MHT CET · Maths · Application of Derivatives
The approximate value of \(\cot ^{-1}(1 \cdot 001)\) is
- A \(\frac{\pi}{4}-0 \cdot 0005\)
- B \(\frac{\pi}{4}+0 \cdot 005\)
- C \(\frac{\pi}{4}+0 \cdot 0005\)
- D \(\frac{\pi}{4}-0 \cdot 005\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{4}-0 \cdot 0005\)
Step-by-step Solution
Detailed explanation
Let \(f(x)=\cos ^{-1} x \Rightarrow f^{\prime}(x)=\frac{-1}{1+x^{2}}\)
Let, \(\mathrm{a}=1, \mathrm{~h}=0.001\)
Now \(\mathrm{f}(\mathrm{a})=\cot ^{-1} 1=\frac{\pi}{4}\) and \(\mathrm{f}^{\prime}(\mathrm{a})=\frac{-1}{1+1}=-\frac{1}{2}\)
We know that, \(f(a+h) \doteqdot f(a)+h f^{\prime}(a)\)
\(=\frac{\pi}{4}+(0.001)\left(-\frac{1}{2}\right)=\frac{\pi}{4}-0.0005\)
Let, \(\mathrm{a}=1, \mathrm{~h}=0.001\)
Now \(\mathrm{f}(\mathrm{a})=\cot ^{-1} 1=\frac{\pi}{4}\) and \(\mathrm{f}^{\prime}(\mathrm{a})=\frac{-1}{1+1}=-\frac{1}{2}\)
We know that, \(f(a+h) \doteqdot f(a)+h f^{\prime}(a)\)
\(=\frac{\pi}{4}+(0.001)\left(-\frac{1}{2}\right)=\frac{\pi}{4}-0.0005\)
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