Application of Derivatives Test

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Application of Derivatives Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

The values of $$x$$ satisfying $$\left| sinx \right| ^{\left| cosx \right|} +log_\left| cosx \right| \left| sinx \right| =2,$$ where $$x\epsilon (0,\dfrac{\pi}{2}),$$ is

A
$${\{\dfrac{\pi}{2}}\} $$

B
$${\{\dfrac{\pi}{3}}\} $$

C
$$(0, \pi 4) $$

D
$$( \pi 4, \pi 2) $$

Solution
Question 2
1 / -0

The slope of the straight line which is both tangent and normal to the curve $$4x^3=27y^2$$ is

A
$$\underline { + } 1$$

B
$$\underline { + } \dfrac { 1 }{ 2 } $$

C
$$\underline { + } \dfrac { 1 }{ \sqrt { 2 } } $$

D
$$\underline { + } \sqrt { 2 } $$

Solution

$$ Given: $$
$$ 4{{x}^{3}}=27{{y}^{2}} $$
$$ differentiating\,w.r.t.x $$
$$ 12{{x}^{2}}=54yy' $$
$$ t $$
$$ The\text{ line is tangent at}\,\text{P}\left( 3{{t}^{2}},2{{t}^{3}} \right)and\,normal\,at\left( 3t_{1}^{2},2t_{1}^{3} \right) $$
$$ \frac{dy}{dx}=\frac{2\left( 9{{t}^{4}} \right)}{9\left( 2{{t}^{3}} \right)}=1 $$
$$ equation\,to\,\tan gent\,P\left( t \right)is $$
$$ y-2{{t}^{3}}=t\left( x-3{{t}^{2}} \right) $$
$$ \Rightarrow tx-y={{t}^{3}}\,\,\,\,\,........\left( i \right) $$
$$ Equation\,to\,normal\,atQ\left( t_{1}^{{}} \right)is $$
$$ y-2t_{1}^{3}=-\frac{1}{t_{1}^{{}}}\left( x-3t_{1}^{2} \right) $$
$$ \Rightarrow x+t_{1}^{{}}y=2t_{1}^{4}+3t_{1}^{2}\,\,\,\,\,\,\,\,...........\left( ii \right) $$
$$ \left( i \right)\,and\left( ii \right)are\,identical $$
$$ \frac{t}{1}=\frac{-1}{t_{1}^{{}}}=\frac{{{t}^{3}}}{2t_{1}^{4}+3t_{1}^{2}} $$
$$ \Rightarrow {{t}^{2}}=\frac{2}{{{t}^{4}}}+\frac{3}{{{t}^{2}}}\,\,\,\,\,\,\,\,or\,\,t_{1}^{{}}=-\frac{1}{t} $$
$$ \Rightarrow {{t}^{6}}=2+3{{t}^{2}} $$
$$ \Rightarrow {{t}^{6}}-3{{t}^{2}}-2=0 $$
$$ \Rightarrow {{t}^{2}}=2 $$
$$ t=\pm \sqrt{2} $$
Question 3
1 / -0

The function $$f(x)=log x$$

A
has maxima at x=e

B
has minima at x=e

C
has neither maxima nor minima

D
all of these

Solution

Given, $$ f(x) = \log x \\ \therefore f^{'}(x)= \dfrac{1}{x} \, \textrm{and} \, f^{"}(x) = \dfrac{1}{x^2}$$
For maxima or minima $$ f^{'}(x) = 0 $$
Hence,$$ \dfrac{1}{x} = 0 \\ $$
$$ \therefore $$ x is undefined.
Therefore this function has neither maxima nor minima.
Question 4
1 / -0

Let $$g(x)=\displaystyle \int _{1-x}^{1+x}t|f'(t)|dt$$, where $$f(x)$$ does not behave like a constant function in any interval $$(a,b)$$ and the graph of $$y=f'(x)$$ is symmetric about the line $$x=1$$, then

A
$$g(x)$$ is increasing $$\forall \ x\ \in\ R$$

B
$$g(x)$$ is increasing only if $$x<1$$

C
$$g(x)$$ is increasing only if $$f(x)$$ is increasing

D
$$g(x)$$ is decreasing $$\forall \ x\ \in \ R$$
Question 5
1 / -0

Three normals are drawn from the point $$\left(c,0\right)$$ to the curve $${y}^{2}=x.$$If two of the normals are perpendicular to each other,then $$c=$$

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{2}$$

C
$$\dfrac{3}{4}$$

D
$$1$$

Solution

$$ Let\,the\text{ equation of normal to }{{\text{y}}^{2}}=4ax\,is $$
$$ y=mx-2am-a{{m}^{2}} $$
$$ \therefore equation\,of\,normal\,for\,{{y}^{2}}=x\,is $$
$$ y=mx-\frac{m}{2}-\frac{1}{4}{{m}^{3}}which\,passes\,through\left( c,0 \right) $$
$$ \therefore 0=m\left( c-\frac{1}{2}-\frac{{{m}^{2}}}{4} \right) $$
$$ \Rightarrow m=0\,\,and\,\frac{{{m}^{2}}}{4}=c-\frac{1}{2} $$
$$ \Rightarrow m=\pm 2\sqrt{c-\frac{1}{2}} $$
$$ \text{which gives a normal as x-axis and for other two normal} $$
$$ c-\frac{1}{2}>0 $$
$$ \Rightarrow c>\frac{1}{2} $$
$$ N\text{ow, if normals are perpendicular}\text{.} $$
$$ \left( 2\sqrt{c-\frac{1}{2}} \right)\left( -2\sqrt{c-\frac{1}{2}} \right)=-1 $$
$$ \Rightarrow c-\frac{1}{2}=\frac{1}{4} $$
$$ \Rightarrow c=\frac{3}{4} $$
Question 6
1 / -0

