Solution : The equation of the two curves are xy = 6ย ย ย ย ย โฆโฆ.(i) and, \(x^2 y\) = 12ย ย ย ย ย ย โฆโฆโฆโฆ(ii) from (i) , we obtain y = \(6\over x\). Putting this value of y in (ii), we obtain \(x^2\) \((6\over x)\) = 12 \(\implies\) 6x = 12 โฆ
Find the angle between the curves xy = 6 and \(x^2 y\) =12. Read More ยป
Solution : Solving the curves simultaneously we get points of intersection as (1, 1) and (0, 0). At (1, 1) for first curve \(2y({dy\over dx})_1\) = 1ย \(\implies\)ย \(m_1\) = \(1\over 2\) & for second curve 2x = \(({dy\over dx})_2\) \(\implies\)ย \(m_2\) = 2 \(m_1m_2\) = -1 at (1, 1). But at (0, 0) clearly โฆ
Check the orthogonality of the curves \(y^2\) = x and \(x^2\) = y. Read More ยป
Solution : \(x^2\) = 32yย \(\implies\)ย \(dy\over dx\) = \(x\over 16\)ย \(\implies\)ย \(y^2\) = 4x \(\implies\)ย \(dy\over dx\) = \(2\over y\) \(\therefore\)ย atย (16, 8), \((dy\over dx)_1\) = 1, \((dy\over dx)_2\) = \(1\over 4\) So, required angle = \(tan^{-1}({1 โ {1\over 4}\over 1 + 1({1\over 4})})\) = \(tan^{-1}({3\over 5})\) Similar Questions Find the equations of โฆ
Solution : We have,ย f(x) = \(-x^2 โ 2x + 15\) \(\implies\) f'(x) = -2x โ 2 = -2(x + 1) for f(x) to be increasing, we must have f'(x) > 0 -2(x + 1) > 0 \(\implies\) x + 1 < 0 \(\implies\) x < -1 \(\implies\) x \(\in\) \((-\infty, -1)\). Thus f(x) is โฆ
Find the interval in which f(x) = \(-x^2 โ 2x + 15\) is increasing or decreasing. Read More ยป
Solution : We have, \(f(\theta)\) = \({4sin \theta\over 2 + cos\theta} โ \theta\) \(\implies\) \(f'(\theta)\) = \((2 + cos\theta)(4 cos\theta) + 4 sin^2\theta\over (2 + cos\theta)^2\) โ 1 \(\implies\) \(f'(\theta)\)ย = \(8 cos\theta + 4\over (2 + cos\theta)^2\) โ 1 \(\implies\) \(f'(\theta)\) = \(4\cos\theta โ cos^2\theta\over (2 + cos\theta)^2\) \(\implies\) \(f'(\theta)\) = \(cos\theta(4 โ cos\theta)\over โฆ
Solution : We have, f(x) = \(x^3 โ 3x^2 + 3x โ 100\) \(\implies\)ย f'(x) = \(3x^2 โ 6x + 3\) = \(3(x โ 1)^2\) Now, x \(\in\) R \(\implies\)ย \((x โ 1)^2\)ย \(\ge\)ย 0ย \(\implies\)ย f'(x)ย \(\ge\) 0. Thus, f'(x) \(\ge\) 0 for all x \(\in\) R. Hence, f(x) is increasing on R. Similar โฆ
Prove that the function f(x) = \(x^3 โ 3x^2 + 3x โ 100\) is increasing on R Read More ยป
Solution : We have, f(x) = sin 3x \(\therefore\)ย ย f'(x) = 3cos 3x Now,ย 0 < x < \(pi\over 2\)ย ย \(\implies\)ย 0 < 3x < \(3\pi\over 2\) Since cosine function is positive in first quadrant and negative in the second and third quadrants. Therefore, we consider the following cases. Case 1 : When 0 < โฆ
Solution : f(x) = \(x^4\over 12\) โ \(5x^3\over 6\) + \(3x^2\) + 7. f'(x) = \(x^3\over 3\) โ \(5x^2\over 2\) + 6x fโ(x) = \(x^2\) โ 5x + 6 Since, fโ(x) = 0 at point of inflection. \(\implies\) \(x^2\) โ 5x + 6 = 0 \(\implies\) x = 2 and x = 3 Hence, points โฆ
Solution : y = \(x^3 โ 6x^2 + 12x + 5\) yโ = \(3x^2 โ 12x + 12\) yโ = \(6x โ 12\) yโ = 0 \(\implies\) 6x โ 12 = 0 \(\implies\)ย x = 2 Since, yโ = 0 at x = 2, Hence the point of inflection is 2. Similar Questions Prove that โฆ
Find the point of inflection for the curve y = \(x^3 โ 6x^2 + 12x + 5\). Read More ยป
Solution : f(x) = \(3x^4 โ 4x^3\) f'(x) = \(12x^3 โ 12x^2\) f'(x) = \(12x^2(x โ 1)\) Now, fโ(x) = \(12(3x^2 โ 2x)\) fโ(x) = 12x(3x โ 2) fโ(x) = 0ย \(\implies\)ย x = 0, 2/3 Here, fโ(x) = 0 Thus, x = 0, 2/3 are the inflection points. Similar Questions Prove that the function โฆ
Find the inflection point of f(x) = \(3x^4 โ 4x^3\). Read More ยป
