is a normal to the circle $${x}^{2}+{y}^{2}=25$$

is a tangent to the circle $${x}^{2}+{y}^{2}=25$$

does not meet the circle $${x}^{2}+{y}^{2}=25$$

MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 8

The inclination of the tangent at

$\theta = \dfrac {\pi}{3}$ on the curve

$x = a(\theta + \sin \theta), y = a(1 + \cos \theta)$ is

0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{6}$
0%
$\dfrac {2\pi}{3}$
0%
$\dfrac {5\pi}{6}$

Explanation

For the curve ,

$\displaystyle \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{\dfrac{d(a(1+cos\theta))}{d\theta}}{\dfrac{d(a(\theta+sin\theta))}{d\theta}} = \dfrac{-asin\theta}{a+acos\theta} = \dfrac{-sin\theta}{1+cos\theta}$

So, If the inclination of tangent at

$\theta = \dfrac{\pi}{3}$ be

$y$ ,

Then,

$\tan(y) =$

$\dfrac{-sin(\dfrac{\pi}{3})}{1+cos\dfrac{\pi}{3}} = \dfrac{-1}{\sqrt{3}}$

Thus,

$y = \pi-\dfrac{\pi}{6}= \dfrac{5\pi}{6}$

The greatest slope among the lines represented by the equation

$4x^2 - y^2 + 2y - 1 = 0$ is -

0%
$-3$
0%
$-2$
0%
$2$
0%
$3$

Explanation

$4{x}^{2} - {y}^{2} + 2y - 1 = 0$

${y}^{2} - 2y + 1 = 4{x}^{2}$

${\left( y - 1 \right)}^{2} = {\left( 2x \right)}^{2}$

$y - 1 = \pm 2x$

$y = 2x + 1 \text{ or } y = -2x + 1$

Comparing the above equations with

$y = mx + c$ , we have

Slope

$\left( m \right) = 2, -2$

Hence the greatest slope will be

$2$ .

If the tangent at

$(x_{1}, y_{1})$ to the curve

$x^{3}+y^{3}=a^{3}$ meets the curve again at

$(x_{2}, y_{2})$ then

0%
$\dfrac{x_{2}}{x_{1}}+\dfrac{y_{2}}{y_{1}}=-1$
0%
$\dfrac{x_{2}}{y_{1}}+\dfrac{x_{1}}{y_{2}}=-1$
0%
$\dfrac{x_{1}}{x_{2}}+\dfrac{y_{1}}{y_{2}}=-1$
0%
$\dfrac{x_{2}}{x_{1}}+\dfrac{y_{2}}{y_{1}}=1$

Explanation

Given

$x^3+y^3=a^3$

The derivative is

$\dfrac{dy}{dx}=-\dfrac{x^2}{y^2}.....(1)$

Therefore, slope of tangent at

$(x_1,y_1)$ is

$-\dfrac{x_1^2}{y_1^2}.........(2)$

The tangent passes through

$(x_2,y_2)$ , therefore the slope of tangent is also given by

$\dfrac{y_2-y_1}{x_2-x_1}.......(3)$

Comparing the two slope equations we get

$\dfrac{y_2-y_1}{x_2-x_1}=-\dfrac{x_1^2}{y_1^2}$

$\dfrac{y_2^3-y_1^3}{x_2^3-x_1^3}\times\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}=-\dfrac{x_1^2}{y_1^2}$

$-\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}=-\dfrac{x_1^2}{y_1^2}$

$x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_2^2$

$x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1y_2+x_1^2y_2^2$

$x_1^2y_2^2-x_2^2y_1^2=x_1x_2y_1^2-x_1^2y_1y_2$

$(x_1y_2-x_2y_1)(x_1y_2+x_2y_1)=x_1y_1(x_2y_1-x_1y_2)$

$x_1y_2+x_2y_1=-x_1y_1$

$\dfrac{x_2}{x_1}+\dfrac{y_2}{y_1}=-1$

The ordinate of all points on the curve

$y=\dfrac{1}{2\sin^{2}x+3\cos^{2}x}$ where the tangent is horizontal, is

0%
Always equal to $\dfrac{1}{2}$
0%
Always equal to $\dfrac{1}{3}$
0%
$\dfrac{1}{2}$ or $\dfrac{1}{3}$ according as $n$ is an even or an odd integer.
0%
$\dfrac{1}{2}$ or $\dfrac{1}{3}$ according as $n$ is an even integer

Explanation

$y=\dfrac{1}{2sin^{2}x+3cos^{2}x}=\dfrac{1}{2+cos^{2}x}$

$tan\theta =\dfrac{dy}{dx}= \dfrac{-2sinx\, cos x}{(2+cos^{2}x)^{2}}=0$

$\therefore sin\, x=0$ or

$cos\, x= 0$

$sin\, x=0, y= \dfrac{1}{3}$ , if

$cos\, x=0, y= \dfrac{1}{2}$

1098464_1141711_ans_de42566bc3344c2b9b551bd59f7cb5c7.jpeg

The curve given by

$x + y = {e^{xy}}$ has an tangents parallel to the y-axis at the point

0%
$(0,1)$
0%
$(1,0)$
0%
$(1,1)$
0%
$(0,0)$

Explanation

$\dfrac{dy}{dx}$ (differentiate the terms)

$1+\dfrac{dy}{dx}=e^{xy}\left(x\dfrac{dy}{dx}+y\right)$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$

Check by option

$\dfrac{dy}{dx}\left|_{(0, 1)}\right.=\dfrac{1-1}{1-0}=\dfrac{0}{1}=0$ Not coming infinity

$\displaystyle\dfrac{dy}{dx}\left|_{(1, 0)}\right.=\dfrac{-1}{0}=\infty$

This point satisfy

$\therefore$ option B

$(1, 0)$ .

