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।
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 is \(\frac{5}{4}\) and the product is \(\frac{3}{8}\). Multiply \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) by (8).
Step 2
Why this answer is correct
The correct answer is A. \(8x^2-10x+3=0\). The sum is \(\frac{5}{4}\) and the product is \(\frac{3}{8}\). Multiply \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) by (8).
Step 3
Exam Tip
जड़ों का योग \(\frac{5}{4}\) और गुणनफल \(\frac{3}{8}\) है। समीकरण \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) को (8) से गुणा करें।
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 \(k^2\le16\)।
C. एक धनात्मक और एक ऋणात्मक/One positive and one negative
Step 1
Concept
A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.
Step 2
Why this answer is correct
The correct answer is C. एक धनात्मक और एक ऋणात्मक / One positive and one negative. A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.
Step 3
Exam Tip
ऋणात्मक गुणनफल तभी मिलता है जब एक मूल धनात्मक और दूसरा ऋणात्मक हो। \(\alpha\beta<0\) संकेतों की जांच का छोटा संकेत है।
The monic equation is \(x^2-x-6=0\). Since the coefficient of \(x^2\) must be (2), multiply the whole equation by (2).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-2x-12=0\). The monic equation is \(x^2-x-6=0\). Since the coefficient of \(x^2\) must be (2), multiply the whole equation by (2).
Step 3
Exam Tip
मॉनिक समीकरण \(x^2-x-6=0\) है। \(x^2\) का गुणांक (2) चाहिए, इसलिए पूरे समीकरण को (2) से गुणा करें।
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।
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।
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।
After rationalising, the sum of roots is \(\frac{3}{2}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{3}{2}\). After rationalising, the sum of roots is \(\frac{3}{2}\). In \(x^2+ax+b=0\), the sum is (-a), so \(a=-\frac{3}{2}\).
Step 3
Exam Tip
रैशनलाइज करने पर जड़ों का योग \(\frac{3}{2}\) मिलता है। \(x^2+ax+b=0\) में जड़ों का योग (-a) होता है, इसलिए \(a=-\frac{3}{2}\)।
We have \(\tan\theta\cdot\cot\theta=1\) and \(\tan\theta+\cot\theta=s\). For real values, \(s^2-4\ge0\), so \(s^2\ge4\).
Step 2
Why this answer is correct
The correct answer is C. \(s^2\ge4\). We have \(\tan\theta\cdot\cot\theta=1\) and \(\tan\theta+\cot\theta=s\). For real values, \(s^2-4\ge0\), so \(s^2\ge4\).
Step 3
Exam Tip
\(\tan\theta\cdot\cot\theta=1\) और \(\tan\theta+\cot\theta=s\) है। वास्तविक मानों के लिए \(s^2-4\ge0\), इसलिए \(s^2\ge4\)।
The sum of these two roots is \(\frac{4t+2}{3}\), and the product is (\frac{t(t+2)}{3}). These match \(-\frac{b}{a}\) and \(\frac{c}{a}\) of the given equation.
Step 2
Why this answer is correct
The correct answer is C. हर (t) पर / For every (t). The sum of these two roots is \(\frac{4t+2}{3}\), and the product is (\frac{t(t+2)}{3}). These match \(-\frac{b}{a}\) and \(\frac{c}{a}\) of the given equation.
Step 3
Exam Tip
इन दोनों जड़ों का योग \(\frac{4t+2}{3}\) और गुणनफल (\frac{t(t+2)}{3}) है। ये दिए गए समीकरण के \(-\frac{b}{a}\) और \(\frac{c}{a}\) से मेल खाते हैं।
The sum (5) is positive and product (c>0) is needed for both roots. For real roots, \(25-4c\ge0\), so \(0<c\le\frac{25}{4}\).
Step 2
Why this answer is correct
The correct answer is B. \(0<c\le\frac{25}{4}\). The sum (5) is positive and product (c>0) is needed for both roots. For real roots, \(25-4c\ge0\), so \(0<c\le\frac{25}{4}\).
Step 3
Exam Tip
योग (5) धनात्मक है और दोनों जड़ों के लिए गुणनफल (c>0) चाहिए। वास्तविक जड़ों के लिए \(25-4c\ge0\), इसलिए \(0<c\le\frac{25}{4}\)।
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)।
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।
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।
