Application of Derivatives Test

Result

Application of Derivatives Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

The curve $$y-{e}^{xy}+x=0$$ has a vertical tangent at the point

A
$$(1,1)$$

B
At no point

C
$$(0,1)$$

D
$$(1,0)$$

Solution
Question 2
1 / -0

The slope of the curve $$y=\sin { x } +\cos ^{ 2 }{ x }$$ is zero at a point , whose x-coordinate can be ?

A
$$\dfrac { \pi }{ 4 } $$

B
$$\dfrac { \pi }{ 2 } $$

C
$${ \pi }$$

D
$$\dfrac { \pi }{ 3 } $$

Solution
Question 3
1 / -0

If the slope of one of the lines given by $${a^2}{x^2} + 2hxy+by^2 = 0$$ be three times of the other , then h is equal to

A
$$2\sqrt 3 ab$$

B
$$-2\sqrt 3 ab$$

C
$$\frac{2}{{\sqrt 3 }}ab$$

D
$$-\frac{2}{{\sqrt 3 }}ab$$

Solution
Question 4
1 / -0

If the curves $$\dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{4} = 1$$ and $$y^{3} = 16x$$ intersect at right angles, then $$a^{2}$$ is equal to

A
$$5/3$$

B
$$4/3$$

C
$$6/11$$

D
$$3/2$$

Solution

$$y^{3}=16x$$
$$3y^{2}\dfrac{dy}{dx}=16$$
$$\dfrac{dy}{dx}=m_{1}=\dfrac{16}{3y^{2}}$$

$$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{4}=1$$
$$\dfrac{2x}{a^{2}}+\dfrac{2y}{4}\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}=\dfrac{-x}{a^{2}}\times \dfrac{4}{y}=m_{2}$$

$$m_{1}.m_{2}=-1$$
$$\dfrac{16}{3y^{2}}\left(\dfrac{-4x}{a^{2}y}\right)=-1$$
$$\dfrac{16\times 4\times x}{3a^{2}y^{3}}=1\Rightarrow 64x=3a^{2}y^{3}$$
$$a^{2}=\dfrac{64 x}{3y^{3}}$$ Now $$y^{3}=16 x$$
$$a^{2}=\dfrac{64 x}{3\times 16 x}\Rightarrow a^{2}=\dfrac{4}{3}$$
Question 5
1 / -0

If for a curve represented parametrically by $$x={ sec }^{ 2 }t,\quad y=cot\quad t\quad $$ , the tangent at a point $$P(t=\frac { \pi }{ 4 } )$$ meets the curve again at the point Q, then $$\begin{vmatrix} PQ \end{vmatrix}$$is equal to

A
$$\frac { 2\sqrt { 5 } }{ 3 } $$

B
$$\frac { 3\sqrt { 5 } }{ 2 } $$

C
$$\frac { 5\sqrt { 3 } }{ 3 } $$

D
$$\frac { 5\sqrt { 5 } }{ 4 } $$

Solution
Question 6
1 / -0

$$\dfrac { d } { d x } \left( \sin ^ { 5 } x \cdot \sin 5 x \right) =$$

A
$$\sin ^ { 4 } x \cdot \sin 5 x$$

B
$$5\sin ^ { 4 } x \cdot \sin 6 x$$

C
$$5\sin ^ { 4 } x \cdot \sin 5 x$$

D
$$- 5 \sin ^ { 4 } x \cdot \sin 6 x$$

Solution

$$\frac { d }{ dx } \left( { sin }^{ 5 }x.sin5x \right) $$

$$={ sin }^{ 5 }x\frac { d }{ dx } \left( sin5x \right) +sin5x\frac { d }{ dx } \left( { sin }^{ 5 }x \right) $$

$$={ sin }^{ 5 }x\left[ 5cos5x \right] +sin5x\left[ 5{ sin }^{ 4 }x.cosx \right] $$

$$=5{ sin }^{ 5 }xcos5x+5sin5x{ sin }^{ 4 }x.cosx$$

$$=5{ sin }^{ 4 }x\left[ sinx.cos5x+cosx.sin5x \right] $$

$$=5{ sin }^{ 4 }x\left[ sin\left( x+5x \right) \right] $$

$$=5{ sin }^{ 4 }xsin\left( 6x \right) $$
Question 7
1 / -0

Let f and g be non-increasing and non-decreasing functions respectively from $$[0,\infty ]$$ vto $$[0,\infty ]$$ and $$h(x)=f(g(x)),h(0)=0$$, then in $$[0,\infty ]$$, $$h(x)-h(1)$$ is

A
<0

B
>0

C
=0

D
none of these

Solution
Question 8
1 / -0

The function $$f\left( x \right) = \frac{{\left| {x - 1} \right|}}{{{x^2}}}$$ is

A
One-One in $$\left( { - \infty ,1 }\right)$$

B
One-One in $$\left( {0,\infty} \right) $$

C
One-One in $$\left( {0,1} \right) $$

D
One-One in $$\left( { - \infty ,0 }\right)$$

Solution
Question 9
1 / -0

$$y = 6\tan \,x\left( {{e^x} - x - 1} \right) - 3{x^3} - {x^4} - \frac{5}{4}{x^5},\,$$ if $${n^{th}}$$ derivative at x=0 is non zero then least value of n is

A
3

B
4

C
5

D
6

Solution
Question 10
1 / -0

If $$(x+{ y }^{ 3 })\dfrac { dy }{ dx } $$=y and y(0)=2. then sum of all possible value(s) of y(1) is ________________.

A
-4

B
4

C
0

D
2

Solution