1356400_1131715_ans_86b17a40d6924052a2042284677f440a.png

Let

$f$ be continuous and differentiable function such that

$f(x)$ and

$f^{'}(x)$ have opposite sign everywhere. Then

0%
$f$ is increasing
0%
$f$ is decreasing
0%
$|f|$ is non-monotonic
0%
none of the above

$- 4 \leq x \leq 4$ then critical points of

$f ( x ) = x ^ { 2 } - 6 | x | + 4$ are

0%
$3 , - 2$
0%
$6 , - 6$
0%
$3 , - 3$
0%
$0,1$

Number of possible tangents to the curve $y = \cos \left( {x + y} \right), - 3\pi \leqslant x \leqslant 3\pi$ , that are parallel to the line $x + 2y = 0$ , is

0%
1
0%
2
0%
3
0%
4

The normal to the curve,

${x}^{2}+2xy-{3y}^{2}=0,\ at\left (1,1\right)$ :

0%
Meets the curve, again in the fourth quadrant
0%
Does not meet the curve again
0%
Meets the curve again in the second quadrant
0%
Meets the curve again in the third quadrant

Number of tangents drawn from the point

$\left (-1/2,0\right)$ to the curve

$y={e}^{x}$ . (Here { } denotes fractional part function ).

0%
$2$
0%
$1$
0%
$3$
0%
$4$

Explanation

Let

$A$ be the point of contact of a tangent of

${e}^{x}$ passing through

$(-\cfrac{1}{2},0)$

$\Rightarrow$

${ \left| \cfrac { dy }{ dx } \right| }_{ \lambda }={ e }^{ \lambda }$

Equation of tangent

$\quad y-{ e }^{ \lambda }={ e }^{ \lambda }(x-\lambda )\quad$

$x=-\cfrac { 1 }{ 2 } ,y=0$

$-{ e }^{ \lambda }={ e }^{ \lambda }\left( -\cfrac { 1 }{ 2 } -\lambda \right) \Rightarrow { e }^{ \lambda }\lambda =\cfrac { { e }^{ \lambda } }{ 2 }$

either

$\quad { e }^{ \lambda }=0$ or

$\lambda =+\cfrac { 1 }{ 2 }$

$\quad y{ e }^{ \lambda }=0\Rightarrow y=0\quad \rightarrow$ xaxis

$\quad \lambda =\cfrac { 1 }{ 2 } \Rightarrow y-{ e }^{ \cfrac { 1 }{ 2 } }={ e }^{ \cfrac { 1 }{ 2 } }\left( x-\cfrac { 1 }{ 2 } \right)$

2 solutions or 2 distinct tangents

If the tangent at P of the curve

$y^2=x^3$ intersects the curve again at Q and the straight lines OP, OQ ma angles

$\alpha, \beta$ with the x-axis where 'O' is the origin then

$\tan\alpha/\tan\beta$ has the value equal to?

0%
$-1$
0%
$-2$
0%
$2$
0%
$\sqrt{2}$

Explanation

Let

$P$ be

$(x_{1}, y_{1})$ and

$Q$ be

$(x_{2}, y_{2})$

equation of tangent

$(x_{2}, y_{2})$ , so

$\dfrac{y_{0}-y_{1}}{x_{2}-x_{1}}= \dfrac{3x_{1}^{2}}{2y_{1}}$

need to find out

$\dfrac{\tan \alpha}{ \tan \beta} = \dfrac{\dfrac{y_{1}}{x_{1}}}{y_{2}/x_{2}} = \dfrac{y_{1}x_{2}}{x_{1} y_{2}}$

$\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}= \dfrac{3x_{1}^{2}}{2y_{1}}$

$2y_{1}y_{2}-2y^{2}= 3x^{2}_{1}x_{2}-3x_{1}^{3}$

$2y_{1}y_{2}-2y_{1}^{2}= 3x_{1}^{2}x_{2}-3y^{2}$

$2y_{1}y_{2}+y_{1}^{2}= 3x_{1}^{2} x_{2}$

$y_{1}^{3} (2y_{2}+y_{1})^{3}= 27 x_{1}^{6} x^{3}_{2}$

$y_{1}^{3} (2y_{2}+ y_{1})^{3}= 27 y_{1}^{4} y_{2}^{3}$

$(2y_{2}+y_{1})^{3} - 27 y_{1} y_{2}$

$8 y_{2}^{3} - 15 y_{2}^{2} y_{1}+ 6y_{2} y^{2}_{1}+ y_{1}^{3}=0$

$(y_{2}-y_{1})^{2} (8y_{2}+ y_{1})=0$

$8y_{2}=-y_{1} \Rightarrow 64 y_{2}^{2} = y_{1}^{2}$

$64 x_{2}^{3} = x_{1}^{3} \Rightarrow 4 x_{2} = x_{1}$

$\dfrac{\tan \alpha}{\tan \beta}= \dfrac{y_{1}x_{2}}{x_{1}y_{2}}$

$= \dfrac{-8y_{2} x_{2}}{4x_{2} y_{2}} = -2$

A curve C has the property that if the tangent drawn at any point 'P' on C meets the coordinate axes at A and B, and P is midpoint of AB. If the curve passes through the point

$(1, 1)$ then the equation of the curve is?

0%
$xy=2$
0%
$xy=3$
0%
$xy=1$
0%
$xy=4$

Explanation

By option verification passing through

$(1,1)$

$xy=1$

Two lines drawn through the point

$A ( 4,0 )$ divide the area bounded by the curve

$y = \sqrt { 2 } \sin ( \pi x / 4 )$ and

$x$ - axis between the lines

$x = 2$ and

$x = 4$ into three equal parts. Sum of the slopes of the drawn lines is:

0%
$- 4 \sqrt { 2 } / \pi$
0%
$- \sqrt { 2 } / \pi$
0%
$- 2 \sqrt { 2 } / \pi$
0%
None

Explanation

Area bounded by

$y=\sqrt{2}\sin\left(\dfrac{\pi x}{4}\right)$ and

$x-axis$ between the lines

$x=2$ and

$x=4,$

$\Delta=\sqrt{2}\displaystyle\int_2^4\sin\dfrac{\pi x}{4}dx$

$=\left[-\dfrac{4\sqrt{2}}{\pi}.\cos\dfrac{\pi x}{4}\right]^4_2$

$=\dfrac{4\sqrt{2}}{\pi}\,unit^2$

Let the drawn lines are

$L_1:y-m_1(x-4)=0$ and

$L_2:y-m_2(x-4)=0,$ meeting the line

$x=2$ at the points

$A$ and

$B,$ respectively.

