We will will looking at the types of numbers like natural numbers, whole numbers, integers, prime and composite numbers etc. and basic concepts of number system. Let’s begin – Concepts of Number System Natural Numbers and Whole Numbers Natural numbers, also called counting numbers, are the first numbers we learnt. They are the number set […]
Concepts of Number System Read More »
Solution : We know that the vector equation of line passing through two points with position vectors \(\vec{a}\) and \(\vec{b}\) is, \(\vec{r}\) = \(\lambda\) \((\vec{b} – \vec{a})\) Here \(\vec{a}\) = \(3\hat{i} + 4\hat{j} – 7\hat{k}\) and \(\vec{b}\) = \(\hat{i} – \hat{j} + 6\hat{k}\). So, the vector equation of the required line is \(\vec{r}\) = (\(3\hat{i}
Solution : We have, \(\vec{a}\) = \(4\hat{i} – 3\hat{j} + 5\hat{k}\) and \(\vec{b}\) = \(3\hat{i} + 4\hat{j} + 5\hat{k}\) Let \(\theta\) is the angle between the given vectors. Then, cos\(\theta\) = \(\vec{a}.\vec{b}\over |\vec{a}||\vec{b}|\) \(\implies\) cos\(\theta\) = \(12 – 12 + 25\over \sqrt{16 + 9 + 25} \sqrt{16 + 9 + 25}\) = \(1\over 2\) \(\implies\)
Solution : We have \(\vec{a}\) = \(2\hat{i}+2\hat{j}-\hat{k}\) and \(\vec{b}\) = \(6\hat{i}-3\hat{j}+2\hat{k}\) \(\vec{a}\).\(\vec{b}\) = (\(2\hat{i}+2\hat{j}-\hat{k}\)).(\(6\hat{i}-3\hat{j}+2\hat{k}\)) = (2)(6) + (2)(-3) + (-1)(2) = 12 – 6 – 2 = 4 Similar Questions Find the angle between the vectors with the direction ratios proportional to 4, -3, 5 and 3, 4, 5. Find the vector equation of a
Solution : We have 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\) = 1 + \(1\over sqrt{2}\)\((cos\theta + cos\theta)\) + \(\sqrt{2}\)\((cos\theta + cos\theta)\) = 1 + \(({1\over \sqrt{2}} + \sqrt{2})\) + \((cos\theta + cos\theta)\) = 1 + \(({1\over \sqrt{2}} + \sqrt{2})\).\(\sqrt{2}cos(\theta – {pi\over 4})\) \(\therefore\) Maximum Value = 1 + \(({1\over \sqrt{2}}
Solution : The expression = (sin78 – sin42) – (sin66 – sin6) = 2cos(60)sin(18) – 2cos36.sin30 = sin18 – cos36 = \(({\sqrt{5} – 1\over 4})\) – \(({\sqrt{5} + 1\over 4})\) = -\(1\over 2\) Similar Questions Find the maximum value of 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\) If A + B
Evaluate sin78 – sin66 – sin42 + sin6. Read More »
Solution : Here, tanx + secx = 2cosx \(\implies\) sinx + 1 = \(2cos^2x\) \(\implies\) \(2sin^2x\) + sinx – 1 = 0 \(\implies\) sinx = \(1\over 2\), -1 But sinx = -1 \(\implies\) x = \(3\pi\over 2\) for which tanx + secx = 2cosx is not defined. Thus, sinx =
Find the number of solutions of tanx + secx = 2cosx in [0, \(2\pi\)]. Read More »
Solution : Since, \({1\over 6}sin\theta\), \(cos\theta\) and \(tan\theta\) are in G.P. \(\implies\) \(cos^2\theta\) = \({1\over 6}sin\theta\).\(cos\theta\) \(\implies\) \(6cos^3\theta\) + \(cos^2\theta\) – 1 = 0 \(\therefore\) (\(2cos\theta – 1\))(\(3cos^2\theta\) + \(2cos\theta\) + 1) = 0 \(cos\theta\) = \(1\over 2\) (other values are imaginary) \(cos\theta\) = \(cos\pi\over 3\) \(\theta\) = \(2n\pi \pm {\pi\over 3}\), n
Solution : \(2log{2\over 5}\) + \(3log{25\over 8}\) + \(log{128\over 625}\) = \(log{2^2\over 5^2}\) + \(log({5^2\over 2^3})^3\) + \(log{2^7\over 5^4}\) = \(log({2^2\over 5^2}{5^6\over 2^9}{2^7\over 5^4})\) = log 1 = 0 Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Evaluate the given log : \(81^{l\over {log_5 3}}\) + \(27^{log_9 36}\) + \(3^{4\over
Find the value of \(2log{2\over 5}\) + \(3log{25\over 8}\) – \(log{625\over 128}\). Read More »
Solution : We have \(2^{x + 2}\) > \(2^{2/x}\). Since the base 2 > 1, we have x + 2 > -\(2\over x\) (the sign of the inequality is retained) Now x + 2 + \(2\over x\) \(\implies\) \(x^2 + 2x + 2\over x\) > 0 \(\implies\) \((x + 1)^2 + 1\over x\) > 0
Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Read More »
