MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 4

The curve given by the equation

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

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

Explanation

$y-e^{xy}+x=0$

$\therefore \dfrac{dy}{dx}-e^{xy}\left ( x\dfrac{dy}{dx}+y \right )+1=0$

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

$\therefore$ if the tangent is vertical at

$(x,y),xe^{xy}=1$ 1 which is possible only if

$x=1,y=0.$

The sum of the intercepts made on the axes of coordinates by any tangent to the curve $\sqrt{x}+\sqrt{y}=2$ is equal to

0%
$4$
0%
$2$
0%
$8$
0%
None of these

Explanation

Curve equation $\sqrt{x}+\sqrt{y}=2 \Rightarrow x > 0, y>0$
$\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0$
$\dfrac{dy}{dx}=-\sqrt{\dfrac {y}{x}}$
and since in question its asking in general not for a particular point on curve. Hence it we find it for any point on curve then it will be true for any point.
So as use can see (1, 1) lies on curve
So, $\dfrac{dy}{dx} \int_{1, 1}=-1$
So tangent equation at (1, 1) is $y= (-1)x+c$
and it will pass through (1, 1)
So, $c=2$
tangent equation : $y= -x +2$
$x+y=2$
$\dfrac{x}{2}+\dfrac{y}{2} =1$
So x - intercept $=2$
y- intercept $=2$
So sum $=4$

If the parabola

$y = ax^2 - 6x + b$ passes through

$(0, 2)$ and has its tangent at

$x = \dfrac{3}{2}$ parallel to the

$x-$ axis then

0%
$a = 2, b = -2$
0%
$a = 2, b = 2$
0%
$a = - 2, b = 2$
0%
$a = - 2, b = - 2$

Explanation

Given equation of parabola is

$y = ax^2 - 6x + b$
Since it passes through

$(0, 2)$

$2 = 0 + b$

$\Rightarrow b = 2$
Also,

$\displaystyle \frac{dy}{dx} = 2 ax - 6$
So, slope of tangent at

$\displaystyle (x=\frac {3}{2}) = 2a \left ( \frac{3}{2} \right ) - 6$

$= 3a - 6$
Since the tangent at

$x=\dfrac{3}{2}$ is parallel to

$x-$ axis

$\Rightarrow 3a-6=0$

$\Rightarrow a = 2$
Hence,

$a=2,b=2$

At what points of curve

$y= \displaystyle \frac {2}{3} x^3 + \displaystyle \frac{1}{2} x^2$ , the tangent makes equal angle with the axis

0%
$\left( \displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 5 }{ 24 } \right)$ and $\left( -1,-\displaystyle \frac { 1 }{ 6 } \right)$
0%
$\left(\displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 4 }{ 9 } \right)$ and $\left( -1,0 \right)$
0%
$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 1 }{ 7 } \right)$ and $\left( -3,\displaystyle \frac { 1 }{ 2 } \right)$
0%
$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 4 }{ 47 } \right)$ and $\left( -1,-\displaystyle \frac { 1 }{ 3 } \right)$

Explanation

Given equation of curve is

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

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

Since ,the tangent makes equal angle with x-axis,

$\dfrac{dy}{dx}=1$

$\Rightarrow 2x^{2}+x-1=0$

$\Rightarrow x=-1,\dfrac{1}{2}$

$x=-1\Rightarrow y=-\dfrac{1}{6}$

$x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{24}$

The point on the curve

$3y=6x-5x^3$ , the normal at which passes through the origin is

0%
$\left( 1,\displaystyle \frac { 1 }{ 3 } \right)$
0%
$\left( \displaystyle \frac { 1 }{ 3 } ,1 \right)$
0%
$\left( 2,\displaystyle \frac { -28 }{ 3 } \right)$
0%
none of these

Explanation

Let

$P(x_1,y_1)$ be the required point .

Given equation of curve is

$3y=6x-5x^{3}$

$\dfrac{dy}{dx}=2-5x^{2}$

Slope of tangent at P is

$2-5{x_{1}}^{2}$

Slope of normal at P is

$\displaystyle\dfrac{1}{5{x_{1}}^{2}-2}$

Equation of normal at P is

$y-y_{1}=\dfrac{1}{5{x_{1}}^{2}-2}(x-x_1)$

Since it passes through origin,

$\displaystyle\Rightarrow y_{1}=\dfrac{x_1}{5{x_{1}}^{2}-2}$

Here , we can solve by check options.

So ,option A satisfies above equation.

Hence

$P(1,\dfrac{1}{3})$ is the required point.

If at each point of the curve

$y=x^3-ax^2 +x+1$ , the tangent is inclined at an acute angle with the positive direction of the

$x$ -axis, then

0%
$a>0$
0%
$a\leq \sqrt 3$
0%
$-\sqrt 3 \leq a \leq \sqrt 3$
0%
none of these

Explanation

$y=x^3-ax^2 +x+1$

$\dfrac{dy}{dx}=3x^{2}-2ax+1$
Since, the tangent is inclined at an acute angle with the positive direction of x-axis.