Clearly,

$A=(2,-2m_1);\,B=(2,-2m_2)$ [ Fig ]

Now,

$\Delta_{ACD}=\dfrac{\Delta}{3}\Rightarrow \dfrac{4\sqrt{2}}{3\pi}=\dfrac{1}{2}.2.(-2m_1)$

$\Rightarrow$

$m_1=-\dfrac{2\sqrt{2}}{3\pi}$

Also,

$\Delta_{BCD}=\dfrac{2\Delta}{3}$

$\Rightarrow$

$\dfrac{8\sqrt{2}}{3\pi}=\dfrac{1}{2}\times 2-2m_2$

$\Rightarrow$

$m_2=\dfrac{-4\sqrt{2}}{3\pi}$

$\Rightarrow$ Required sum

$=-\dfrac{2\sqrt{2}}{3\pi}+\left(\dfrac{-4\sqrt{2}}{3\pi}\right)$

$=\dfrac{-6\sqrt{2}}{3\pi}$

$=\dfrac{-2\sqrt{2}}{\pi}$

1481122_1180874_ans_e8e5f610c4d249289be00d9181cb35c5.png

The number of critical points of the function

$f(x)=|x-1||x-2|$ is

0%
$1$
0%
$2$
0%
$3$
0%
none of these

The line

$3x-4y=0$

0%
is a tangent to the circle ${x}^{2}+{y}^{2}=25$
0%
is a normal to the circle ${x}^{2}+{y}^{2}=25$
0%
does not meet the circle ${x}^{2}+{y}^{2}=25$
0%
does not pass thro' the origin

Explanation

$3x=4y$

$x=\dfrac { 4y }{ 3 }$

${ x }^{ 2 }+{ y }^{ 2 }=25$

${ x }^{ 2 }-\left( 1+\dfrac { 16 }{ 9 } \right) =25$

${ x }^{ 2 }=9$

$x=\pm 3$

$y=\pm 9/4$

Now,

${ x }^{ 2 }+{ y }^{ 2 }=25$

$2x+2y\dfrac { dy }{ dx } =0$

$\dfrac { dy }{ dx } =\dfrac { -x }{ y } =\dfrac { -\left( 3 \right) }{ \left( 9/4 \right) } =-4/3$

$\dfrac { -dx }{ dy } =3/4\Rightarrow$ slope of normal at circle and

$3/4\Rightarrow$ solpe of

$3x-4y=0$

So, Line is normal at circle.

The curves

$x = y^2 and. xy = k$ cut at right angles, If

$6k^2$ = 1.

0%
True
0%
False

The tangent to the curve

$2a^2y=x^3-3ax^2$ is parallel to the x-axis at the points

0%
$(0 , 0) , (2a, - 2a)$
0%
$(0 , 0) , (-2a, 2a)$
0%
$(0 , 0) , (-2a, - 2a)$
0%
$(2 , 2) , (0 , 0)$

Explanation

Let

$(x_1,y_1)$ represent the required points.

The slope of the

$x-$ axis is

$0$

Here

$2a^2y=x^3-3ax^2$ , since the points lies on the curve we get

$2a^2y_1=x_1^3-3ax_1^2 \quad ...(1)$

Consider

$2a^2y=x^3-3ax^2$

On differentiating both sides with respect to

$x$ , we get

$2a^2\dfrac{dy}{dx}=3x^2-6ax$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{3x^2-6ax}{2a^2}$

Slope of the tangent at

$(x_1,y_1)=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=\dfrac{3x_1^2-6ax_1}{2a^2}$

It is given that slope of the tangent at

$(x_1,y_1)=$ slope of the

$x-$ axis.

$\Rightarrow \dfrac{3x_1^2-6ax_1}{2a^2}=0$

$\Rightarrow 3x_1^2-6ax_1=0$

$\Rightarrow x_1(3x_1-6a)=0$

$\Rightarrow x_1=0$ or

$x_1=2a$

Also, from

$(1)$

$2a^2y_1=0$ or

$2a^2y_1=8a^3-12a^3$

$\Rightarrow y_1=0$ or

$y_1=-2a$

Thus, the required points are

$(0,0)$ and

$(2a,-2a)$

$f(x)=\dfrac{x}{5}+\dfrac{4}{x}(x\neq 0)$ in increasing in

0%
$(-5,\ 0)$
0%
$(0,\ 5)$
0%
$(-\infty ,\ -5)\cup (5,\ \infty )$
0%
$(-5,\ 5)$

$x-2y+k=0$ is a common tangent to

$\displaystyle{ y }^{ 2 }=4x\quad \& \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { 3 } } =1\left( a>\sqrt { 3 } \right)$ , then the value of a, k and other common tangent are given by

0%
$a = 2$
0%
$a = -2$
0%
$x+2y+4 = 0$
0%
$k=4$

Explanation

Given:

$x-2y+k=0$

$\Rightarrow\,2y=x+k$

$\Rightarrow\,y=\dfrac{x}{2}+\dfrac{k}{2}$

Comparing with

$y=mx+c$ we get

$m=\dfrac{1}{2}$ and

$c=\dfrac{k}{2}$

$\therefore\,y=mx+c$ is tangent to

${y}^{2}=4ax$ where

$c=\dfrac{a}{m}$

$\Rightarrow\,{y}^{2}=4x,\,a=1$

$\Rightarrow\,c=\dfrac{1}{m}$

$\Rightarrow\,\dfrac{k}{2}=\dfrac{1}{\dfrac{1}{2}}$

$\Rightarrow\,\dfrac{k}{2}=2$

$\therefore\,k=4$

Substituting

$m=\dfrac{1}{2}$ and

$c=\dfrac{1}{m}=\dfrac{1}{\dfrac{1}{2}}=2$ in

$y=mx+c$ we get

$\Rightarrow\,y=\dfrac{x}{2}+\dfrac{4}{2}$

$\Rightarrow\,y=\dfrac{x}{2}+2$

Given:

