If the roots of the equation \( a x^{2}+b x+c=0 \) are equal then \( c=? \) \( 58 \quad \) 1) \( -\frac{b}{2 a} \) 2) \( \frac{b}{2 a} \) 3) \( -\frac{b^{2}}{4 a} \) 4) \( \frac{b^{2}}{4 a} \)
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 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(1) f(x) = \(\frac{x}{1-x^3}\) f(-x) = \(\frac{-x}{1-(-x)^3}\) = \(\frac{-x}{1+x^3}\) ≠ -f(x) or f(x) ∴ f(x) is neither even nor odd function. (2) f(x) = \(\frac{x^2}{1+x}\) ∴ f(-x) = \(\frac{(-x^2)}{1+(-x)}\) = \(\frac{x^2}{1-x}\) ≠ -f(x) or f(x). (3) f(x) = x-|x| ∴ f(-x) = -x-|-x| = -x-|x| = -(x+|x|) ≠ -f(x)...
2 Answers 1 viewsGiven differential equation is, \(\frac{d^2y}{dx^2}+\frac{dy}{dx}+x=\sqrt{1+\frac{d^2y}{dx^2}}\) By Squaring both sides, we get \((\frac{d^2y}{dx^2}+\frac{dy}{dx}+x)^2=1+\frac{d^2y}{dx^2}\) \(\Rightarrow\) \((\frac{d^2y}{dx^2})^ 2-\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2+2\,\frac{d^2y}{dx^2}\frac{dy}{dx}+2x\frac{d^2y}{dx^2}+2x\frac{dy}{dx}+x^2=1\) which is differential equation of order 2 and degree 2.
2 Answers 1 views