The function $$f(x)=\dfrac{x}{x^2+1}$$ increasing, if

A
$$-1 < x$$

B
$$x > 1$$

C
$$-1 < x$$ or $$x > 1$$

D
$$-1 < x < 1$$

Solution
Question 7
1 / -0

The approximate value of $$\sqrt[10]{0.999}$$ is

A
0.0998

B
0.9998

C
0.0999

D
0.9999

Solution
Question 8
1 / -0

If the line $$x+y=0$$ touches the curve $$2y^2=\alpha x^2+\beta $$ at $$(1,-1),$$ then $$(\alpha ,\beta )=$$

A
$$(-2,4)$$

B
$$(-1,3)$$

C
$$(4,-2)$$

D
$$(2,0)$$

Solution

Given equation of curve is,
$$2{ y }^{ 2 }=\alpha { x }^{ 2 }+\beta $$

Point $$\left( 1,-1 \right) $$ lies on the curve. Thus, it must satisfy given equation of curve.

$$\therefore 2{ \left( -1 \right) }^{ 2 }=\alpha { \left( 1 \right) }^{ 2 }+\beta $$

$$\therefore 2\left( 1 \right) =\alpha { \left( 1 \right) }+\beta $$

$$\therefore 2=\alpha +\beta $$ (1)

Now, equation of tangent to curve is,
$$x+y=0$$
$$\therefore y=-x$$

Thus, slope of tangent is, $${ m }_{ 1 }=-1$$ (2)

Equation of the curve is,
$$2{ y }^{ 2 }=\alpha { x }^{ 2 }+\beta $$
Differentiate w.r.t. x, we get,

$$2\times 2y\frac { dy }{ dx } =2\alpha x+0$$

$$\therefore 4y\frac { dy }{ dx } =2\alpha x$$

$$\therefore \frac { dy }{ dx } =\frac { 2\alpha x }{ 4y } $$

Slope of curve at $$\left( 1,-1 \right) $$ is,
$${ m }_{ 2 }=\frac { 2\alpha \times 1 }{ 4\times -1 } $$

$${ m }_{ 2 }=\frac { 2\alpha }{ -4 } $$

$$\therefore { m }_{ 2 }=\frac { \alpha }{ -2 } $$ (3)

At point $$\left( 1,-1 \right) $$, slope of tangent and normal will be same.
Thus, from equation (2) and (3),

$$\frac { \alpha }{ -2 } =-1$$

$$\therefore \alpha =2$$

Put this value in equation (1), we get,

$$2+\beta =2$$

$$\therefore \beta =0$$
$$\therefore \left( \alpha ,\beta \right) =\left( 2,0 \right) $$
Question 9
1 / -0

Let $$f ( x ) = \frac { \csc x + \cot x - 1 } { 1 + \cot x - \csc x }$$. The primitive of $$f ( x )$$ with respect to $$x$$ is equal to (Where $$C$$ is constant of integration.)

A
$$\ln \left( \sin \frac { x } { 2 } \right) + C$$

B
$$2 \ln \left( \cos \frac { x } { 2 } \right) + C$$

C
$$\ln ( 1 - \sin x ) + C$$

D
$$\ln ( 1 - \cos x ) + C$$

Solution
Question 10
1 / -0

Let A = {1, 2, ......... 10} and B = {1, 2, .......... 5}
f : A $$\rightarrow$$ B is a non-decreasing into function, then number of such function is

A
1001

B
1876

C
205

D
875

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 46

Result

Application of Derivatives Test - 46

Score

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SHARING IS CARING

Question 1

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

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

$$(0, \pi 4) $$

$$( \pi 4, \pi 2) $$

Solution

Question 2

$$\underline { + } 1$$

$$\underline { + } \dfrac { 1 }{ 2 } $$

$$\underline { + } \dfrac { 1 }{ \sqrt { 2 } } $$

$$\underline { + } \sqrt { 2 } $$

Solution

Question 3

has maxima at x=e

has minima at x=e

has neither maxima nor minima

all of these

Solution

Question 4

$$g(x)$$ is increasing $$\forall \ x\ \in\ R$$

$$g(x)$$ is increasing only if $$x<1$$

$$g(x)$$ is increasing only if $$f(x)$$ is increasing

$$g(x)$$ is decreasing $$\forall \ x\ \in \ R$$

Question 5

$$\dfrac{1}{4}$$

$$\dfrac{1}{2}$$

$$\dfrac{3}{4}$$

$$1$$

Solution

Question 6

$$-1 < x$$

$$x > 1$$

$$-1 < x$$ or $$x > 1$$

$$-1 < x < 1$$

Solution

Question 7

0.0998

0.9998

0.0999

0.9999

Solution

Question 8

$$(-2,4)$$

$$(-1,3)$$

$$(4,-2)$$

$$(2,0)$$

Solution

Question 9

$$\ln \left( \sin \frac { x } { 2 } \right) + C$$

$$2 \ln \left( \cos \frac { x } { 2 } \right) + C$$

$$\ln ( 1 - \sin x ) + C$$

$$\ln ( 1 - \cos x ) + C$$

Solution

Question 10

1001

1876

205

875

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