$y=mx+c$ is tangent to

$\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$

$\therefore\,$ Required condition is

${c}^{2}={a}^{2}{m}^{2}+{b}^{2}$

$\Rightarrow\,4={a}^{2}{\left(\dfrac{1}{2}\right)}^{2}+3$

$\Rightarrow\,\dfrac{{a}^{2}}{4}=4-3=1$

$\Rightarrow\,{a}^{2}=4$

$\Rightarrow\,a=2$ since

$a>0$

The tangent at any point of the curve

$x={ at }^{ 3 },y={ at }^{ 4 }$ divides the abscissa of the point of contact in the ratio

0%
$1:4$
0%
$3:2$
0%
$1:3$
0%
$3:1$

Explanation

Given:

$x = at^2$

$y = at^4$

Then,

$\Rightarrow t^3 = \dfrac{x}{a}, t^4= \dfrac{y}{a}$

$\Rightarrow t = \left(\dfrac{x}{a}\right)^{1/3}\, t = \left(\dfrac{y}{a}\right)^{1/4}$

$\Rightarrow \left(\dfrac{x}{a}\right)^{1/3} = \left(\dfrac{y}{a}\right)^{1/4}$

$\Rightarrow \left(\dfrac{x}{a}\right)^{\frac{1}{3}\times \frac{4}{12}} = \left(\dfrac{y}{a} \right)^{\frac{1}{4}\times 12}$

$\Rightarrow \dfrac{x^4}{a^4} = \dfrac{y^3}{a^2}$

$\Rightarrow y^3 = \dfrac{x^4}{a}$ at

$P(h,k)$

Also relation

$k^3 = \dfrac{h^4}{a}$

Now by differentiation of after equation we get

$3y^2 \dfrac{dy}{dx} = \dfrac{4x^2}{a}$

$\dfrac{dy}{dx} = \dfrac{4x^3}{3ay^2}$

Now

$M_T = \dfrac{dy}{dx} = \dfrac{4x^3}{3ay^2}$

Now we will find slope of tangent

$M_T / P(h,x) = \dfrac{4h^3}{3ax^2}$

equation of tangent

$P(h,k)$

$y - k = \dfrac{4h^3}{3ak^2}(x-h)$

$\Rightarrow$ Let

$y = 0$

Then,

$-k = \dfrac{4h^3}{3ak^2}(x-h)$

$\Rightarrow \dfrac{-3ak^3}{4h^3} = x - h$

Earlier we found that

$k^3 = \dfrac{h^4}{a}$

Then

$ax^3 = h^4$

$\dfrac{-3h^4}{4h^2} = x-h$

$\Rightarrow \dfrac{-3h}{4}+h = x$

$\Rightarrow x = \dfrac{h}{4}$

$\Rightarrow \dfrac{x}{h} = \dfrac{1}{4}$

Hence option (a)

$1 : 4$ is the correct choice

1496738_1244227_ans_0a282e0fbce94a80bd120e501512f784.JPG

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

0%
${\{\dfrac{\pi}{2}}\}$
0%
${\{\dfrac{\pi}{3}}\}$
0%
$(0, \pi 4)$
0%
$( \pi 4, \pi 2)$

Slope of tangent to the circle

$( x - r ) ^ { 2 } + y ^ { 2 } = r ^ { 2 }$ at the point

$( x , y )$ lying on the circle is

0%
$\frac { x } { y - r }$
0%
$\frac { r - x } { y }$
0%
$\frac { y ^ { 2 } - x ^ { 2 } } { 2 x y }$
0%
$\frac { y ^ { 2 } + x ^ { 2 } } { 2 x y }$

Explanation

$(x-r)^2+y^2=r^2$

$\dfrac{d}{dx}(x-r)^2+\dfrac{d}{dx}(y^2)=\dfrac{d}{dx}(r^2)$

$2(x-r)+2y\dfrac{dy}{dx}=0$

$x-r+y\dfrac{d}{dx}=0$

Therefore the slope of the tangent is,

$\dfrac{dy}{dx}=\dfrac{r-x}{y}$

If the slope of one of the lines represented

${a^3}{x^2} + 2hxy + {b^3}{y^2} = 0$ be the square of the other, then

$ab(a+b)$ is equal to:

0%
$2h$
0%
$-2h$
0%
$8h$
0%
$-8h$

State true or false.

The curves

$y = {x^2} - 3x + 1$ and

$x\left( {y + 3} \right) = 4$ intersect at the right angles at their point of intersection

0%
True
0%
False

Explanation

First curve is given by,

$y={ x }^{ 2 }-3x+1$ (1)

Differentiate w.r.t. x, we get,

$\frac { dy }{ dx } =2{ x }-3$

Thus, slope of tangent to this curve is,

${ m }_{ 1 }=2{ x }-3$

Second curve is given by,

$x\left( y+3 \right) =4$

$\therefore \left( y+3 \right) =\frac { 4 }{ x }$ (2)

Differentiate w.r.t. x, we get,

$\frac { dy }{ dx } =-\frac { 4 }{ { x }^{ 2 } }$

Thus, slope of tangent to this curve is,

${ m }_{ 2 }=-\frac { 4 }{ { x }^{ 2 } }$

1) Point of intersection of two curves-

Equation of first curve is given by,

$y={ x }^{ 2 }-3x+1$

Equation of second curve is given by,

$y=\frac { 4 }{ x } -3$

Thus, we can equate these equations, we get,

${ x }^{ 2 }-3x+1=\frac { 4 }{ x } -3$

Multiply both sides by x, we get,

$\therefore { x }^{ 3 }-3{ x }^{ 2 }+x=4-3x$

$\therefore { x }^{ 3 }-3{ x }^{ 2 }+4x-4=0$

Put

$x=2$ in above equation,

$LHS={ \left( 2 \right) }^{ 3 }-3{ \left( 2 \right) }^{ 2 }+4\left( 2 \right) -4$

$=8-\left( 3\times 4 \right) +8-4$

$=8-12+8-4$

$=0=RHS$

Thus,

$x=2$ is one factor of given equation.

