A. \(\alpha\leq3\) और \(\alpha\neq-3\)/\(\alpha\leq3\) and \(\alpha\neq-3\)
Step 1
Concept
Here (D=4\alpha-2-4\(\alpha+3\)\(\alpha-2\)=24-4\alpha). For real roots \(\alpha\leq3\), and for a quadratic \(\alpha\neq-3\).
Step 2
Why this answer is correct
The correct answer is A. \(\alpha\leq3\) और \(\alpha\neq-3\) / \(\alpha\leq3\) and \(\alpha\neq-3\). Here (D=4\alpha-2-4\(\alpha+3\)\(\alpha-2\)=24-4\alpha). For real roots \(\alpha\leq3\), and for a quadratic \(\alpha\neq-3\).
Step 3
Exam Tip
यहाँ (D=4\alpha-2-4\(\alpha+3\)\(\alpha-2\)=24-4\alpha) है। वास्तविक मूलों के लिए \(\alpha\leq3\) और द्विघात के लिए \(\alpha\neq-3\)।
A. \(\theta<0\) या \(\theta>3\)/\(\theta<0\) or \(\theta>3\)
Step 1
Concept
Here (D=4\theta-2-12\theta=4\theta\(\theta-3\)). From (D>0), \(\theta<0\) or \(\theta>3\).
Step 2
Why this answer is correct
The correct answer is A. \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\). Here (D=4\theta-2-12\theta=4\theta\(\theta-3\)). From (D>0), \(\theta<0\) or \(\theta>3\).
Step 3
Exam Tip
यहाँ (D=4\theta-2-12\theta=4\theta\(\theta-3\)) है। (D>0) से \(\theta<0\) या \(\theta>3\)।
A. \(p\leq5\) और \(p\neq2\)/\(p\leq5\) and \(p\neq2\)
Step 1
Concept
Here (D=4(p+2)2-4(p-2)(p+6)=40-8p). For real roots \(p\leq5\), and for a quadratic \(p\neq2\).
Step 2
Why this answer is correct
The correct answer is A. \(p\leq5\) और \(p\neq2\) / \(p\leq5\) and \(p\neq2\). Here (D=4(p+2)2-4(p-2)(p+6)=40-8p). For real roots \(p\leq5\), and for a quadratic \(p\neq2\).
Step 3
Exam Tip
यहाँ (D=4(p+2)2-4(p-2)(p+6)=40-8p) है। वास्तविक मूलों के लिए \(p\leq5\) और द्विघात के लिए \(p\neq2\)।
A. \(\alpha\leq2\) और \(\alpha\neq-2\)/\(\alpha\leq2\) and \(\alpha\neq-2\)
Step 1
Concept
Here (D=4\alpha-2-4\(\alpha+2\)\(\alpha-1\)=8-4\alpha). For real roots \(\alpha\leq2\), and for a quadratic \(\alpha\neq-2\).
Step 2
Why this answer is correct
The correct answer is A. \(\alpha\leq2\) और \(\alpha\neq-2\) / \(\alpha\leq2\) and \(\alpha\neq-2\). Here (D=4\alpha-2-4\(\alpha+2\)\(\alpha-1\)=8-4\alpha). For real roots \(\alpha\leq2\), and for a quadratic \(\alpha\neq-2\).
Step 3
Exam Tip
यहाँ (D=4\alpha-2-4\(\alpha+2\)\(\alpha-1\)=8-4\alpha) है। वास्तविक मूलों के लिए \(\alpha\leq2\) और द्विघात के लिए \(\alpha\neq-2\)।
A. \(p\leq2\) और \(p\neq1\)/\(p\leq2\) and \(p\neq1\)
Step 1
Concept
Here (D=4(p+1)2-4(p-1)(p+3)=16-4p). For real roots \(p\leq2\), and for a quadratic \(p\neq1\).
Step 2
Why this answer is correct
The correct answer is A. \(p\leq2\) और \(p\neq1\) / \(p\leq2\) and \(p\neq1\). Here (D=4(p+1)2-4(p-1)(p+3)=16-4p). For real roots \(p\leq2\), and for a quadratic \(p\neq1\).
