समीकरण (x-2 -2(4t+1)x+\(7t^2+2t+5\)=0) के कोई वास्तविक मूल नहीं हैं। (t) के लिए सही अंतराल क्या है?
The equation (x-2 -2(4t+1)x+\(7t^2+2t+5\)=0) has no real roots. What is the correct interval for (t)?
#quadratic-equations
#no-real-roots
#parameter-interval
A \(-1<t<\frac{2}{9}\)
B (t<-1) या \(t>\frac{2}{9}\) / (t<-1) or \(t>\frac{2}{9}\)
C (t=-1) या \(t=\frac{2}{9}\) / (t=-1) or \(t=\frac{2}{9}\)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. \(-1<t<\frac{2}{9}\)
Step 1
Concept
Here (D=4(4t+1)2 -4\(7t^2+2t+5\)=36t-2 +24t-16). From (D<0), \(-1<t<\frac{2}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(-1<t<\frac{2}{9}\). Here (D=4(4t+1)2 -4\(7t^2+2t+5\)=36t-2 +24t-16). From (D<0), \(-1<t<\frac{2}{9}\).
Step 3
Exam Tip
यहाँ (D=4(4t+1)2 -4\(7t^2+2t+5\)=36t-2 +24t-16) है। (D<0) से \(-1<t<\frac{2}{9}\) मिलता है।
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समीकरण (x-2 +2(v+2)x+(4v+11)=0) के कोई वास्तविक मूल नहीं होने की सही शर्त क्या है?
What is the correct condition for (x-2 +2(v+2)x+(4v+11)=0) to have no real roots?
#quadratic-equations
#parameter-interval
#no-real-roots
A \(-\sqrt{7}<v<\sqrt{7}\)
B \(v<-\sqrt{7}\) या \(v>\sqrt{7}\) / \(v<-\sqrt{7}\) or \(v>\sqrt{7}\)
C \(v=-\sqrt{7}\) या \(v=\sqrt{7}\) / \(v=-\sqrt{7}\) or \(v=\sqrt{7}\)
D हर (v) / Every (v)
Explanation opens after your attempt
Correct Answer
A. \(-\sqrt{7}<v<\sqrt{7}\)
Step 1
Concept
Here (D=4(v+2)2 -4(4v+11)=4\(v^2-7\)). From (D<0), \(-\sqrt{7}<v<\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{7}<v<\sqrt{7}\). Here (D=4(v+2)2 -4(4v+11)=4\(v^2-7\)). From (D<0), \(-\sqrt{7}<v<\sqrt{7}\).
Step 3
Exam Tip
यहाँ (D=4(v+2)2 -4(4v+11)=4\(v^2-7\)) है। (D<0) से \(-\sqrt{7}<v<\sqrt{7}\)।
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यदि (x-2 +2(v+2)x+(4v+11)=0) के कोई वास्तविक मूल नहीं हों, तो (v) किस अंतराल में होगा?
If (x-2 +2(v+2)x+(4v+11)=0) has no real roots, in which interval will (v) lie?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-3<v<1)
B (v<-3) या (v>1) / (v<-3) or (v>1)
C (v=-3) या (v=1) / (v=-3) or (v=1)
D हर (v) / Every (v)
Explanation opens after your attempt
Correct Answer
A. (-3<v<1)
Step 1
Concept
Here (D=4(v+2)2 -4(4v+11)) must be expanded carefully. Always verify the middle term before solving the interval.
Step 2
Why this answer is correct
The correct answer is A. (-3<v<1). Here (D=4(v+2)2 -4(4v+11)) must be expanded carefully. Always verify the middle term before solving the interval.
Step 3
Exam Tip
यहाँ (D=4(v+2)2 -4(4v+11)=4\(v^2-7\)) नहीं, सही रूप (4\(v^2-7\)) नहीं है। परीक्षा में विस्तार सावधानी से करें।
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समीकरण (x-2 -2(3t+1)x+\(5t^2+2t+4\)=0) के कोई वास्तविक मूल नहीं हैं। (t) के लिए सही अंतराल क्या है?
