TS EAMCET · Maths · Quadratic Equation
\(f(x)=x^2-2(4 \mathrm{~K}-1) x+g(\mathrm{~K})>0 \forall x \in \mathbb{R}\) and for \(\mathrm{K} \in(\mathrm{a}, \mathrm{b})\). If \(g(\mathrm{~K})=15 \mathrm{~K}^2-2 \mathrm{~K}-7\), then
- A \(g(\mathrm{~K})\) attains its maximum at the midpoint of \((\mathrm{a}, \mathrm{b})\)
- B \(g(\mathrm{~K})\) attains its minimum at two points in \((\mathrm{a}, \mathrm{b})\)
- C \(g(\mathrm{~K})\) attains its both maximum and minimum in \((\mathrm{a}, \mathrm{b})\)
- D \(g(\mathrm{~K})\) attains no maximum and no minimum in \((\mathrm{a}, \mathrm{b})\)
Answer & Solution
Correct Answer
(D) \(g(\mathrm{~K})\) attains no maximum and no minimum in \((\mathrm{a}, \mathrm{b})\)
Step-by-step Solution
Detailed explanation
\(\Delta \([-2(4K-1)]^2 - 4(1)(15K^2-2K-7) \(4(16K^2-8K+1) - 60K^2+8K+28 \(64K^2-32K+4 - 60K^2+8K+28 \(4K^2-24K+32 \(K^2-6K+8 \((K-2)(K-4) \(2 \(g(K) = 15K^2-2K-7\) Vertex of \(g(K)\) is at \(K = -(-2)/(2 \times 15) = 1/15\). Since \(1/15 \notin (2,4)\) and \(1/15…
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