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Application of Derivatives

Multiple Choice Questions

621.

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)

622.

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

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

623.

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)

624.

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

625.
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

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

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

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

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

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

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

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

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

626.

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

$\infty$
1
3
2

627.

The values of a and b for which the function y = alog_e(x ) + bx² + x, has extremum at the points x₁ = 1 and x₂ = 2 are

$a = \frac{2}{3}, b = - \frac{1}{6}$
$a = - \frac{2}{3}, b = - \frac{1}{6}$
$a = - \frac{2}{3}, b = \frac{1}{6}$
$a = - \frac{1}{3}, b = - \frac{1}{6}$

628.

The value of maxima of ${(\frac{1}{x})}^{x}$ is

${(\frac{1}{e})}^{e}$
e^e
e
e^1/e

629.

A point particle moves along a straight line such that x = $\sqrt{t}$ , where t is time. Then, ratio of acceleration to cube of the velocity is

- 1
- 0.5
- 3
- 2

630.

The tangents to curve y = x³ - 2x² + x - 2 which are parallel to straight line y = x, are

$x + y = 2 and x - y = \frac{86}{27}$
$x - y = 2 and x - y = \frac{86}{27}$
$x - y = 2 and x + y = \frac{86}{27}$
$x + y = 2 and x + y = \frac{86}{27}$

Previous Year Papers

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Multiple Choice Questions

625. 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

625.
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

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

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

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

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