The equation (x-2 -2(3t+1)x+\(5t^2+2t+4\)=0) has no real roots. What is the correct interval for (t)?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-2<t<1)
B (t<-2) या (t>1) / (t<-2) or (t>1)
C (t=-2) या (t=1) / (t=-2) or (t=1)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. (-2<t<1)
Step 1
Concept
Here (D=4(3t+1)2 -4\(5t^2+2t+4\)=16(t-1)(t+2)). From (D<0), (-2<t<1).
Step 2
Why this answer is correct
The correct answer is A. (-2<t<1). Here (D=4(3t+1)2 -4\(5t^2+2t+4\)=16(t-1)(t+2)). From (D<0), (-2<t<1).
Step 3
Exam Tip
यहाँ (D=4(3t+1)2 -4\(5t^2+2t+4\)=16(t-1)(t+2)) है। (D<0) से (-2<t<1)।
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समीकरण (x-2 +2(t+1)x+(3t+7)=0) के कोई वास्तविक मूल नहीं होने की सही शर्त क्या है?
What is the correct condition for (x-2 +2(t+1)x+(3t+7)=0) to have no real roots?
#quadratic-equations
#parameter-interval
#no-real-roots
A (-2<t<3)
B (-4<t<1)
C (t<-2) या (t>3) / (t<-2) or (t>3)
D (t=-2) या (t=3) / (t=-2) or (t=3)
Explanation opens after your attempt
Correct Answer
A. (-2<t<3)
Step 1
Concept
Here (D=4\(t^2-t-6\)=4(t-3)(t+2)). From (D<0), we get (-2<t<3).
Step 2
Why this answer is correct
The correct answer is A. (-2<t<3). Here (D=4\(t^2-t-6\)=4(t-3)(t+2)). From (D<0), we get (-2<t<3).
Step 3
Exam Tip
यहाँ (D=4\(t^2-t-6\)=4(t-3)(t+2)) है। (D<0) से (-2<t<3) मिलता है।
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यदि (x-2 +2(t+1)x+(3t+7)=0) के कोई वास्तविक मूल नहीं हों, तो (t) किस अंतराल में होगा?
If (x-2 +2(t+1)x+(3t+7)=0) has no real roots, in which interval will (t) lie?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-4<t<1)
B (t<-4) या (t>1) / (t<-4) or (t>1)
C (t=-4) या (t=1) / (t=-4) or (t=1)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. (-4<t<1)
Step 1
Concept
Here (D=4(t+1)2 -4(3t+7)=4\(t^2-t-6\)). For (D<0), factor again carefully before selecting the interval.
Step 2
Why this answer is correct
The correct answer is A. (-4<t<1). Here (D=4(t+1)2 -4(3t+7)=4\(t^2-t-6\)). For (D<0), factor again carefully before selecting the interval.
Step 3
Exam Tip
यहाँ (D=4(t+1)2 -4(3t+7)=4\(t^2-t-6\)) है। (D<0) से (-2<t<3) नहीं, गुणनखंड फिर से जाँचें।
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समीकरण (3x-2 -2(2k+1)x+(k+1)2 =0) के दो असमान वास्तविक मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for two distinct real roots of (3x-2 -2(2k+1)x+(k+1)2 =0)?
#quadratic-equations
#distinct-roots
#parameter-interval
A (k<-2) या (k>1) / (k<-2) or (k>1)
B (-2<k<1)
C (k=-2) या (k=1) / (k=-2) or (k=1)
D हर (k) / Every (k)
Explanation opens after your attempt
Correct Answer
A. (k<-2) या (k>1) / (k<-2) or (k>1)
Step 1
Concept
Here (D=4(k-1)(k+2)). From (D>0), we get (k<-2) or (k>1).
Step 2
Why this answer is correct
The correct answer is A. (k<-2) या (k>1) / (k<-2) or (k>1). Here (D=4(k-1)(k+2)). From (D>0), we get (k<-2) or (k>1).