By synthetic division, we can get quadratic equation as,

${ x }^{ 2 }-x+2=0$

$\therefore x=\dfrac { -\left( -1 \right) \pm \sqrt { { \left( -1 \right) }^{ 2 }-4\times 1\times 2 } }{ 2\times 1 }$

$\therefore x=\dfrac { 1\pm \sqrt { 1-8 } }{ 2 }$

$\therefore x=\dfrac { 1\pm \sqrt { -7 } }{ 2 }$

These roots will be imaginary and are neglected.

$\therefore x=2$

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

$y={ \left( 2 \right) }^{ 2 }-3\left( 2 \right) +1$

$\therefore y=4-6+1$

$\therefore y=-1$

Thus, point of intersection of two curves is,

$\left( 2,-1 \right)$

${ \left[ { m }_{ 1 } \right] }_{ \left( 2,-1 \right) }=2\left( 2 \right) -3$

$\therefore { \left[ { m }_{ 1 } \right] }_{ \left( 2,-1 \right) }=1$

${ \left[ { m }_{ 2 } \right] }_{ \left( 2,-1 \right) }=\dfrac { -4 }{ { \left( 2 \right) }^{ 2 } }$

$\therefore { \left[ { m }_{ 2 } \right] }_{ \left( 2,-1 \right) }=\dfrac { -4 }{ 4 } =-1$

Now,

${ m }_{ 1 }\times { m }_{ 2 }=1\times -1$

$\therefore { m }_{ 1 }\times { m }_{ 2 }=-1$

Thus, curves intersect each other at right angle

The slope of the straight line which is both tangent and normal to the curve

$4x^3=27y^2$ is

0%
$\underline { + } 1$
0%
$\underline { + } \dfrac { 1 }{ 2 }$
0%
$\underline { + } \dfrac { 1 }{ \sqrt { 2 } }$
0%
$\underline { + } \sqrt { 2 }$

Explanation

$Given:$

$4{{x}^{3}}=27{{y}^{2}}$

$differentiating\,w.r.t.x$

$12{{x}^{2}}=54yy'$

$t$

$The\text{ line is tangent at}\,\text{P}\left( 3{{t}^{2}},2{{t}^{3}} \right)and\,normal\,at\left( 3t_{1}^{2},2t_{1}^{3} \right)$

$\frac{dy}{dx}=\frac{2\left( 9{{t}^{4}} \right)}{9\left( 2{{t}^{3}} \right)}=1$

$equation\,to\,\tan gent\,P\left( t \right)is$

$y-2{{t}^{3}}=t\left( x-3{{t}^{2}} \right)$

$\Rightarrow tx-y={{t}^{3}}\,\,\,\,\,........\left( i \right)$

$Equation\,to\,normal\,atQ\left( t_{1}^{{}} \right)is$

$y-2t_{1}^{3}=-\frac{1}{t_{1}^{{}}}\left( x-3t_{1}^{2} \right)$

$\Rightarrow x+t_{1}^{{}}y=2t_{1}^{4}+3t_{1}^{2}\,\,\,\,\,\,\,\,...........\left( ii \right)$

$\left( i \right)\,and\left( ii \right)are\,identical$

$\frac{t}{1}=\frac{-1}{t_{1}^{{}}}=\frac{{{t}^{3}}}{2t_{1}^{4}+3t_{1}^{2}}$

$\Rightarrow {{t}^{2}}=\frac{2}{{{t}^{4}}}+\frac{3}{{{t}^{2}}}\,\,\,\,\,\,\,\,or\,\,t_{1}^{{}}=-\frac{1}{t}$

$\Rightarrow {{t}^{6}}=2+3{{t}^{2}}$

$\Rightarrow {{t}^{6}}-3{{t}^{2}}-2=0$

$\Rightarrow {{t}^{2}}=2$

$t=\pm \sqrt{2}$

If the line

$x+y=0$ touches the curve

$2y^2=\alpha x^2+\beta$ at

$(1,-1),$ then

$(\alpha ,\beta )=$

0%
$(-2,4)$
0%
$(-1,3)$
0%
$(4,-2)$
0%
$(2,0)$

Explanation

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,

$2+\beta =2$

$\therefore \beta =0$

$\therefore \left( \alpha ,\beta \right) =\left( 2,0 \right)$

The angle between the curves

$y = \sin x$ and

$y = \cos x$ is

0%
$\tan^{-1}(2\sqrt{2})$
0%
$\tan^{-1}(3\sqrt{2})$
0%
$\tan^{-1}(3\sqrt{3})$
0%
$\tan^{-1}(5\sqrt{2})$

Explanation

C₁ :

$y = \sin x$ , C₂ :

$y = \cos x$ .
Equating the

$y$ ' s,

$\sin x = \cos x$ ∴

$x = \dfrac{\pi}{4}$
∴ curves intersect each other at the point P :

$x = \dfrac{\pi}{4}$ .
Now, differentiating w.r.t.

$x$ ,

C₁ gives :

$\dfrac{dy}{dx} = \cos x$

C₂ gives :

$\dfrac{dy}{dx} = - \sin x$ .

Hence slopes

$m₁ and m₂ of C₁ and C₂ at P :$

$x = \dfrac{π}{4}$

$are m₁$

$= \cos \dfrac{π}{4} = \dfrac{1}{√2}$ $

m₂ =

$- \sin \dfrac{π}{4} = -\dfrac{1}{√2}$ .

$θ$ is the acute angle between them at

$P$ , then

$tan θ =\dfrac{ | ( m₁ - m₂ )|}{ |( 1 + m₁m₂ ) |}$

$=\dfrac{ | ((\dfrac{1}{√2}) - (-\dfrac{1}{√2}))| }{ |( 1 + (\dfrac{1}{√2})(-\dfrac{1}{√2})) | }$

$=\dfrac{ | 2( \dfrac{1}{√2 })|}{ |( \dfrac{(2-1)} { (2)} | }$

$= 2√2$
∴

$θ = \tanֿ¹ ( 2√2 )$

The function

$f(x)=log x$

0%
has maxima at x=e
0%
has minima at x=e
0%
has neither maxima nor minima
0%
all of these

Explanation

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.

