Here \(\alpha+\beta=\frac{8}{3}\) and \(\alpha\beta=\frac{4}{3}\). Using (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)), we get \(\frac{224}{27}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{224}{27}\). Here \(\alpha+\beta=\frac{8}{3}\) and \(\alpha\beta=\frac{4}{3}\). Using (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)), we get \(\frac{224}{27}\).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{8}{3}\) और \(\alpha\beta=\frac{4}{3}\) है। (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)) से \(\frac{224}{27}\) मिलता है।
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 \(k^2\le25\)।
We use (\frac{1}{\alpha-2}+\frac{1}{\beta-2}=\frac{\alpha-2+\beta-2}{\(\alpha\beta\)2}). Since \(\alpha^2+\beta^2=30\) and (\(\alpha\beta\)2=9), the value is \(\frac{10}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{10}{3}\). We use (\frac{1}{\alpha-2}+\frac{1}{\beta-2}=\frac{\alpha-2+\beta-2}{\(\alpha\beta\)2}). Since \(\alpha^2+\beta^2=30\) and (\(\alpha\beta\)2=9), the value is \(\frac{10}{3}\).
Step 3
Exam Tip
(\frac{1}{\alpha-2}+\frac{1}{\beta-2}=\frac{\alpha-2+\beta-2}{\(\alpha\beta\)2}) होता है। \(\alpha^2+\beta^2=30\) और (\(\alpha\beta\)2=9), इसलिए मान \(\frac{10}{3}\) है।
Putting (x=6) gives (36-6(a+6)+6a=0). Hence (6) is always one root and the other root is (a).
Step 2
Why this answer is correct
The correct answer is B. (6) हमेशा जड़ है / (6) is always a root. Putting (x=6) gives (36-6(a+6)+6a=0). Hence (6) is always one root and the other root is (a).
Step 3
Exam Tip
(x=6) रखने पर (36-6(a+6)+6a=0) मिलता है। इसलिए (6) हमेशा एक जड़ है और दूसरी जड़ (a) 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।
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।
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।
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 \(\alpha+\beta=5\) 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)।
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 \(0<\lambda<1\)।
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।
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।
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।