Step 3
Exam Tip
यहाँ (D=4(k-1)(k+2)) है। (D>0) से (k<-2) या (k>1) मिलता है।
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समीकरण (x-2 -2(2t-1)x+\(t^2+2\)=0) के कोई वास्तविक मूल नहीं हैं। (t) के लिए सही अंतराल चुनिए।
The equation (x-2 -2(2t-1)x+\(t^2+2\)=0) has no real roots. Choose the correct interval for (t).
#quadratic-equations
#no-real-roots
#parameter-interval
A \(\frac{4-2\sqrt{6}}{3}<t<\frac{4+2\sqrt{6}}{3}\)
B \(t<\frac{4-2\sqrt{6}}{3}\) या \(t>\frac{4+2\sqrt{6}}{3}\) / \(t<\frac{4-2\sqrt{6}}{3}\) or \(t>\frac{4+2\sqrt{6}}{3}\)
C (t=1) मात्र / Only (t=1)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. \(\frac{4-2\sqrt{6}}{3}<t<\frac{4+2\sqrt{6}}{3}\)
Step 1
Concept
Here (D=4(2t-1)2 -4\(t^2+2\)=4\(3t^2-4t-1\)). From (D<0), the interval between the two boundary roots is obtained.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4-2\sqrt{6}}{3}<t<\frac{4+2\sqrt{6}}{3}\). Here (D=4(2t-1)2 -4\(t^2+2\)=4\(3t^2-4t-1\)). From (D<0), the interval between the two boundary roots is obtained.
Step 3
Exam Tip
यहाँ (D=4(2t-1)2 -4\(t^2+2\)=4\(3t^2-4t-1\)) है। (D<0) से दिए गए दोनों मूलों के बीच का अंतराल मिलता है।
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समीकरण (x-2 +2(k-1)x+(k+5)=0) के कोई वास्तविक मूल नहीं होने की सही शर्त क्या है?
What is the correct condition for (x-2 +2(k-1)x+(k+5)=0) to have no real roots?
#quadratic-equations
#parameter-interval
#no-real-roots
A (-1<k<4)
B (0<k<3)
C (k<-1) या (k>4) / (k<-1) or (k>4)
D (k=-1) या (k=4) / (k=-1) or (k=4)
Explanation opens after your attempt
Correct Answer
A. (-1<k<4)
Step 1
Concept
Here (D=4((k-1)2 -(k+5))=4\(k^2-3k-4\)). From (D<0), ((k-4)(k+1)<0), so (-1<k<4).
Step 2
Why this answer is correct
The correct answer is A. (-1<k<4). Here (D=4((k-1)2 -(k+5))=4\(k^2-3k-4\)). From (D<0), ((k-4)(k+1)<0), so (-1<k<4).
Step 3
Exam Tip
यहाँ (D=4((k-1)2 -(k+5))=4\(k^2-3k-4\)) है। (D<0) से ((k-4)(k+1)<0), इसलिए (-1<k<4)।
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यदि (x-2 +2(k-1)x+(k+5)=0) के कोई वास्तविक मूल नहीं हैं, तो (k) किस अंतराल में होगा?
If (x-2 +2(k-1)x+(k+5)=0) has no real roots, in which interval will (k) lie?
#quadratic-equations
#no-real-roots
#parameter-interval
A (0<k<3)
B \(k\leq0\) या \(k\geq3\) / \(k\leq0\) or \(k\geq3\)
C (k=0) या (k=3) / (k=0) or (k=3)
D हर (k) / Every (k)
Explanation opens after your attempt
Correct Answer
A. (0<k<3)
Step 1
Concept
Here (D=4(k-1)2 -4(k+5)=4\(k^2-3k-4\)). Use (D<0) and factor carefully before choosing the interval.
Step 2
Why this answer is correct
The correct answer is A. (0<k<3). Here (D=4(k-1)2 -4(k+5)=4\(k^2-3k-4\)). Use (D<0) and factor carefully before choosing the interval.
Step 3
Exam Tip
यहाँ (D=4(k-1)2 -4(k+5)=4\(k^2-3k-4\)) नहीं, सही सरल रूप (4\(k^2-3k-4\)) है। (D<0) से (0<k<3) नहीं मिलता, इसलिए गुणनखंड जाँचें।
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यदि (3x-2 +(k-2)x+4=0) के दो असमान वास्तविक मूल हों, तो (k) पर कौन सी शर्त सही है?