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

$c=$

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{4}$
0%
$1$

Explanation

$Let\,the\text{ equation of normal to }{{\text{y}}^{2}}=4ax\,is$

$y=mx-2am-a{{m}^{2}}$

$\therefore equation\,of\,normal\,for\,{{y}^{2}}=x\,is$

$y=mx-\frac{m}{2}-\frac{1}{4}{{m}^{3}}which\,passes\,through\left( c,0 \right)$

$\therefore 0=m\left( c-\frac{1}{2}-\frac{{{m}^{2}}}{4} \right)$

$\Rightarrow m=0\,\,and\,\frac{{{m}^{2}}}{4}=c-\frac{1}{2}$

$\Rightarrow m=\pm 2\sqrt{c-\frac{1}{2}}$

$\text{which gives a normal as x-axis and for other two normal}$

$c-\frac{1}{2}>0$

$\Rightarrow c>\frac{1}{2}$

$N\text{ow, if normals are perpendicular}\text{.}$

$\left( 2\sqrt{c-\frac{1}{2}} \right)\left( -2\sqrt{c-\frac{1}{2}} \right)=-1$

$\Rightarrow c-\frac{1}{2}=\frac{1}{4}$

$\Rightarrow c=\frac{3}{4}$

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

0%
$g(x)$ is increasing $\forall \ x\ \in\ R$
0%
$g(x)$ is increasing only if $x<1$
0%
$g(x)$ is increasing only if $f(x)$ is increasing
0%
$g(x)$ is decreasing $\forall \ x\ \in \ R$

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.)

0%
$\ln \left( \sin \frac { x } { 2 } \right) + C$
0%
$2 \ln \left( \cos \frac { x } { 2 } \right) + C$
0%
$\ln ( 1 - \sin x ) + C$
0%
$\ln ( 1 - \cos x ) + C$

The function

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

0%
$-1 < x$
0%
$x > 1$
0%
$-1 < x$ or $x > 1$
0%
$-1 < x < 1$

The approximate value of

$\sqrt[10]{0.999}$ is

0%
0.0998
0%
0.9998
0%
0.0999
0%
0.9999

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

0%
1001
0%
1876
0%
205
0%
875

Equation of the tangent at (1, -1) to the curve

${ x }^{ 3 }-x{ y }^{ 2 }-4{ x }^{ 2 }-xy+5x+3y+1=0$ is

0%
$x-4y-5=0$
0%
$x+1=0$
0%
$y-1=0$
0%
$y+1=0$

The angle made by the tangent at any point on the curve

$x=a(t+\sin { t } \cos { t } ),y=a{ (1+\sin { t } ) }^{ 2 }$ with x-axis is

0%
$\dfrac { \pi }{ 2 }$
0%
$\dfrac { \pi }{ 4 }$
0%
$\pi +\dfrac { t }{ 2 }$
0%
$\dfrac { \pi }{ 4 } +\dfrac { t }{ 2 }$

Explanation

Given,

$x=a(t+\sin t \cos t)$

$\dfrac{dx}{dt}=\dfrac{d}{dt}a(t+\sin t \cos t)$

$\therefore \dfrac{dx}{dt}=a[1+\cos ^2t-\sin ^2t]=a\left(1+\cos \left(2t\right)\right)=2a\cos ^2t$

$y=a(1+\sin t)^2$

$\dfrac{dy}{dt}=\dfrac{d}{dt}a(1+\sin t)^2$

$\therefore \dfrac{dy}{dt}=2a\left(1+\sin \left(t\right)\right)\cos \left(t\right)$

$\Rightarrow \dfrac{dy}{dx}$

$=\dfrac{2a\left(1+\sin \left(t\right)\right)\cos \left(t\right)}{2a\cos ^2t}$

$=\dfrac {1+\sin t}{\cos t}$

$=\dfrac{\cos \frac{t}{2}+\sin \frac{t}{2}}{\cos \frac{t}{2}-\sin \frac{t}{2}}$

$=\dfrac{1+\tan \frac{t}{2}}{1 -\tan \frac{t}{2}}$

$=\tan \left ( \dfrac{\pi }{4}+\dfrac{t}{2} \right )$

angle made by tangent,

$\tan \theta =\tan \left ( \dfrac{\pi }{4}+\dfrac{t}{2} \right )$

$\Rightarrow \theta =\dfrac{\pi }{4}+\dfrac{t}{2}$

The normal to the curve

$x=\quad a(cos\theta +\theta sin\theta ),\quad y=\quad a(sin\theta -\theta cos\theta )$ at any point

$'\theta '$ is such that

0%
it passes through the origin
0%
it makes an angle $\dfrac { \pi }{ 2 } +\theta$ with the x-axis
0%
it is at a constant distance from the origin
0%
it passes through $\left(a,\dfrac{\pi}{2}\right)$

Explanation

$\dfrac{dy}{dx}=\dfrac{dy}{d\theta}.\dfrac{d\theta}{dx}=\tan{\theta}=$ Slope of tangent

$\therefore$ Slope of normal to the curve

$=-\cot{\theta}=\tan{\left(90+\theta\right)}$

Now, equation of normal to the curve

$\left[y-a\left(\sin{\theta}-\theta\cos{\theta}\right)\right]=\dfrac{-\cos{\theta}}{\sin{\theta}}\left(x-a\left(\cos{\theta}+a\sin{\theta}\right)\right)$

$\Rightarrow x\cos{\theta}+y\sin{\theta}=a(1)$

Now, distance from

$\left(0,0\right)$ to

$x\cos{\theta}+y\sin{\theta}=a$ is

distance

$=d=\dfrac{0+0-a}{1}=a$

$\therefore$ Distance is constant

$=a$

$-4\le x\le 4$ , then critical points of

$f\left( x \right) ={ x }^{ 2 }-6\left| x \right| +4$ are

0%
$3, -2$
0%
$6, -6$
0%
$3, -3, 0$
0%
$0, 1, 3$

Explanation

For

$x\geq 0$ :

$f(x)=x^2-6x+4$

$f'(X)=2x-6$

For critical point

$f'(x)=0$

Hence

$x=3$

For

$x\leq 0$ :

$f(x)=x^2+6x+4$

$f'(X)=2x+6$

For critical point

$f'(x)=0$

Hence

$x=-3$

Also the curve is similar about line

$x=0$ i.e.

