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) \( \frac{b^{2}}{4 a} \)
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4. The value of \( \left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2} \) is (a) \( \frac{61}{144} \) (b) \( \frac{144}{61} \) (c) 29 (d) none of these
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
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12 In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by \( \hat{ k } \) and \( 2 \hat{ i }-2 \hat{ j } \), respectively. What is the unit vector along direction of propagation of the wave? (a) \( \frac{1}{\sqrt{2}}(\hat{ i }+\hat{ j }) \) (b) \( \frac{1}{\sqrt{2}}(\hat{ j }+\hat{ k }) \) (c) \( \frac{1}{\sqrt{5}}(\hat{ i }+2 \hat{ j }) \) (d) \( \frac{1}{\sqrt{5}}(2 \hat{ i }+\hat{ j }) \)
Correct 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...
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Consider a particle of mass \( m \) having linear momentum \( P \) at position \( r \) relative to the origin \( O \). Which of the following equations correctly relates \( r , P , L \) ? a) \( \left[\left(\frac{d L}{d t}\right)+r x\left(\frac{d P}{d t}\right)\right]=0 \) b) \( \left[\left(\frac{d L}{d t}\right)-r x\left(\frac{d P}{d t}\right)\right]=0 \) c) \( \left[\left(\frac{d L}{d t}\right) \times\left(\frac{d r}{d t}\right) \times P\right]=0 \) d) \( \left[\left(\frac{d L}{d t}\right)-\left(\frac{d r}{d t}\right) \times P\right]=0 \)
Correct answer is (b) We know that \(\tau\) = r × F ∵ \(\tau\) \(= \frac{dL}{dt},\) \(F = \frac{dP}{dt}\) \(\left(\frac{dL}{dt}\right) = r \times \left(\frac{dP}{dt}\right)\) \(\left(\frac{dL}{dt}\right) - r \times \left(\frac{dP}{dt}\right) = 0\)
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\( \lim _{n \rightarrow \infty}\left(\frac{1}{n^{3}+1}+\frac{4}{n^{3}+1}+\frac{9}{n^{3}+1}+\ldots .+\frac{n^{2}}{n^{3}+1}\right) \) is equal to (A) 1 (B) \( 2 / 3 \) (C) \( 1 / 3 \) (D) 0
1/1+n^3 +4/n^3+1 put n equal to 1/0 then 1/1+{1/0}^3 0/1 then answer is D
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Evaluate : \( \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}} \)
\(\lim\limits_{x \to π/4}\frac{sinx - cosx}{x - \frac{\pi}4}\) (0/0 type) \(=\lim\limits_{x \to \pi/4}\frac{cosx + sinx}1\) \(=cos\frac{\pi}4+sin\frac{\pi}4\) \(=\frac{1}{\sqrt2}+\frac1{\sqrt2}=\frac2{\sqrt2}\) = \(\sqrt2\)
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Use the definition of Odd and Even functions to determine whether each of the following functions is Odd, Even, or neither. Must show your work for credit II! 1) \( f(x)=\frac{x}{1-x^{3}} \) 2) \( f(x)=\frac{x^{2}}{1+x} \) 3) \( \quad f(x)=x-|x| \)
(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)...
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The order and degree the differential equation \( \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{2} y}{d x^{2}}} \) are pespectively.
Given 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.
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