$\dfrac{dy}{dx}>0$

$\Rightarrow 3x^{2}-2ax+1>0$

$\Rightarrow D<0$

$\Rightarrow 4(a^2-3)<0$

$\Rightarrow (a-\sqrt{3})(a+\sqrt{3})<0$

$\Rightarrow -\sqrt{3}<a<\sqrt{3}$

Consider a curve

$y=f(x)$ in

$xy$ - plane. The curve passes through

$(0,0)$ and has the property that a segment of tangent drawn at any point

$P(x, f(x))$ and the line

$y=3$ gets bisected by the line

$x+y=1$ , then the equation of the curve is

0%
$y^2=9(x-y)$
0%
$(y-3)^2=9(1-x-y)$
0%
$(y+3)^2=9(1-x-y)$
0%
$(y-3)^2-9(1+x+y)$

The value of

$'c'$ such that the line joining

$(0,3),(5,-2)$ is a tangent to the curve

$y=\dfrac{c}{x+1}$

0%
$4$
0%
$3$
0%
$23$
0%
$1$

Explanation

Given equation of curve is

$y=\dfrac{c}{x+1}$

$\displaystyle \frac{dy}{dx}=-\dfrac{c}{(x+1)^{2}}$
Slope of tangent to curve

$=-\dfrac{c}{(x+1)^{2}}$
Equation of tangent line joining (0,3) and (5,-2) is

$y=-x+3$ .

$\Rightarrow -\dfrac{c}{(x+1)^{2}}=-1$

$\Rightarrow c=(x+1)^{2}$
Equation of curve becomes

$y=x+1$

$-x+3=x+1$

$\Rightarrow x=1$

$\Rightarrow c=4$

The equation of the tangent to the curve

$y=be^{-x/a}$ at the point where it crosses the

$y$ -axis is

0%
$\displaystyle \frac {x}{a}-\displaystyle \frac{y}{b} =1$
0%
$ax+by=1$
0%
$ax-by=1$
0%
$\displaystyle \frac {x}{a}+\displaystyle \frac{y}{b} =1$

Explanation

Given equation of curve is

$y=be^{-x/a}$

Let

$P(0,y_1)$ be the point where tangent to curve crosses y-axis .

$\Rightarrow y_1=b$

So, the point P is (0,b)

$\dfrac{dy}{dx}=-\dfrac{y}{a}$

Slope of tangent at P(0,b) is

$-\dfrac{b}{a}$

Equation of tangent through P(0,b) is

$y-b=-\dfrac{b}{a}(x-0)$

$\Rightarrow bx+ay=ab$

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

The distance between the origin and the tangent to the curve

$y = e^{2x} + x^{2}$ drawn at the point

$x = 0$ is

0%
$\dfrac {1}{\sqrt 5}$
0%
$\dfrac {2}{\sqrt 5}$
0%
$\dfrac {-1}{\sqrt 5}$
0%
$\dfrac {2}{\sqrt 3}$

Explanation

$y=e^{2x} + x^2$
At

$x=0 \Rightarrow y=1$

$\displaystyle \frac{dy}{dx}=2e^{2x}+2x$
Slope of tangent at (0,1)

$=2$
Equation of tangent through (0,1) is

$y-1=2(x-0)$

$\Rightarrow y-2x-1=0$
Distance of tangent from origin

$\displaystyle=|\frac{-1}{\sqrt{1+4}}|=\frac{1}{\sqrt{5}}$

The slope of the tangent to the curve

$y= \sqrt {4-x^2}$ at the point where the ordinate and the abscissa are equal is

0%
-1
0%
1
0%
0
0%
none of these

Explanation

Given equation of curve is

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

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

$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{x}{y}$
Let

$P(x_1,y_1)$ be the point on the curve such that

$x_1=y_1$
So, slope of tangent to curve at P is

$-1$ .

The curve given by

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

0%
(0,1)
0%
(1,0)
0%
(1,1)
0%
none of these

Explanation

Given equation of curve is

$x+y=e^{xy}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{y-1}{1-xe^{xy}}$
Since, the tangent is parallel to y-axis,

$\displaystyle\frac{dy}{dx}=\frac{1}{0}$

$\Rightarrow xe^{xy}=1$
Also,

$y=0$ on the line parallel to y-axis

$\Rightarrow x=1$
Hence, the required point is (1,0)

The normal to the curve

$2x^2 +y^2=12$ at the point

$(2,2)$ cuts the curve again at

0%
$\left (-\displaystyle \frac {22}{9}, -\displaystyle \frac {2}{9}\right )$
0%
$\left (\displaystyle \frac {22}{9}, \displaystyle \frac {2}{9}\right )$
0%
$(-2,-2)$
0%
none of these

Explanation

Given equation of curve is

$2x^2 +y^2=12$

$\dfrac{dy}{dx}=-\dfrac{2x}{y}$

Slope of tangent at (2,2) is -2.