Step 3
Exam Tip
यहाँ (D=4(p+1)2-4(p-1)(p+3)=16-4p) है। वास्तविक मूलों के लिए \(p\leq2\) और द्विघात के लिए \(p\neq1\)।
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 \(\lambda>1\)।
Here (D=(2k)2-4(k-2)(k+3)=4(6-k)). For real roots we need \(k\leq6\), so check simplification carefully.
Step 2
Why this answer is correct
The correct answer is A. \(k\geq\frac{3}{2}\). Here (D=(2k)2-4(k-2)(k+3)=4(6-k)). For real roots we need \(k\leq6\), so check simplification carefully.
Step 3
Exam Tip
यहाँ (D=(2k)2-4(k-2)(k+3)=4(6-k)) नहीं, सही सरल रूप (4(6-k)) है। वास्तविक मूलों के लिए \(k\leq6\) चाहिए।
A. \(r^2<\frac{25}{4}\) और \(r\neq0\)/\(r^2<\frac{25}{4}\) and \(r\neq0\)
Step 1
Concept
For two distinct real roots, (D>0), so \(25-4r^2>0\). Also \(r\neq0\) is needed because the equation must remain quadratic.
Step 2
Why this answer is correct
The correct answer is A. \(r^2<\frac{25}{4}\) और \(r\neq0\) / \(r^2<\frac{25}{4}\) and \(r\neq0\). For two distinct real roots, (D>0), so \(25-4r^2>0\). Also \(r\neq0\) is needed because the equation must remain quadratic.
Step 3
Exam Tip
दो भिन्न वास्तविक मूलों के लिए (D>0), इसलिए \(25-4r^2>0\)। साथ में \(r\neq0\) चाहिए क्योंकि समीकरण द्विघात होना चाहिए।
A. \(k\leq-3\) या \(k\geq1\)/\(k\leq-3\) or \(k\geq1\)
Step 1
Concept
Here (D=4(k+1)2-4(k+5)). \(D\geq0\) gives \(k^2+k-4\geq0\), so solve the resulting inequality carefully.
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-3\) या \(k\geq1\) / \(k\leq-3\) or \(k\geq1\). Here (D=4(k+1)2-4(k+5)). \(D\geq0\) gives \(k^2+k-4\geq0\), so solve the resulting inequality carefully.
Step 3
Exam Tip
यहाँ (D=4(k+1)2-4(k+5)) है। \(D\geq0\) से \(k^2+k-4\geq0\) नहीं, सही सरल रूप \(k^2+k-4\geq0\) देता है।
A. \(k\leq\frac{-1-2\sqrt{10}}{2}\) या \(k\geq\frac{-1+2\sqrt{10}}{2}\)/\(k\leq\frac{-1-2\sqrt{10}}{2}\) or \(k\geq\frac{-1+2\sqrt{10}}{2}\)
Step 1
Concept
For real roots, ((2k+1)2-40\geq0) is needed. Hence \(2k+1\leq-2\sqrt{10}\) or \(2k+1\geq2\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq\frac{-1-2\sqrt{10}}{2}\) या \(k\geq\frac{-1+2\sqrt{10}}{2}\) / \(k\leq\frac{-1-2\sqrt{10}}{2}\) or \(k\geq\frac{-1+2\sqrt{10}}{2}\). For real roots, ((2k+1)2-40\geq0) is needed. Hence \(2k+1\leq-2\sqrt{10}\) or \(2k+1\geq2\sqrt{10}\).
Step 3
Exam Tip
वास्तविक मूलों के लिए ((2k+1)2-40\geq0) चाहिए। इसलिए \(2k+1\leq-2\sqrt{10}\) या \(2k+1\geq2\sqrt{10}\)।
A. \(k\leq-\sqrt{10}\) या \(k\geq\sqrt{10}\)/\(k\leq-\sqrt{10}\) or \(k\geq\sqrt{10}\)
Step 1
Concept
Here (D=(2k)2-4(5)(2)=4\(k^2-10\)). From \(D\geq0\), we get \(k^2\geq10\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-\sqrt{10}\) या \(k\geq\sqrt{10}\) / \(k\leq-\sqrt{10}\) or \(k\geq\sqrt{10}\). Here (D=(2k)2-4(5)(2)=4\(k^2-10\)). From \(D\geq0\), we get \(k^2\geq10\).
