Mathemerize

Solution : We have, \(f(\theta)\) = \({4sin \theta\over 2 + cos\theta} – \theta\) \(\implies\) \(f'(\theta)\) = \((2 + cos\theta)(4 cos\theta) + 4 sin^2\theta\over (2 + cos\theta)^2\) – 1 \(\implies\) \(f'(\theta)\)  = \(8 cos\theta + 4\over (2 + cos\theta)^2\) – 1 \(\implies\) \(f'(\theta)\) = \(4\cos\theta – cos^2\theta\over (2 + cos\theta)^2\) \(\implies\) \(f'(\theta)\) = \(cos\theta(4 – cos\theta)\over …

Prove that \(f(\theta)\) = \({4sin \theta\over 2 + cos\theta} – \theta\) is an increasing function of \(\theta\) in \([0, {\pi\over 2}]\). Read More »

Solution : We have, f(x) = \(x^3 – 3x^2 + 3x – 100\) \(\implies\)  f'(x) = \(3x^2 – 6x + 3\) = \(3(x – 1)^2\) Now, x \(\in\) R \(\implies\)  \((x – 1)^2\)  \(\ge\)  0  \(\implies\)  f'(x)  \(\ge\) 0. Thus, f'(x) \(\ge\) 0 for all x \(\in\) R. Hence, f(x) is increasing on R. Similar …

Prove that the function f(x) = \(x^3 – 3x^2 + 3x – 100\) is increasing on R Read More »

Solution : We have, f(x) = sin 3x \(\therefore\)   f'(x) = 3cos 3x Now,  0 < x < \(pi\over 2\)   \(\implies\)  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 < …

Separate \([0, {\pi\over 2}]\) into subintervals in which f(x) = sin 3x is increasing or decreasing. Read More »

Solution : f(x) = \(x^4\over 12\) – \(5x^3\over 6\) + \(3x^2\) + 7. f'(x) = \(x^3\over 3\) – \(5x^2\over 2\) + 6x f”(x) = \(x^2\) – 5x + 6 Since, f”(x) = 0 at point of inflection. \(\implies\) \(x^2\) – 5x + 6 = 0 \(\implies\) x = 2 and x = 3 Hence, points …

Find the point of inflection for f(x) = \(x^4\over 12\) – \(5x^3\over 6\) + \(3x^2\) + 7. Read More »

Solution : y = \(x^3 – 6x^2 + 12x + 5\) y’ = \(3x^2 – 12x + 12\) y” = \(6x – 12\) y” = 0 \(\implies\) 6x – 12 = 0 \(\implies\)  x = 2 Since, y” = 0 at x = 2, Hence the point of inflection is 2. Similar Questions Prove that …

Find the point of inflection for the curve y = \(x^3 – 6x^2 + 12x + 5\). Read More »

Solution : f(x) = \(3x^4 – 4x^3\) f'(x) = \(12x^3 – 12x^2\) f'(x) = \(12x^2(x – 1)\) Now, f”(x) = \(12(3x^2 – 2x)\) f”(x) = 12x(3x – 2) f”(x) = 0  \(\implies\)  x = 0, 2/3 Here, f”(x) = 0 Thus, x = 0, 2/3 are the inflection points. Similar Questions Prove that the function …

Find the inflection point of f(x) = \(3x^4 – 4x^3\). Read More »