Slope of normal at (2,2) is

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

Equation of normal at (2,2) is

$y-2=\dfrac{1}{2}(x-2)$

$\Rightarrow 2y-x=2$

Here , we can solve by checking options.

So ,option A satisfies above equation.

Hence

$(-\dfrac{22}{9},-\dfrac{2}{9})$ is the required point.

If a variable tangent to the curve

$x^2y=c^3$ makes intercepts

$a , b$ on

$X$ -axes and

$Y$ - axes, respectively, then the value of

$a^2b$ is

0%
$27c^3$
0%
$\displaystyle \frac {4}{27}c^3$
0%
$\displaystyle \frac {27}{4}c^3$
0%
$\displaystyle \frac {4}{9} c^3$

Explanation

Let a variable point

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ on the curve

${ x }^{ 2 }y={ c }^{ 3 }$
Then slope

$\displaystyle\frac { dy }{ dx } =-\frac { 2{ y }_{ 1 } }{ { x }_{ 1 } }$
So, equation of tangent is

$\displaystyle y-{ y }_{ 1 }=-\frac { 2{ y }_{ 1 } }{ { x }_{ 1 } } \left( x-{ x }_{ 1 } \right)$

$\Rightarrow y{ x }_{ 1 }-2x{ y }_{ 1 }=3{ x }_{ 1 }{ y }_{ 1 }$
Using

${ { x }_{ 1 } }^{ 2 }{ y }_{ 1 }={ c }^{ 3 }$
Equation of tangent becomes

$\displaystyle y{ x }_{ 1 }^{ 2 }+\frac { 2{ c }^{ 3 }x }{ { x }_{ 1 } } =3{ c }^{ 3 }$
Therefore, X intercept is

$\displaystyle a=\frac { 3{ x }_{ 1 } }{ 2 }$
And Y intercept is

$\displaystyle b=\frac { 3{ c }^{ 3 } }{ { { x }_{ 1 } }^{ 2 } }$
Now

$\displaystyle { a }^{ 2 }b={ \left( \frac { 3{ x }_{ 1 } }{ 2 } \right) }^{ 2 }\left( \frac { 3{ c }^{ 3 } }{ { { x }_{ 1 } }^{ 2 } } \right) =\frac { 27{ c }^{ 3 } }{ 4 }$
Hence, option 'C' is correct.

$y = 4x - 5$ is a tangent to the curve

$\displaystyle y^2 = px^3 + q$ at

$(2, 3)$ , then

0%
$p = 2, q = -7$
0%
$p = -2, q = 7$
0%
$p = -2, q = -7$
0%
$p = 2, q = 7$

Explanation

Given equation of curve

$\displaystyle y^2 = px^3 + q\\$

$\displaystyle \dfrac{dy}{dx}=\dfrac{3px^{2}}{y}\\$
Slope of tangent at (2,3)

$= \displaystyle (\dfrac{dy}{dx})_{(2,3)}=2p$

Given equation of tangent is

$y=4x-5$
Slope of tangent =4

$\Rightarrow 2p=4$

$\Rightarrow p=2$

Since, (2,3) lies on

$y^{2} = px^{3} + q$

$9=16+q$

$\Rightarrow q=-7$

The point(s) at each of which the tangents to the curve

$\displaystyle y = x^3 - 3x^2 - 7x + 6$ cut off on the positive semi axis

$OX$ a line segment half that on the negative semi axis

$OY$ , then the co-ordinates of the point(s) is/are give by:

0%
$(-1, 9)$
0%
$(3, -15)$
0%
$(1, -3)$
0%
none

Explanation

$y={ x }^{ 3 }-3{ x }^{ 2 }-7x+6\\ \dfrac { dy }{ dx } =3{ x }^{ 2 }-6x-7$
Let coordinate of X be

$\left( a,0 \right)$ then coordinate of Y is

$\left( 0,-2a \right)$
Therefore slope of XY is

$\cfrac { 0+2a }{ a-0 } =2$
Equating slope

$3{ x }^{ 2 }-6x-7=2\\ 3{ x }^{ 2 }-6x-9=0$
Solving we get

$x=-1,y=9\quad \\ x=3,y=-15$
Hence, option 'B' is correct.