Step 3
Exam Tip
यहाँ (D=(2k)2-4(5)(2)=4\(k^2-10\)) है। \(D\geq0\) से \(k^2\geq10\) मिलता है।
A. \(k\leq-3\) या \(k\geq3\)/\(k\leq-3\) or \(k\geq3\)
Step 1
Concept
For real roots, \(D\geq0\) is needed. Here \(4k^2-36\geq0\) gives \(k\leq-3\) or \(k\geq3\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-3\) या \(k\geq3\) / \(k\leq-3\) or \(k\geq3\). For real roots, \(D\geq0\) is needed. Here \(4k^2-36\geq0\) gives \(k\leq-3\) or \(k\geq3\).
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\geq0\) चाहिए। यहाँ \(4k^2-36\geq0\) से \(k\leq-3\) या \(k\geq3\) मिलता है।
A. \(k\leq-1\) या \(k\geq4\)/\(k\leq-1\) or \(k\geq4\)
Step 1
Concept
Here (D=4(k-1)2-4(k+2)). From \(D\geq0\), \(k^2-3k-4\geq0\), so \(k\leq-1\) or \(k\geq4\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-1\) या \(k\geq4\) / \(k\leq-1\) or \(k\geq4\). Here (D=4(k-1)2-4(k+2)). From \(D\geq0\), \(k^2-3k-4\geq0\), so \(k\leq-1\) or \(k\geq4\).
Step 3
Exam Tip
यहाँ (D=4(k-1)2-4(k+2)) है। \(D\geq0\) से \(k^2-3k-4\geq0\), इसलिए \(k\leq-1\) या \(k\geq4\)।
A. \(k\leq\frac{1-2\sqrt{3}}{2}\) या \(k\geq\frac{1+2\sqrt{3}}{2}\)/\(k\leq\frac{1-2\sqrt{3}}{2}\) or \(k\geq\frac{1+2\sqrt{3}}{2}\)
Step 1
Concept
For real roots, ((2k-1)2-12\geq0) is needed. Hence \(2k-1\leq-2\sqrt{3}\) or \(2k-1\geq2\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq\frac{1-2\sqrt{3}}{2}\) या \(k\geq\frac{1+2\sqrt{3}}{2}\) / \(k\leq\frac{1-2\sqrt{3}}{2}\) or \(k\geq\frac{1+2\sqrt{3}}{2}\). For real roots, ((2k-1)2-12\geq0) is needed. Hence \(2k-1\leq-2\sqrt{3}\) or \(2k-1\geq2\sqrt{3}\).
Step 3
Exam Tip
वास्तविक मूलों के लिए ((2k-1)2-12\geq0) चाहिए। इसलिए \(2k-1\leq-2\sqrt{3}\) या \(2k-1\geq2\sqrt{3}\)।
A. \(k\leq-\frac{3}{2}\) या \(k\geq\frac{3}{2}\)/\(k\leq-\frac{3}{2}\) or \(k\geq\frac{3}{2}\)
Step 1
Concept
Here (D=(4k)2-4(4)(9)=16\(k^2-9\)). For real roots \(k^2\geq9\), so \(k\leq-\frac{3}{2}\) or \(k\geq\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-\frac{3}{2}\) या \(k\geq\frac{3}{2}\) / \(k\leq-\frac{3}{2}\) or \(k\geq\frac{3}{2}\). Here (D=(4k)2-4(4)(9)=16\(k^2-9\)). For real roots \(k^2\geq9\), so \(k\leq-\frac{3}{2}\) or \(k\geq\frac{3}{2}\).
Step 3
Exam Tip
यहाँ (D=(4k)2-4(4)(9)=16\(k^2-9\)) है। वास्तविक मूलों के लिए \(k^2\geq9\) यानी \(k\leq-\frac{3}{2}\) या \(k\geq\frac{3}{2}\)।