Solution : \(\int\) \(x^2 + x – 1\over x^2 – 1\) dx = \(\int\) (\(x^2 – 1\over x^2 – 1\) + \(x\over x^2 – 1\))dx = \(\int\) 1 dx + \(\int\) \(x\over x^2 – 1\)) dx Let \(x^2 – 1\) = t \(\implies\) 2x dx = dt = x + \(\int\) \(dt\over 2t\) = x […]
Integrate \(x^2 + x – 1\over x^2 – 1\) with respect to x. Read More »
Solution : Let r be the radius of a sphere and \(\delta\)r be the error in measuring the radius. Then, r = 9 cm and \(\delta\)r = 0.03 cm. Let V be the volume of the sphere. Then, V = \({4\over 3}\pi r^3\) \(\implies\) \(dV\over dr\) = \(4\pi r^2\) \(\implies\) \(({dV\over dr})_{r = 9}\) =
Solution : Let y = f(x), x = 3 and x + \(\delta x\). Then, \(\delta x\).= 0.02. For x = 3, we get y = f(3) = 45 Now, y = f(x) \(\implies\) y = \(3x^2 + 5x + 3\) \(\implies\) \(dy\over dx\) = 6x + 5 \(\implies\) \(({dy\over dx})_{x = 3}\) = 23
Find the approximate value of f(3.02), where f(x) = \(3x^2 + 5x + 3\). Read More »
Solution : Since a polynomial function is everywhere differentiable and so continuous also. Therefore, f(x) is continuous on [2, 3] and differentiable on (2, 3). Also, f(2) = \(2^2\) – 5 \(\times\) 2 + 6 = 0 and f(3) = \(3^2\) – 5 \(\times\) 3 + 6 = 0 \(\therefore\) f(2) = f(3) Thus, all
Verify Rolle’s theorem for the function f(x) = \(x^2\) – 5x + 6 on the interval [2, 3]. Read More »
Solution : We are given that f(1) = f(3) = 0. \(\therefore\) \(1^3 – 6 \times 1 + a + b\) = \(3^3 – 6 \times 3^2 + 3a + b\) = 0 \(\implies\) a + b = 5 and 3a + b = 27 Solving these two equations for a and b, f'(c) is
Solution : Let f(x) = cos x – 1, Clearly f(x) is continous on \([{\pi\over 2}, {3\pi\over 2}]\) and differentiable on \(({\pi\over 2}, {3\pi\over 2})\). Also, f\((\pi\over 2)\) = \(cos {\pi\over 2}\) – 1 = -1 = f\((3\pi\over 2)\). Thus, all the conditions of rolle’s theorem are satisfied. Consequently,there exist at least one point c
Solution : We have, x = \(a cos^3\theta\), y = \(a sin^3\theta\) \(\implies\) \(dx\over d\theta\) = \(-3 a cos^2\theta sin\theta\), \(dy\over d\theta\) = \(3 a sin^2\theta cos\theta\) Now, \(dy\over dx\) = \(dy/d\theta\over dx/d\theta\) \(\implies\) \(dy\over dx\) = -\(tan\theta\) \(\therefore\) Slope of the normal at any point on the curve = \(-1\over dy/dx\) = \(cot\theta\) Hence, the
Solution : We have, x = 1 – \(a sin\theta\), y = \(b cos^2\theta\) \(\implies\) \(dx\over d\theta\) = \(-a cos\theta\) and \(dy\over d\theta\) = \(-2b cos\theta sin\theta\) \(\therefore\) \(dy\over dx\) = \(dy/d\theta\over dx/d\theta\) = \(2b\over a\) \(sin\theta\) \(\implies\) \(dy\over dx\) at \(\pi\over 2\) = \(2b\over a\) Hence, Slope of normal at \(\theta\) = \(\pi\over 2\)
Solution : The equation of the curve is y = \(2x^3 – 3\) Differentiating with respect to x, we get \(dy\over dx\) = \(6x^2\) Now, \(m_1\) = (Slope of the tangent at x = 2) = \(({dy\over dx})_{x = 2}\) = \(6 \times (2)^2\) = 24 and, \(m_2\) = (Slope of the tangent at x
Solution : The equation of the given curve is y = \(-5x^2 + 6x + 7\) \(\implies\) \(dy\over dx\) = -10x + 6 \(\implies\) \(({dy\over dx})_{(1/2, 35/4)}\) = \(-10\over 4\) + 6 = 1 The required equation at (1/2, 35/4) is y – \(35\over 4\) = \(({dy\over dx})_{(1/2, 35/4)}\) \((x – {1\over 2})\) \(\implies\) y
