\( \lim _{n \rightarrow \infty}\left(\frac{1}{n^{3}+1}+\frac{4}{n^{3}+1}+\frac{9}{n^{3}+1}+\ldots .+\frac{n^{2}}{n^{3}+1}\right) \) is equal to (A) 1 (B) \( 2 / 3 \) (C) \( 1 / 3 \) (D) 0
Correct answer is:- (c) 29 Explanation:- 1/2-2 + 1/3-2 + 1/4-2 = 22 + 32 + 42 (x-1 = 1/x) = 4 + 9 + 16 = 29
2 Answers 1 viewsCorrect answer is (a) We know that propagation wave vector ∵ \(\vec{E} = \hat{k}\) \(\vec{B} = 2\hat{i} - 2\hat{j}\) \(\vec{C} = \vec{E} \times \vec{B}\) \(\vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1...
2 Answers 1 viewsCorrect answer is (b) We know that \(\tau\) = r × F ∵ \(\tau\) \(= \frac{dL}{dt},\) \(F = \frac{dP}{dt}\) \(\left(\frac{dL}{dt}\right) = r \times \left(\frac{dP}{dt}\right)\) \(\left(\frac{dL}{dt}\right) - r \times \left(\frac{dP}{dt}\right) = 0\)
2 Answers 1 viewsb^2 -4ac = 0 (as roots are equal ) b^2 =4ac b^2/4a = c
2 Answers 1 views1/1+n^3 +4/n^3+1 put n equal to 1/0 then 1/1+{1/0}^3 0/1 then answer is D
2 Answers 1 viewsGiven series is sin(p + x) sin(p + q) + x2/2! sin(p + 2q) + .... We know that eiα = cos α + i sin α. We develop new series cos p + x cos(p +...
2 Answers 1 views\(\lim\limits_{x \to π/4}\frac{sinx - cosx}{x - \frac{\pi}4}\) (0/0 type) \(=\lim\limits_{x \to \pi/4}\frac{cosx + sinx}1\) \(=cos\frac{\pi}4+sin\frac{\pi}4\) \(=\frac{1}{\sqrt2}+\frac1{\sqrt2}=\frac2{\sqrt2}\) = \(\sqrt2\)
2 Answers 1 views\(\lim\limits_{x \to 0} \frac{1-cos3x}{x^2}\) = \(\lim\limits_{x \to 0} \)\(\frac{1-(1-\frac{(3x)^2}{2!}+\frac{(3x)^4}{4!}- .....)}{x^2}\) (∵ cosx = 1 - \(\frac{x^2}{2!}+\frac{x^4}{4!}-...)\) = \(\lim\limits_{x \to 0} \) \(\frac{\frac{9x^2}{2}-\frac{81x^4}{24}+....}{x^2}\) = \(\lim\limits_{x \to 0} \) \((\frac{9}{2}-\frac{81}{24}x^2+....)\) = \(\frac{9}{2}\) (By taking limit)
2 Answers 1 views