If the tangent to the curve

$xy + ax + by = 0$ at (1, 1) makes an angle

$\displaystyle \tan ^{-1}(2)$ with x-axis, then

$\displaystyle a + 2b$ is equal to

0%
$\displaystyle \frac {1}{2}$
0%
$\displaystyle - \frac {1}{2}$
0%
$3$
0%
$-3$

Explanation

$xy'+y+a+by'=0$

$(x+b)y'+a+y=0$

$y'=\dfrac{-(a+y)}{x+b}$
Now

$y'_{1,1}=\dfrac{-(a+1)}{b+1}$

$=2$
Or

$2b+2=-a-1$

$2b+a=-3$

A line L is perpendicular to the curve

$\displaystyle y = \dfrac {x^2}{4} - 2$ at its point P and passes through (10, -1). The coordinates of the point P are

0%
(2, -1)
0%
(6, 7)
0%
(0, -2)
0%
(4, 2)

Explanation

Let point

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ on curve

$y=\cfrac { { x }^{ 2 } }{ 4 } -2$
Slope of tangent is

$\dfrac { dy }{ dx } =\dfrac { { x }_{ 1 } }{ 2 }$
So slope of perpendicular line is

$-\dfrac { 2 }{ { x }_{ 1 } }$ ...(1)
That perpendicular line also passes through

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( 10,-1 \right)$
Slope of line from these point is

$\dfrac { { y }_{ 1 }+1 }{ { x }_{ 1 }-10 }$ ...(2)
Equating (1) and (2) and solving we get,

${ x }_{ 1 }=4$ and

${ y }_{ 1 }=2$
Thus point is

$\left( 4,2 \right)$
Hence, option 'D' is correct.

If the curve

$y=ax^3+bx^2+cx$ is inclined at

$45^0$ to the x-axis at

$(0,0)$ but it touches the x-axis at

$(1,0)$ then the value of

$a,b$ and

$c$ are given by

0%
$a=-1,b=2,c=1$
0%
$a=1,b=-2,c=1$
0%
$a=1,b=1,c=-2$
0%
$a=-2,b=1,c=1$

Explanation

$\displaystyle \frac { dy }{ dx } =3a{ x }^{ 2 }+2bx+c$ (given)

$\displaystyle \left( 0,0 \right) ,\frac { dy }{ dx } =c=\tan { { 45 }^{ 0 } } =1\Rightarrow c=1$

$\displaystyle \left( 1,0 \right) ,\frac { dy }{ dx } =3a+2b+c$

Tangent at

$(1,0)$ is

$y-0=(3a+2b+c).(x-1)$

This is x-axis if

$3a+2b+c=0$

which is true if

$a=1,b=-2$

Hence

$a=1,b=-2,c=1$

If the tangent at each point of the curve

$\displaystyle y=\frac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$ makes an acute angle with the positive direction of x-axis, then

0%
$a\ge 1$
0%
$-1\le a\le 1$
0%
$a\le -1$
0%
none of these

Explanation

We have,

$\displaystyle y=\dfrac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =2{ x }^{ 2 }-4ax+2.$

Since, the tangent makes an acute angle with the positive direction of x-axis, therefore,

$\displaystyle \dfrac { dy }{ dx } \ge 0\Rightarrow 2{ x }^{ 2 }-4ax+2\ge 0$ for all

$x$

$\Rightarrow 16{ a }^{ 2 }-16\le 0$

$(\because$ Disc.

$={ \left( 4a \right) }^{ 2 }-4\left( 2 \right) \left( 2 \right) \le 0)$

$\Rightarrow { a }^{ 2 }-1\le 0$ i.e.,

$\left( a-1 \right) \left( a+1 \right) \le 0$

$\Rightarrow -1\le a\le 1.$

The angle formed by the positive

$y-axis$ and the tangent to

$y=x^2+4x-17$ at

$(5/2,-3/4)$ is

0%
${\tan^{-1}(9)}$
0%
$\displaystyle \frac{\pi}{2}-{\tan^{-1}(9)}$
0%
$\displaystyle \frac{\pi}{2}+{\tan^{-1}(9)}$
0%
none of these

Explanation

Slope of tangent to given curve at

$\dfrac{5}{2},-\dfrac{3}{4}$ is 9.

$\theta$ is the angle between the tangent and x-axis , then

$\theta=\pm tan^{-1}{9}$

Angle between y-axis and x-axis is

$\dfrac{\pi}{2}$

So, the angle between y-axis and tangent is

$\dfrac{\pi}{2} \pm tan^{-1}{9}$

The coordinates of the point(s) on the graph of the function

$f(x)=\displaystyle \frac {x^3}{3}-\displaystyle \frac{5x^2}{2}+7x-4$ , where the tangent drawn cuts off intercepts from the coordinate axes which are equal in magnitude but opposite in sign, are

0%
$(2,8/3)$
0%
$(3,7/2)$
0%
$(1,5/6)$
0%
none of these

Explanation

$f(x)=\dfrac { x^{ 3 } }{ 3 } -\dfrac { 5x^{ 2 } }{ 2 } +7x-4$
since, tangent have intercepts equal in magnitude but opposite in sign
then let equation of tangent be

