KCET · Maths · Trigonometric Ratios & Identities
If \( x \) is real, then the minimum value of \( x^{2}-8 x+17 \) is
- A \( 11 \)
- B \( 02 \)
- C \( 03 \)
- D \( 04 \)
Answer & Solution
Correct Answer
(A) \( 11 \)
Step-by-step Solution
Detailed explanation
Given that, \(f(x)=x^{2}-8 x+17\)
So, \(f^{\prime}(x)=2 x-8\)
Now, \(f^{\prime}(x)=0\)
\(2 x-8=0 \Rightarrow x=4\)
Now, \(f^{\prime \prime}(x)=2>0\)
Therefore, \(f(4)=4^{2}-8(4)+17=1\)
So, \(f^{\prime}(x)=2 x-8\)
Now, \(f^{\prime}(x)=0\)
\(2 x-8=0 \Rightarrow x=4\)
Now, \(f^{\prime \prime}(x)=2>0\)
Therefore, \(f(4)=4^{2}-8(4)+17=1\)
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