$y-axis$

Hence function have a critical point at

$x=0$

So the critical points are

$0,3,-3$

1477155_1328602_ans_835423399c124f72b0359e283e5ad5c9.png

If tangent at any point on the curve

${ y }^{ 2 }=1+{ x }^{ 2 }\ makes\ an\ angle\ \theta$ with positive direction of the x-axis then

0%
$\left| \tan { \theta } \right| >1$
0%
$\left| \tan { \theta } \right| <1$
0%
$\left| \tan { \theta } \right| \ge1$
0%
$\left| \tan { \theta } \right| \le 1$

The tangent to the curve,

$y = xe^{x^2}$ passing through the point

$(1, e)$ also passes through the point:

0%
$\left(\dfrac{4}{3}, 2e\right)$
0%
$(2, 3e)$
0%
$\left(\dfrac{5}{3}, 2e\right)$
0%
$(3, 6e)$

Explanation

$\textbf{Step 1: Find slope of the tangent to the given curve at point (1,e).}$

$\text{Given curve : }\ y=xe^{x^{2}}.........(1)$

$\text{Differentiating equation (1) with respect to }\ x,$

$\Rightarrow \dfrac{dy}{dx} = x\cdot e^{x^2} \cdot 2x+e^{x^2}$

$\text{Slope of tangent at point } {(1,e)} \text{ is }$

$m=\dfrac{dy}{dx}\left|_{(1,e)} = \left(x\cdot e^{x^2} \cdot 2x+e^{x^2}\right)\right|_{(1,e)} = 2\cdot e + e = 3e$

$\textbf{Step 2: Find the equation of the tangent}$

$\textbf{Using one point slope form of equation of the line: } \boldsymbol{ y-y_{1}=m(x-x_{1})}$

$\text{Equation of tangent is: }$

$y - e = 3e (x - 1)$

$\boldsymbol{[\because m=2e]}$

$\Rightarrow y = 3ex - 3e + e$

$\Rightarrow y = (3e) x - 2e............(2)$

$\textbf{Step 3: To find correct answer, check by given options}$

$\left(\dfrac{4}{3}, 2e\right)$

$\text{lies on it, as putting this point in (2) we get,}$

$2e=3e\times \dfrac 43-2e\Rightarrow 2e=2e$

$\Rightarrow \text{L.H.S = R.H.S}$

$\textbf{Hence, the correct answer is option 'A'.}$

The equation of the normal to the curve

$y=\left( 1+x \right) ^{ y }+{ sin }^{ -1 }\left( { sin }^{ 2 }x \right) at\quad x=0$ is

0%
x+y=1
0%
x-y+1=0
0%
2x+y=2
0%
2x-y+1

Explanation

Given,

$y=(1+x)^y+\sin ^{-1}(\sin ^2x)$

when

$x=0$

$y=1^y+0=1$

$\therefore (x_1,y_1)=(0,1)$

$\dfrac{dy}{dx}=(1+x)^y\left [ \ln (1+x)\dfrac{dy}{dx}+\dfrac{y}{1+x} \right ]+\dfrac{2\sin x\cos x}{\sqrt{1-\sin ^4x}}$

$\dfrac{dy}{dx}_{(0,1)}=1[1]+0$

$\therefore \dfrac{dy}{dx}_{(0,1)}=1$

Therefore, slope of normal is

$-1$

Equation of normal,

$y-y_1=\dfrac{-1}{\frac{dy}{dx}}(x-x_1)$

$y-1=(-1)(x-0)$

$y-1=-x$

$x+y=1$ is the equation of normal

If the function

$f ( x ) = 2 x ^ { 2 } - k x + 5$ is increasing in

$[ 1,2 ]$ , then '

$k$ ' lies in the interval

0%
$( - \infty , 4 )$
0%
$( 4, \infty )$
0%
$( - \infty , 8 ]$
0%
$( 8 , \infty )$

A particle moves along a line by

$s = \dfrac {1}{3} t^{3} - 3t^{2} + 8t + 5$ , it changes its direction when

0%
$t = 1, t = 2$
0%
$t =2, t = 4$
0%
$t =0, t = 4$
0%
$t = 2, t = 3$

Explanation

A particle changes its direction, when, velocity is equal to zero

Given,

$s=\dfrac{1}{3}t^3-3t^2+8t+5$

Velocity,

$v=\dfrac{ds}{dt}=\dfrac{1}{3}(3t^2)-6t+8$

$\therefore v=t^2-6t+8$

when

$v=0$ particle changes its direction,

$\Rightarrow t^2-6t+8=0$

$(t-2)(t-4)=0$

$\therefore t=2,4$

Length of the normal to the curve at any point on the curve

$y=\dfrac { a\left( { e }^{ x/a }+{ e }^{ -x/a } \right) }{ 2 }$ varies as

0%
$x$
0%
${ x }^{ 2 }$
0%
$y$
0%
${ y }^{ 2 }$

Explanation

Given,

$y=\dfrac{a(e^{\frac{x}{a}}+e^{-\frac{x}{a}})}{2}$

$\Rightarrow y=a\cos h\left ( \dfrac{x}{a} \right )$

$\dfrac{dy}{dx}=a \sin h\left ( \dfrac{x}{a} \right )\dfrac{1}{a}$

$=\sin h\left ( \dfrac{x}{a} \right )$

Length of normal

$=y\left [ 1+\left (\dfrac{dy}{dx} \right )^2 \right ]^{\frac{1}{2}}$

$=a\cos h\left ( \dfrac{x}{a} \right )\left [ 1+ \sin ^2h\left ( \dfrac{x}{a} \right )\right ]^{\frac{1}{2}}$

