Evaluate : \( \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}} \)
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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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If the roots of the equation \( a x^{2}+b x+c=0 \) are equal then \( c=? \) \( \quad \) 1) \( -\frac{b}{2 a} \) 2) \( \frac{b}{2 a} \) 3) \( -\frac{b^{2}}{4 a} \) 4) \( \frac{b^{2}}{4 a} \)
b^2 -4ac = 0 (as roots are equal ) b^2 =4ac b^2/4a = c
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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 : \[ \text { (i) } \lim _{x \rightarrow 0} \frac{1-\cos 3 x}{x^{2}} \] without L Hospital rule
\(\lim\limits_{x \to 0} \frac{1-cos3x}{x^2}\) = \(\lim\limits_{x \to 0} \)\(\frac{1-(1-\frac{(3x)^2}{2!}+\frac{(3x)^4}{4!}- .....)}{x^2}\) (∵ 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)
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39) Simplify: \( \frac{a^{2}-16}{a^{3}-8} \times \frac{2 a^{2}-3 a-2}{2 a^{2}+9 a+4} \div \frac{3 a^{2}-11 a-4}{a^{2}+2 a+4} \)
\(\frac{a^2+6}{a^3-8}\) x \(\frac{2a^2-3a-2}{2a^2+9a+4}\) ÷ \(\frac{3a^2-11a-4}{a^2+2a+4}\) = \(\frac{(a+4)(a-4)}{(a-2)(a^2+2a+4)}\) x \(\frac{(2a+1)(a-2)}{(2a+1)(a+4)}\) x \(\frac{(a+2)^2}{(3a+1)(a-4)}\) = \(\frac{(a+2)^2}{(a^2+2a+4)(3a+1)}\)
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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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