$x-y=a$
Slope of tangent

$=1=f'(x)={ x }^{ 2 }-5x+7$

$\Rightarrow { x }^{ 2 }-5x+6=0\quad \quad \quad \Rightarrow x=2,3$
Therefore, co-ordinates are

$(2,f(2))=(2,8/3)$ &

$(3,f(3))=(3,7/2)$

Ans: A,B

If a variable tangent to the curve

$\displaystyle x^2y = c^3$ makes intercepts a, b on x and y axis respectively, then the value of

$\displaystyle a^2b$ is

0%
$\displaystyle 27 c^3$
0%
$\displaystyle \frac {4}{27} c^3$
0%
$\displaystyle \frac {27}{4} c^3$
0%
$\displaystyle \frac {4}{9} c^3$

Explanation

Let a variable point

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ on the curve

${ x }^{ 2 }y={ c }^{ 3 }$
Then slope

$\cfrac { dy }{ dx } =-\cfrac { 2{ y }_{ 1 } }{ { x }_{ 1 } }$
So, equation of tangent is

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

$y{ x }_{ 1 }-2x{ y }_{ 1 }=3{ x }_{ 1 }{ y }_{ 1 }$
Using

${ { x }_{ 1 } }^{ 2 }{ y }_{ 1 }={ c }^{ 3 }$
Equation of tangent becomes

$y{ x }_{ 1 }^{ 2 }+\cfrac { 2{ c }^{ 3 }x }{ { x }_{ 1 } } =3{ c }^{ 3 }$
Therefore, X intercept is

$a=\cfrac { 3{ x }_{ 1 } }{ 2 }$
And Y intercept is

$b=\cfrac { 3{ c }^{ 3 } }{ { { x }_{ 1 } }^{ 2 } }$

Now

${ a }^{ 2 }b={ \left( \cfrac { 3{ x }_{ 1 } }{ 2 } \right) }^{ 2 }\left( \cfrac { 3{ c }^{ 3 } }{ { { x }_{ 1 } }^{ 2 } } \right) =\cfrac { 27{ c }^{ 3 } }{ 4 }$

The line

$ax + by = 1$ is tangent to the curve

$\displaystyle ax^2 + by^2 = 1$ , if

$(a, b)$ can be equal to

0%
$\displaystyle (\frac {1}{2}, \frac {1}{2})$
0%
$\displaystyle (\frac {1}{4}, \frac {3}{4})$
0%
$\displaystyle (\frac {1}{2}, \frac {3}{4})$
0%
$\displaystyle (\frac {1}{4}, \frac {1}{2})$

Explanation

$\displaystyle ax^2 + by^2 = 1$

$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{ax}{by}$
Slope of tangent at

$(x_1, y_1)$ is

$-\dfrac{ax_1}{by_1}$

Equation of tangent is

$ax+by=1$
Slope of tangent is

$\displaystyle -\frac{a}{b}$

$\Rightarrow \displaystyle -\frac{a{x}_{1}}{b{y}_{1}}=-\frac{a}{b}$

$\Rightarrow x_{1}=y_{1}$

substitute in

$ax^2+by^2=1, ax+by=1$

$\Rightarrow x_1=y_1 = \dfrac1{a+b}$ and we get

$a+b=1$

The slope of the tangent to the curve

$y=x^{2}-x$ at the point where the line

$y=2$ cuts the curve in the first quadrant is

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

Explanation

For

$y=2, { x }^{ 2 }-x-2=0$ gives

$x=-1$ and

$x=2$ as point is in first quadrant.
Therefore point is

$\left( 2,2 \right)$
Now

$\cfrac { dy }{ dx } =2x-1=4-1=3$

The slope of the tangent to the locus

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

$x=\displaystyle \frac{\pi }{4}$ is

0%
$1$
0%
$0$
0%
$2$
0%
$-1$

Explanation

For

$x=\cfrac { \pi }{ 4 } \quad y=\cos^{ -1 }\cos\left( \cfrac { \pi }{ 4 } \right) =\cfrac { \pi }{ 4 }$

Therefore point is

$\left( \cfrac { \pi }{ 4 } ,\cfrac { \pi }{ 4 } \right)$

Now

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

Slope is

$\cfrac { dy }{ dx } =\cfrac { -\sin\left( x \right) }{ -\sin\left( y \right) } =1$

The point of contact of a tangent from the point

$(1, 2)$ to the circle

$x^{2}+y^{2}=1$ has the coordinates

0%
$(1,0)$
0%
$(-1,0)$
0%
$\displaystyle (\frac{3}{5},-\frac{4}{5})$
0%
$\displaystyle (-\frac{3}{5},\frac{4}{5})$

Explanation

Let the point be

$(h,k)$
Hence

$h^{2}+k^{2}=1$ ...(i)

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

$\dfrac{dy}{dx}_{h,k}=\dfrac{-h}{k}$
Hence

$\dfrac{y-k}{x-h}=\dfrac{-h}{k}$

$ky-k^{2}=-hx+h^{2}$

$ky+hx=h^{2}+k^{2}$

$ky+hx=1$ ...(i)
This is the equation of tangent.
Since it passes through

$(1,2)$ hence the point must satisfy the above equation.
Hence

$2k+h=1$ ...(ii)
Now considering the above options, only

$A$ and

$C$ satisfy equation ii as well as the equation of the given circle.
Hence answers are Option A and Option C.