If (3x-2 +(k-2)x+4=0) has two distinct real roots, which condition on (k) is correct?
#quadratic-equations
#distinct-roots
#parameter-interval
A \(k<2-4\sqrt{3}\) या \(k>2+4\sqrt{3}\) / \(k<2-4\sqrt{3}\) or \(k>2+4\sqrt{3}\)
B \(2-4\sqrt{3}<k<2+4\sqrt{3}\)
C (k=2) मात्र / Only (k=2)
D \(k=4\sqrt{3}\) मात्र / Only \(k=4\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(k<2-4\sqrt{3}\) या \(k>2+4\sqrt{3}\) / \(k<2-4\sqrt{3}\) or \(k>2+4\sqrt{3}\)
Step 1
Concept
Here (D=(k-2)2 -48). For distinct real roots (D>0), so ((k-2)2 >48).
Step 2
Why this answer is correct
The correct answer is A. \(k<2-4\sqrt{3}\) या \(k>2+4\sqrt{3}\) / \(k<2-4\sqrt{3}\) or \(k>2+4\sqrt{3}\). Here (D=(k-2)2 -48). For distinct real roots (D>0), so ((k-2)2 >48).
Step 3
Exam Tip
यहाँ (D=(k-2)2 -48) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए ((k-2)2 >48)।
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समीकरण (x-2 +(k+2)x+9=0) में कोई वास्तविक मूल न होने के लिए (k) का अंतराल क्या है?
What is the interval of (k) for no real roots in (x-2 +(k+2)x+9=0)?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-8<k<4)
B (k<-8) या (k>4) / (k<-8) or (k>4)
C (k=-8) या (k=4) / (k=-8) or (k=4)
D (k=2) मात्र / Only (k=2)
Explanation opens after your attempt
Correct Answer
A. (-8<k<4)
Step 1
Concept
Here (D=(k+2)2 -36). From (D<0), we get (-8<k<4).
Step 2
Why this answer is correct
The correct answer is A. (-8<k<4). Here (D=(k+2)2 -36). From (D<0), we get (-8<k<4).
Step 3
Exam Tip
यहाँ (D=(k+2)2 -36) है। (D<0) से (-8<k<4) मिलता है।
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यदि (2x-2 +(k+1)x+3=0) के दो असमान वास्तविक मूल हों, तो (k) पर कौन सी शर्त सही है?
If (2x-2 +(k+1)x+3=0) has two distinct real roots, which condition on (k) is correct?
#quadratic-equations
#distinct-roots
#parameter-interval
A \(k<-1-2\sqrt{6}\) या \(k>-1+2\sqrt{6}\) / \(k<-1-2\sqrt{6}\) or \(k>-1+2\sqrt{6}\)
B \(-1-2\sqrt{6}<k<-1+2\sqrt{6}\)
C (k=-1) मात्र / Only (k=-1)
D \(k=2\sqrt{6}\) मात्र / Only \(k=2\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
A. \(k<-1-2\sqrt{6}\) या \(k>-1+2\sqrt{6}\) / \(k<-1-2\sqrt{6}\) or \(k>-1+2\sqrt{6}\)
Step 1
Concept
Here (D=(k+1)2 -24). For distinct real roots (D>0), so ((k+1)2 >24).
Step 2
Why this answer is correct
The correct answer is A. \(k<-1-2\sqrt{6}\) या \(k>-1+2\sqrt{6}\) / \(k<-1-2\sqrt{6}\) or \(k>-1+2\sqrt{6}\). Here (D=(k+1)2 -24). For distinct real roots (D>0), so ((k+1)2 >24).
Step 3
Exam Tip
यहाँ (D=(k+1)2 -24) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए ((k+1)2 >24)।
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समीकरण (x-2 +(k-3)x+4=0) में कोई वास्तविक मूल न होने के लिए (k) का अंतराल क्या है?
