोई वास्तविक \(\theta\) नहीं / No such real \(\theta\). We would need \(\sin \theta+\cos \theta=4\), but its maximum is \(\sqrt{2}\). Therefore no real \(\theta\) is possible.
Step 3
Exam Tip
\(\sin \theta+\cos \theta=4\) होना पड़ेगा, पर इसका अधिकतम \(\sqrt{2}\) है। इसलिए ऐसा वास्तविक \(\theta\) संभव नहीं है।
Let the roots be (2r) and (3r). From (5r=12), \(r=\frac{12}{5}\), so the product is \(6r^2=\frac{864}{25}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{864}{25}\). Let the roots be (2r) and (3r). From (5r=12), \(r=\frac{12}{5}\), so the product is \(6r^2=\frac{864}{25}\).
Step 3
Exam Tip
जड़ें (2r) और (3r) मानें। (5r=12) से \(r=\frac{12}{5}\), इसलिए गुणनफल \(6r^2=\frac{864}{25}\) है।
The prime pairs with sum (10) are ((3,7)) and ((5,5)). Hence (m=21) or (m=25), and their sum is (46), so option (B) should be correct.
Step 2
Why this answer is correct
The correct answer is A. (42). The prime pairs with sum (10) are ((3,7)) and ((5,5)). Hence (m=21) or (m=25), and their sum is (46), so option (B) should be correct.
Step 3
Exam Tip
योग (10) वाली अभाज्य जोड़ियाँ ((3,7)) और ((5,5)) हैं। इसलिए (m=21) या (m=25), और योग (46) है, अतः विकल्प (B) सही होना चाहिए।
Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). With \(\alpha+\beta=\frac{7}{2}\) and \(\alpha\beta=\frac{3}{2}\), the positive difference is \(\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{2}\). Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). With \(\alpha+\beta=\frac{7}{2}\) and \(\alpha\beta=\frac{3}{2}\), the positive difference is \(\frac{5}{2}\).
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) लगाएँ। \(\alpha+\beta=\frac{7}{2}\) और \(\alpha\beta=\frac{3}{2}\), इसलिए धनात्मक अंतर \(\frac{5}{2}\) है।
The product is (a-2+3a+2=(a+1)(a+2)) and the sum is (2a+3). Hence the roots are (a+1) and (a+2).
Step 2
Why this answer is correct
The correct answer is A. (a+1) और (a+2) / (a+1) and (a+2). The product is (a-2+3a+2=(a+1)(a+2)) and the sum is (2a+3). Hence the roots are (a+1) and (a+2).
Step 3
Exam Tip
गुणनफल (a-2+3a+2=(a+1)(a+2)) है और योग (2a+3) है। इसलिए जड़ें (a+1) और (a+2) 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 \(\alpha+\beta=5\) and \(\alpha\beta=6\). Since \(\alpha^2+\beta^2=13\), the value is (13-4\(\alpha+\beta\)=13-20=-7).
Step 2
Why this answer is correct
The correct answer is A. (-7). Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). Since \(\alpha^2+\beta^2=13\), the value is (13-4\(\alpha+\beta\)=13-20=-7).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=6\) है। \(\alpha^2+\beta^2=13\), इसलिए (13-4\(\alpha+\beta\)=13-20=-7)।
A. (a=0) पर समान वास्तविक, अन्यथा वास्तविक नहीं/Equal real at (a=0), otherwise not real
Step 1
Concept
The discriminant is (D=4-4\(a^2+1\)=-4a-2). Thus (D=0) at (a=0), and (D<0) when \(a\neq0\).
Step 2
Why this answer is correct
The correct answer is A. (a=0) पर समान वास्तविक, अन्यथा वास्तविक नहीं / Equal real at (a=0), otherwise not real. The discriminant is (D=4-4\(a^2+1\)=-4a-2). Thus (D=0) at (a=0), and (D<0) when \(a\neq0\).
Step 3
Exam Tip
विविक्तकर (D=4-4\(a^2+1\)=-4a-2) है। इसलिए (a=0) पर (D=0), और \(a\neq0\) पर (D<0)।
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।
Here \(\alpha+\beta=-1\) and \(\alpha\beta=-1\). The power sums give \(S_1=-1\), \(S_2=3\), \(S_3=-4\), \(S_4=7\), \(S_5=-11\).
Step 2
Why this answer is correct
The correct answer is A. (-11). Here \(\alpha+\beta=-1\) and \(\alpha\beta=-1\). The power sums give \(S_1=-1\), \(S_2=3\), \(S_3=-4\), \(S_4=7\), \(S_5=-11\).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=-1\) और \(\alpha\beta=-1\) है। शक्ति योग क्रम से \(S_1=-1\), \(S_2=3\), \(S_3=-4\), \(S_4=7\), \(S_5=-11\) मिलता है।
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।