Let

$\displaystyle f(x)=e\:^{x}\sin x$ be the equation of a curve. If at

$\displaystyle x=a,0\leq a\leq 2\pi$ , the slope of the tangent is the maximum then the value of

$a$ is

0%
$\displaystyle \pi /2$
0%
$\displaystyle 3\pi /2$
0%
$\displaystyle \pi$
0%
$\displaystyle \pi /4$

Explanation

Let

$y={ e }^{ x }sin\left( x \right)$
Slope is

$\cfrac { dy }{ dx } ={ e }^{ x }\left( sin\left( x \right) +cos\left( x \right) \right)$
To maximize slope

$\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ 2e }^{ x }cos\left( x \right) =0$
Critical point is

$x=\cfrac { \pi }{ 2 }$
And at

$x=\cfrac { \pi }{ 2 }$ is the point of maxima

If at each point of the curve

$y=x^{3}-ax^{2}+x+1$ the tangent is inclined at an acute angle with the positive direction of the x-axis then

0%
$4a>0$
0%
$a\leq \sqrt{3}$
0%
$-\sqrt{3}\leq a\leq \sqrt{3}$
0%
none of these

The slope of the tangent to the curve

$y=\sqrt{4-x^{2}}$ at the point where the ordinate and the abscissa are equal, is

0%
$-1$
0%
$1$
0%
$0$
0%
none of these

Explanation

Let abscissa = ordinate =

$a$ , then point is

$\left( a,a \right)$
Now for

$y=\sqrt { 4-{ x }^{ 2 } } \Rightarrow { y }^{ 2 }=4-{ x }^{ 2 }$
Hence slope is

$\cfrac { dy }{ dx } =\cfrac { -2x }{ 2y } =\cfrac { -2a }{ 2a } =-1$

The equation of the curve is given by

$x=e^{t}\sin t$ ,

$y=e^{t}\cos t$ . The inclination of the tangent to the curve at the point

$t=\displaystyle \frac{\pi }{4}$ is

0%
$\displaystyle \frac{\pi }{4}$
0%
$\displaystyle \frac{\pi }{3}$
0%
$\displaystyle \frac{\pi }{2}$
0%
$0$

Explanation

$dx=e^{t}(\cos t+\sin t).dt$
$dy=e^{t}[\cos t-\sin t).dt$
Hence
$\dfrac{dy}{dx}_{t=\tfrac{\pi}{4}}$

$=\dfrac{\cos t-\sin t}{\cos t+\sin t}_{t=\tfrac{\pi}{4}}$

$=0$

The triangle by the tangent to the curve

$f(x) = x^{2} bx -b$ at the point (1, 1) and the co-ordinate axes lies in the first quadrant. If its area is 2, then the value of b is

0%
-1
0%
3
0%
-3
0%
1

If the tangent to the curve

${ 2y }^{ 3 }={ ax }^{ 2 }+{ x }^{ 3 }$ at the point

$(a,a)$ cuts off intercepts

$\alpha$ and

$\beta$ on the coordinate axes such that

${ \alpha }^{ 2 }+{ \beta }^{ 2 }=61,$ then

$a=$

0%
$\pm 30$
0%
$\pm 5$
0%
$\pm 6$
0%
$\pm 61$

Explanation

The slope of the tangent at any point

$(a,a)$ on the curve is

$\displaystyle { m }_{ T }={ \frac { dy }{ dx } }_{ \left( a,a \right) }$

$\displaystyle \therefore { 6y }^{ 2 }\frac { dy }{ dx } =2ax+{ 3x }^{ 2 }$

$\displaystyle \Rightarrow \frac { dy }{ dx } =\frac { 2ax+{ 3x }^{ 2 } }{ { 6y }^{ 2 } }$

$\displaystyle \Rightarrow { m }_{ T }={ \frac { dy }{ dx } }_{ \left( a,a \right) }=\frac { 5 }{ 6 }$

The equation of the tangent at

$(a,a)$ is

$\displaystyle y-a=\frac { 5 }{ 6 } \left( x-a \right) \Rightarrow 5x-6y+a=0$
This cuts off intercepts of lengths

$\displaystyle -\frac { a }{ 5 }$ and

$\displaystyle \frac { a }{ 6 }$ with

$x$ and

$y$ -axis respectively.

