JEE Advanced · Mathematics · 2. Quadratic Equations
The smallest value of \(k\), for which both the roots of the equation \(x^2-8 k x+16\left(k^2-k+1\right)=0\) are real, distinct and have values atleast 4 , is
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
(i) Given, \(x^2-8 k x+16\left(k^2-k+1\right)=0\)
Now, \(\quad D=64\left\{k^2-\left(k^2-k+1\right)\right\}\)
\(=64(k-1)>0 \)
\( \therefore k>1 \)
\( 16-32 k+16\left(k^2-k+1\right) \geq 0 \)
\( \Rightarrow k^2-3 k+2 \geq 0 \)
\( \Rightarrow(k-2)(k-1) \geq 0 \Rightarrow k \leq 1\)\(\text { or } k \geq 2\)
Hence, \(k=2\)
Now, \(\quad D=64\left\{k^2-\left(k^2-k+1\right)\right\}\)
\(=64(k-1)>0 \)
\( \therefore k>1 \)
\( 16-32 k+16\left(k^2-k+1\right) \geq 0 \)
\( \Rightarrow k^2-3 k+2 \geq 0 \)
\( \Rightarrow(k-2)(k-1) \geq 0 \Rightarrow k \leq 1\)\(\text { or } k \geq 2\)
Hence, \(k=2\)
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