By considering 1^{prime} = 0.0175 radians,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $f(x) = \cot x$. Given $1^{\prime} = 0.0175$ radians,then $2^{\prime} = 0.035$ radians. The derivative is $f^{\prime}(x) = -\operatorname{cose

Vedclass · Accepted Answer

$0.93$

Vedclass · Answer

(D) Let $f(x) = \cot x$. Given $1^{\prime} = 0.0175$ radians,then $2^{\prime} = 0.035$ radians. The derivative is $f^{\prime}(x) = -\operatorname{cosec}^2 x$. Using the linear approximation formula $f(a+h) \approx f(a) + h f^{\prime}(a)$: $f(45^{\circ} + 2^{\prime}) \approx \cot(45^{\circ}) + (0.035) 	imes (-\operatorname{cosec}^2(45^{\circ}))$. Since $\cot(45^{\circ}) = 1$ and $\operatorname{cosec}(45^{\circ}) = \sqrt{2}$,then $\operatorname{cosec}^2(45^{\circ}) = 2$. $f(45^{\circ} + 2^{\prime}) \approx 1 - 0.035 	imes 2 = 1 - 0.07 = 0.93$.

By considering $1^{\prime} = 0.0175$ radians,the approximate value of $\cot 45^{\circ} 2^{\prime}$ is

Similar Questions

The approximate value of $\sqrt[3]{28}$ rounded up to $3$ decimal places is

The nearest approximate value of $\sqrt{2023}$ is

The approximate value of $\sqrt[3]{0.026}$ is

Find the approximate value of $f(2.01),$ where $f(x)=4x^{2}+5x+2$.

The approximate value of $\log _{10} 998$ is (given that $\log _{10} e=0.4343$ )

Vedclass Test Series

Exam Paper Generator

Online Exam Module