So,

$\displaystyle \alpha =-\frac { a }{ 5 }$ and

$\displaystyle \beta =\frac { a }{ 6 } .$
Since,

${ \alpha }^{ 2 }+{ \beta }^{ 2 }=61$ {given}

$\displaystyle \therefore \frac { a^{ 2 } }{ 25 } +\frac { { a }^{ 2 } }{ 36 } =61\Rightarrow { a }^{ 2 }=25\times 36\therefore a=\pm 30.$

$m$ be the slope of a tangent to the curve

${ e }^{ 2y }=1+4{ x }^{ 2 }$ , then

0%
$m<1$
0%
$\left| m \right| \le 1$
0%
$\left| m \right| >1$
0%
None of these

Explanation

We have,

$\displaystyle { e }^{ 2y }=1+4{ x }^{ 2 }\quad \Rightarrow { e }^{ 2y }.2\dfrac { dy }{ dx } =8x$

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\dfrac { 4x }{ { e }^{ 2y } } =\dfrac { 4x }{ 1+4{ x }^{ 2 } } .$

$\therefore$ Slope of tangent

$\displaystyle =m=\dfrac { 4x }{ 1+4{ x }^{ 2 } }$

$\displaystyle \Rightarrow \left| m \right| =\dfrac { 4\left| x \right| }{ 1+4{ \left| x \right| }^{ 2 } } \le 1$

$\displaystyle \left[ \because { \left( 1-2\left| x \right| \right) }^{ 2 }\ge 0\quad \quad \Rightarrow 1+4{ \left| x \right| }^{ 2 }-4\left| x \right| \ge 0 \Rightarrow \dfrac { 4\left| x \right| }{ 1+4{ \left| x \right| }^{ 2 } } \le 1 \right]$

$m$ be the slope of tangent to the curve

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

0%
$|m|>1$
0%
$m<1$
0%
$|m|<1$
0%
$|m|\leq 1$

Explanation

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

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

$\dfrac{dy}{dx}$

$=m$

$=\dfrac{2x}{1+x^{2}}$
Now

$(1-x)^{2}\geq0$
Or

$1+x^{2}-2x\geq0$
Or

$1+x^{2}\geq2x$
Or

$\dfrac{2x}{1+x^{2}}\leq 1$
Hence

$|m|\leq 1$

If the tangent to the curve

$\sqrt{x}+\sqrt{y}=\sqrt{a}$ at any points on it cuts the axes

$OX$ and

$OY$ at

$P$ and

$Q$ respectively then

$OP+OQ$ is

0%
$2a$
0%
$a$
0%
$\displaystyle \frac{1}{2}a$
0%
none of these

Explanation

Let point

$\left( p,q \right)$ lies on curve

$\sqrt { x } +\sqrt { y } =\sqrt { a }$
therefore

$\sqrt { p } +\sqrt { q } =\sqrt { a }$
Now slope

$\cfrac { dy }{ dx } =-\sqrt { \cfrac { p }{ q } }$
Equation of tangent is

$y-q=-\sqrt { \cfrac { p }{ q } } (x-p)$
X intercept = OP =

$\sqrt { p } (\sqrt { p } +\sqrt { q } )$
Y intercept = OQ =

$\sqrt { q } (\sqrt { p } +\sqrt { q } )$
Hence,
OP+OQ

$=\sqrt { a } \sqrt { a } =a$

$P(2, 2)$ and

$Q\left ( \displaystyle \frac{1}{2}, -1 \right )$ are two points on the parabola

$y^{2}=2x$ . The coordinates of the point

$R$ on the parabola, where tangent to the curve is parallel to the chord

$PQ$ is

0%
$\left ( \displaystyle \frac{5}{4}, \sqrt{\frac{5}{2}} \right )$
0%
$(2, -1)$
0%
$\left ( \displaystyle \frac{1}{8}, \frac{1}{2} \right )$
0%
none of these

Explanation

Let point

$\left( 2{ t }^{ 2 },2t \right)$ lies on curve

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

Then slope is

$\displaystyle\dfrac { dy }{ dx } =\dfrac { 2 }{ 2y } =\dfrac { 1 }{ 2t }$

And this slope is equal to the the slope of line

$PQ =\displaystyle\dfrac { -1-2 }{ \dfrac { 1 }{ 2 } -2 } =2$

Equating slope we get

$\displaystyle t=\dfrac { 1 }{ 4 }$

Hence point is

$\displaystyle\left( \dfrac { 1 }{ 8 } ,\dfrac { 1 }{ 2 } \right)$

The number of tangents to the curve

$y^{2}-2x^{3}-4y+8=0$ that pass through

$(1, 0)$ is

0%
$3$
0%
$1$
0%
$2$
0%
$6$

Explanation

Point

$(0,1)$ doesn't lie on curve, so let

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ is point of contact
Therefore slope of