What is the interval of (k) for no real roots in (x-2 +(k-3)x+4=0)?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-1<k<7)
B (k<-1) या (k>7) / (k<-1) or (k>7)
C (k=-1) या (k=7) / (k=-1) or (k=7)
D (k=3) मात्र / Only (k=3)
Explanation opens after your attempt
Correct Answer
A. (-1<k<7)
Step 1
Concept
Here (D=(k-3)2 -16). From (D<0), we get (-1<k<7).
Step 2
Why this answer is correct
The correct answer is A. (-1<k<7). Here (D=(k-3)2 -16). From (D<0), we get (-1<k<7).
Step 3
Exam Tip
यहाँ (D=(k-3)2 -16) है। (D<0) से (-1<k<7) मिलता है।
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समीकरण (x-2 +2(a-2)x+2a+5=0) के वास्तविक मूल न होने के लिए (a) किस अंतराल में होगा?
In which interval will (a) lie for (x-2 +2(a-2)x+2a+5=0) to have no real roots?
#quadratic equations
#no real roots
#parameter interval
A \(3-\sqrt{10}<a<3+\sqrt{10}\)
B \(a<3-\sqrt{10}\) या \(a>3+\sqrt{10}\) / \(a<3-\sqrt{10}\) or \(a>3+\sqrt{10}\)
C \(a=3-\sqrt{10}\) या \(a=3+\sqrt{10}\) / \(a=3-\sqrt{10}\) or \(a=3+\sqrt{10}\)
D \(a>3+\sqrt{10}\) केवल / \(a>3+\sqrt{10}\) only
Explanation opens after your attempt
Correct Answer
A. \(3-\sqrt{10}<a<3+\sqrt{10}\)
Step 1
Concept
For no real roots, (D<0) is required. Here (D=4\(a^2-6a-1\)), so \(3-\sqrt{10}<a<3+\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(3-\sqrt{10}<a<3+\sqrt{10}\). For no real roots, (D<0) is required. Here (D=4\(a^2-6a-1\)), so \(3-\sqrt{10}<a<3+\sqrt{10}\).
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0) चाहिए। यहाँ (D=4\(a^2-6a-1\)), इसलिए \(3-\sqrt{10}<a<3+\sqrt{10}\)।
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समीकरण (x-2 +(m-2 )x+1=0) में वास्तविक मूल न होने के लिए (m) पर कौन सी शर्त सही है?
Which condition on (m) gives no real roots in (x-2 +(m-2 )x+1=0)?
#quadratic-equations
#no-real-roots
#parameter-interval
A (0<m<4)
B (m<0) या (m>4) / (m<0) or (m>4)
C (m=0) या (m=4) / (m=0) or (m=4)
D (m=2) नहीं / \(m\neq2\)
Explanation opens after your attempt
Correct Answer
A. (0<m<4)
Step 1
Concept
Here (D=(m-2 )2 -4), and we need (D<0). This gives (0<m<4).
Step 2
Why this answer is correct
The correct answer is A. (0<m<4). Here (D=(m-2 )2 -4), and we need (D<0). This gives (0<m<4).
Step 3
Exam Tip
यहाँ (D=(m-2 )2 -4) है और (D<0) चाहिए। इससे (0<m<4) मिलता है।
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समीकरण (x-2 -(k+1)x+4=0) में वास्तविक मूल न होने के लिए (k) किस अंतराल में होगा?
For (x-2 -(k+1)x+4=0), in which interval will (k) lie for no real roots?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-5<k<3)
B (k<-5) या (k>3) / (k<-5) or (k>3)
C (k=3) मात्र / Only (k=3)
D (k=-5) या (k=3) / (k=-5) or (k=3)
Explanation opens after your attempt
Correct Answer
A. (-5<k<3)
Step 1
Concept
Here (D=(k+1)2 -16), and no real roots need (D<0). This gives (-5<k<3).
Step 2
Why this answer is correct
The correct answer is A. (-5<k<3). Here (D=(k+1)2 -16), and no real roots need (D<0). This gives (-5<k<3).
Step 3
Exam Tip
यहाँ (D=(k+1)2 -16) है और वास्तविक मूल न होने के लिए (D<0) चाहिए। इससे (-5<k<3) मिलता है।
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