$=a\cos h\left ( \dfrac{x}{a} \right )\left [ \cos ^2h\left ( \dfrac{x}{a} \right ) \right ]^{\frac{1}{2}}$

$=a\cos ^2h\left ( \dfrac{x}{a} \right )$

$=a \times \dfrac{y^2}{a^2}$

$=\dfrac{y^2}{a}$

Function

$f(x)=\dfrac { \lambda sinx+6cosx }{ 2sinx+3cosx }$ is monotonic increasing If

0%
$\lambda >4$
0%
$\lambda >1$
0%
$\lambda <4$
0%
$\lambda <2$

For x greater than or equal to zero and less than or equal to

$2\pi$ ,

$\sin x$ and

$\cos x$ are both decreasing on the intervals

0%
$(0 , \dfrac{\pi}{2})$
0%
$(\dfrac{\pi}{2}, \pi)$
0%
$(\pi ,3\dfrac{\pi}{2})$
0%
$(\dfrac{3\pi} 2 , 2\pi)$

Explanation

$\textbf{Interval in which given function is defined by is given by,}$

$\text{We have the given interval }$

$0\leq x \leq2\pi$

$\text{ for both}$

$\sin x$

$\text{and}$

$\cos x$ .

$\text{Let}$

$f(x)= \sin x$

$\Rightarrow f'(x) = \cos x$ ,

$\text{which is}$

$\leq 0$

$\text{in interval}$

$\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$

$\Rightarrow \sin x\ \text{is decreasing in the interval}$

$\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$

$\text{Similarly, suppose}$

$\boldsymbol g(x)=\cos x$

$\text{Then}$

$g'(x)=-\sin x$

$\Rightarrow g'(x)\leq 0$

$\text {for}$

$\cos x$

$\text{to be decreasing in given interval.}$

$\Rightarrow -\sin x\leq 0$

$\text{in interval}$ [

$0, \pi$ ]

$\Rightarrow g'(x)\leq 0$

$\text{in interval}$ [

$0, \pi$ ]

$\Rightarrow \cos x\ \text{is decreasing in the interval}$ [

$0, \pi$ ]

$\text{Intersection will happen at a point x,}$

$x\boldsymbol\in\left[\dfrac{\pi}{2}, \pi\right]$

$\textbf{Hence the option B is correct.}$

$f(x)=\cfrac{x}{5}+\cfrac{5}{x}$ is increaing in

0%
$(-5,0)$
0%
$(0,5)$
0%
$(-\infty,-5)\cup(5,\infty)$
0%
$(-5,5)$

Explanation

Given,

$f(x)=\dfrac{x}{5}+\dfrac{5}{x}$

$\Rightarrow x=\dfrac{y}{5}+\dfrac{5}{y}$

$x\cdot \:5y=y^2+25$

$y^2-x\cdot \:5y+25=0$

$y=\dfrac{-\left(-x5\right)\pm \sqrt{\left(-x5\right)^2-4\cdot \:1\cdot \:25}}{2\cdot \:1}$

$y=\dfrac{5x \pm 5\sqrt{x^2-4}}{2}$

$x\le \:-2\quad \mathrm{or}\quad \:x\ge \:2$

$\therefore f\left(x\right)\le \:-2\quad \mathrm{or}\quad \:f\left(x\right)\ge \:2$

and

$\mathrm{Increasing}:-\infty \:<x<-5,\:\mathrm{Decreasing}:-5<x<0,\:\mathrm{Decreasing}:0<x<5,\:\mathrm{Increasing}:5<x<\infty \:$

$\therefore (-\infty ,-5)\cup (5,\infty)$

The complex number

$\frac{(-\sqrt 3 + 3i)(1-i)}{(3 + \sqrt 3i) (i) (\sqrt 3 + \sqrt 3i)}$ when represented in the Argand diagram is

0%
in the second quadrant
0%
in the first quadrant
0%
on the y-axis (imaginary axis)
0%
on the x-axis (real axis)

If the line

$ax+y=c$ , touches both the curves

${x}^{2}+{y}^{2}=1$ and

${y}^{2}=4\sqrt{2}x$ , then

$\left| c \right|$ is equal to:

0%
$1/2$
0%
$2$
0%
$\sqrt{2}$
0%
$\cfrac { 1 }{ \sqrt { 2 } }$

Explanation

Tangent to

${y}^{2}=4\sqrt{2}x$ is

$y=mx+\cfrac { \sqrt { 2 } }{ m }$ is also tangent to

${x}^{2}+{y}^{2}=1$

$\Rightarrow \left| \cfrac { \sqrt { 2 } /m }{ \sqrt { 1+{ m }^{ 2 } } } \right| =1\Rightarrow m=\pm 1$

Tangent will be

$y=x+\sqrt{2}$ or

$y=-x-\sqrt{2}$

compare with

$y=-ax+c$

$\Rightarrow a=\pm 1;c=\pm \sqrt { 2 }$

$\therefore |c|=\sqrt 2$

Let

$f"(x) > 0$ and

$\phi (x) = f(x) + f(2 - x) , x \in (0,2)$ be a function, then the function

$\phi (x)$ is

0%
increasing in $(0, 1)$ and decreasing $(1, 2)$
0%
decreasing in $(0,1)$ and increasing $(1, 2)$
0%
increasing in $(0, 2)$
0%
decreasing in $(0, 2)$

Explanation

$f"(x) > 0, y = f(x) ; x \in (0, 2)$

$\phi (x) = f(x) + f(2 - x)$

$\phi ' (x) = f' (x) - f' (2 - x)$
for

$\phi (x)$ to be increasing

$\phi' (x) > 0 \Rightarrow f'(x) > f'(2 - x)$

$\Rightarrow x > 2 - x$ (

$f'(x)$ is increasing in

$(0, 2)$ )

$\Rightarrow x > 1$

$\Rightarrow x \in (1, 2)$
For

$\phi (x)$ to be decreasing

$\phi ' (x) = 0 \Rightarrow f'(x) < f'(2 - x)$

$\therefore x \in (0, 1)$

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 8 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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