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Application of Derivatives

Multiple Choice Questions

771.

$The function f (x) = x^{3} + {ax}^{2} + bx + c, a^{2} \leq 3 b has$

one maximum value
one minimum value
no extreme value
one maximum and one minimum value

772.

$The maximum value of \frac{\log (x)}{x}, 0 < x < \infty is$

$\infty$
e
1
e^{- 1}

773.

$z = \tan (y + ax) + \sqrt{y - ax} \Rightarrow z_{xx} - a^{2} z_{yy} = ?$

0
2
$z_{x} + z_{y} = 0$
$z_{x} z_{y}$

774.
The height of the cone of maximum volume inscribed in a sphere of radius R is

$\frac{R}{3}$

$\frac{2 R}{3}$

$\frac{4 R}{3}$

$\frac{4 R}{\sqrt{3}}$

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$\frac{4 R}{3}$

Let the height of the cone = h

and the radius of the cone = r

Given, radius of the sphere = R

Now, In $∆ OPB$

$\Rightarrow R^{2} = r^{2} + {(h - R)}^{2} \Rightarrow r^{2} = R^{2} - {(h - R)}^{2} = (R + h - R) (R - h + R) \Rightarrow r^{2} = h (2 R - h) The volume of the cone is V = \frac{1}{3} {πr}^{2} h \Rightarrow V = \frac{1}{3} πh (2 R - h) h \Rightarrow V = \frac{π}{3} (4 {Rh}^{2} - 3 h^{3}) Differentiating with r to h \frac{dV}{dh} = 0 \Rightarrow \frac{π}{3} (4 Rh - 3 h^{2}) = 0 \Rightarrow h (4 R - 3 h) = 0 \Rightarrow h = 0, h = \frac{4 R}{3} (Not possible) Now, \frac{d^{2} V}{{dh}^{2}} = \frac{π}{3} (4 R - 6 h) {(\frac{d^{2} V}{{dh}^{2}})}_{(at h = \frac{4 R}{3})} = \frac{π}{3} (4 R - 6 . \frac{4 R}{3}) = \frac{π}{3} (4 R - 8 R) = - \frac{4 π}{3} R \Rightarrow Negative ie, maximum Hence, the height og the cone of maximum volume is (\frac{4 R}{3})$

775.

If the distance s travelled by a particle in time t is given by s = t²- 2t + 5, then its acceleration is

776.

The length of the sub tangent at any point (x₁, y₁) on the curve y = 5^x is

$5^{x_{1}}$
$y_{1} 5^{x_{1}}$
$\log_{e} (5)$
$\frac{1}{\log_{e} (5)}$

777.

$If f : R \to R is defined by f (x) = [\frac{4}{5}] for x \in R, where [y] denotes the greatest integer not exceeding y, then \{f (x) : |x| < 71\} is equal to$

$(0, \infty)$
$(1, \infty)$
$(4, \infty)$
$(5, \infty)$

778.

The condition that f(x) = a + bx² + cx + d has no extreme value, is

$b^{2} > 3 ac$
$b^{2} = 4 ac$
$b^{2} = 3 ac$
$b^{2} < 3 ac$

779.

If there is an error of ± 0.04cm in the measurement of the diameter of a sphere, then the approximate percentage error in its volume, when the radius is 10cm, is