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Application of Derivatives

Multiple Choice Questions

181.

The maximum value of $f (x) = \frac{\log (x)}{x} (x \neq 0, x \neq 1)$ is

e
$\frac{1}{e}$
e²
$\frac{1}{e^{2}}$

182.

If the volume of spherical ball is increasing at the rate of 4 $π$ cm³/s, then the rate of change of its surface area when the volume is 288 $π$ cm³, is

$\frac{4}{3} π {cm}^{2} / s$
$\frac{2}{3} π {cm}^{2} / s$
$4 π {cm}^{2} / s$
$2 π {cm}^{2} / s$

183.

The equation of displacement of a particle is x(t) = 5t² - 7t + 3. The acceleration at the moment when its velocity becomes 5 m/sec is

3 m/sec²
7 m/sec²
10 m/sec²
8 m/sec²

184.

The mean value of the function $f (x) = \frac{2}{e^{x} + 1}$ on the interval [0, 2] is

$2 - \log_{e} (\frac{2}{e^{2} + 1})$
$2 + \log_{e} (\frac{2}{e^{2} + 1})$
$2 + \log_{e} (\frac{2}{e^{2} - 1})$
$- 2 + \log_{e} (\frac{2}{e^{2} - 1})$

185.

The function $y = \sqrt{2 x - x^{2}}$

increases in (0, 1) but decreases in (1, 2)
decreases in (0, 2)
increases m (1, 2) but decreases in (0, 1)
increases in (0, 2)

186.

The interval in which the function $y = x - 2 \sin (x)$ ; $0 \leq x \leq 2 π$ increases throughout is

$(\frac{5 π}{3}, 2 π)$
$(0, \frac{π}{3})$
$(\frac{π}{3}, \frac{5 π}{3})$
$(0, \frac{π}{4})$

187.

The points of the curve y = x³ + x - 2 at which its tangent are parallel to the straight line y = 4x - 1 are

(2, 7), (- 2, - 11)
(0, 2), (2^1/3, 2^1/3)
(- 2^1/3, - 2^1/3), (0, - 4)
(1, 0), (- 1, - 4)

188.

The equation of the normal to the curve y = - $\sqrt{x}$ + 2 at the point of its intersection with the bisector of the first quadrant is

4x - y + 16 = 0
4x - y = 16
2x - y - 1 = 0
2x - y + 1 = 0

189.
The angle at which the curve y = x² and the curve x = $\frac{5}{3} \cos (t)$ , y = $\frac{5}{4} \sin (t)$ intersect is

$\tan^{- 1} (\frac{2}{41})$

$\tan^{- 1} (\frac{41}{2})$

$- \tan^{- 1} (\frac{2}{41})$

$2 \tan^{- 1} (\frac{41}{2})$

$\tan^{- 1} (\frac{41}{2})$

$Given, y = x^{2} . . . (i) x = \frac{5}{3} \cos (t), y = \frac{5}{4} \sin (t) . . . (ii)$

Which is parametnc equation, we change this equation is cartesian equation as follows

$\cos (t) = \frac{3}{5} x, \sin (t) = \frac{4}{5} y$

On squaring and adding both i.e. cos(t) and sin(t), we get

$\frac{9}{25} x^{2} + \frac{16}{25} y^{2} = \cos^{2} (t) + \sin^{2} (t) \Rightarrow 9 x^{2} + 16 y^{2} = 25 . . . (iii) [∵ \cos^{2} (θ) + \sin^{2} (θ) = 1]$

$∵$ The intersection points at Eq. (i) and (iii) are (1, 1) and (- 1, 1)

Now, slope of tangent of Eq. (i) at point (1, 1) is

$m_{1} = \frac{dy}{dx} = 2 x ∴ m_{1} = {\frac{dy}{dx}}_{(1, 1)} = 2$

And slope of tangent of Eq (iii), at point (1, 1) is

$m_{2} = \frac{dy}{dx} = \frac{- 9}{16}$

$∴$ Angle at point of intersection of Eqs. (i) and (iii), we get

$θ_{1} = \tan^{- 1} |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| θ_{1} = \tan^{- 1} |\frac{2 + \frac{9}{16}}{1 - \frac{2 \times 9}{16}}| = \tan^{- 1} (\frac{41}{2})$

Similarly, slope of tangent of Eq. (i) at point (- 1, 1)

$m_{1} = {\frac{dy}{dx}}_{(- 1, 1)} = - 2$

And slope of tangent of Eq (iii) at point (-1, 1)

$m_{2} = \frac{dy}{dx} = \frac{- 9}{16}$

$∴$ Angle at point of intersection of Eqs. (i) and (iii), we get

$θ_{2} = \tan^{- 1} |\frac{\frac{- 2 - 9}{16}}{1 - \frac{18}{16}}| = \tan^{- 1} (\frac{41}{2})$

190.

The maximum value of the function $y = 2 \tan (x) - \tan^{2} (x)$ over $[0, \frac{π}{2}]$ is

$\infty$
1
3
2

Previous Year Papers

Test Series

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Multiple Choice Questions

189. The angle at which the curve y = x2 and the curve x = 53cost, y = 54sint intersect is tan-1241 tan-1412 - tan-1241 2tan-1412

189.
The angle at which the curve y = x² and the curve x = $\frac{5}{3} \cos (t)$ , y = $\frac{5}{4} \sin (t)$ intersect is

$\tan^{- 1} (\frac{2}{41})$

$\tan^{- 1} (\frac{41}{2})$

$- \tan^{- 1} (\frac{2}{41})$

$2 \tan^{- 1} (\frac{41}{2})$