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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Functions

NDA 2022 · Mathematics · Functions

What is the minimum value of the function f(x)= _ 10 (x^2+2 x+11) ?

What is the minimum value of the function f(x)=log⁡10(x2+2x+11)?f(x)=\log _{10}\left(x^2+2 x+11\right) ?

Options

  1. A00
  2. B11
  3. C22
  4. D1010

Correct answer

B. 11

Step-by-step solution

 Since, ,f(x)=log⁡10(x2+2x+11)∴f′(x)=2x+2x2+2x+11log⁡10e=0⇒x=−1( critical point )f′′(x)=2(x2+2x+11)−(2x+2)2(x2+2x+11)2⋅log⁡10e∵f′′(−1)>0∴ Minimum value of  f(x)f(−1)=log⁡1010=1\begin{aligned} & \text{ Since, }, f(x)=\log _{10}\left(x^2+2 x+11\right) \\ & \therefore f^{\prime}(x)=\frac{2 x+2}{x^2+2 x+11} \log _{10}^e=0 \\ & \Rightarrow x=-1(\text{ critical point }) \\ & f^{\prime \prime}(x)=\frac{2\left(x^2+2 x+11\right)-(2 x+2)^2}{\left(x^2+2 x+11\right)^2} \cdot \log _{10}^e \\ & \because f^{\prime \prime}(-1) > 0 \\ & \therefore \quad \text{ Minimum value\text{ of } } f(x) \\ & f(-1)=\log _{10} 10=1\end{aligned}
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