${ y }^{ 2 }-2{ x }^{ 3 }-4y+8=0$ is

$\cfrac { dy }{ dx } =\cfrac { 6{ x }^{ 2 } }{ 2y-4 } =\cfrac { 6{ x }_{ 1 }^{ 2 } }{ 2{ y }_{ 1 }-2 }$
And slope of line trough

$\left( 1,0 \right)$ and

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ is

$\\ \cfrac { 0-{ y }_{ 1 } }{ 1-{ x }_{ 1 } }$
Equating slopes we get

$6{ x }_{ 1 }^{ 2 }-6{ x }_{ 1 }^{ 3 }=-2{ y }_{ 1 }^{ 2 }-4{ y }_{ 1 }$
And on solving it with equation of curve we get a quadratic equation. That give two values of slope of tangent.
Hence 2 tangents is possible.

The curve given by

$x+y=e^{xy}$ has a tangent as the

$y$ -axis at the point

0%
$(0, 1)$
0%
$(1, 0)$
0%
$(1, 1)$
0%
none of these

Explanation

We have,

$x+y=e^{xy}\Rightarrow (1)$
Differentiate both sides w.r.t.

$x$ we get,

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

$\Rightarrow \dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$
Now for tangent to be y-axis,

$\left|\dfrac{dy}{dx}\right|\to \infty$

$\Rightarrow 1-xe^{xy}=0\Rightarrow (2)$
Solving (1) and (2) we get,

$(x,y)=(1,0)$ , which is the required point

The equation of tangent to the curve

$y=e^{-\left | x \right |}$ at the point where the curve cuts the line

$x=1$ is

0%
$x+y=e$
0%
$e(x+y)=1$
0%
$y+ex=1$
0%
none of these

Explanation

Substituting

$x=1$ in

$y={ e }^{ -\left| x \right| }$ , we get

$y=\cfrac { 1 }{ e }$
Slope at point

$\left( 1,\cfrac { 1 }{ e } \right)$ is

$\cfrac { dy }{ dx } =-\cfrac { 1 }{ e }$
Hence equation of tangent is

$y-\cfrac { 1 }{ e } =-\cfrac { 1 }{ e } (x-1)\\ ey+x=2$

The equation of tangent to the curve

$y=be^{-x/a}$ where it cuts the

$y$ -axis is

0%
$\displaystyle \frac{x}{a}+\frac{y}{b}=1$
0%
$\displaystyle \frac{x}{a}+\frac{y}{b}=-1$
0%
$\displaystyle \frac{x}{a}-\frac{y}{b}=1$
0%
none of these

Explanation

Let the point be

$\left( 0,k \right)$
Substituting this in equation of curve

$y=b{ e }^{ -\cfrac { x }{ a } }$
we get

$k=b$ so point is

$\left( 0,b \right)$
Slope

$\cfrac { dy }{ dx } =-\cfrac { b }{ a }$
And equation of curve is

$y-b=-\cfrac { b }{ a } (x-0)\\ \cfrac { x }{ a } +\cfrac { y }{ b } =1$

If the line joining the points

$(0, 3)$ and

$(5, -2)$ is the tangent to the curve

$\displaystyle y=\frac{c}{x+1}$ then the value of

$c$ is

0%
$1$
0%
$-2$
0%
$4$
0%
none of these

Explanation

Slop of line joining the points

$(0,3)\quad and\quad (5,2)$ is given by

$\dfrac { 3+2 }{ 0-5 } =-1$

This is equal to the slop of tangent on the curve and that is given by

$\dfrac { dy }{ dx } =\dfrac { -c }{ { \left( x+1 \right) }^{ 2 } } \\ \dfrac { dy }{ dx } =-1\\ \Rightarrow c={ \left( x+1 \right) }^{ 2 }...................(1)$ \

Equation of line joining the points

$(0,3)\quad and\quad (5,2)$ is given by

$(y-3)=-1(x-0)$

Solving equation of tangent and the curve for point of intersection

$\dfrac { c }{ x+1 } +x=3$ .............(2)

Solving (1) and (2)

$x=1$

Putting this in (2), we get c=4

The angle between two tangents to the ellipse

$\displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at the points where the line

$y=1$ cuts the curve is

0%
$\displaystyle \frac{\pi }{4}$
0%
$\tan^{-1}\displaystyle \frac{6\sqrt{2}}{7}$
0%
$\displaystyle \frac{\pi }{2}$
0%
none of these

Explanation

Substituting

$y=1$ in

$\cfrac { { x }^{ 2 } }{ 16 } +\cfrac { { y }^{ 2 } }{ 9 } =1$
We get

$x=\pm \cfrac { 8\sqrt { 2 } }{ 3 }$
And slope of tangents

$\cfrac { dy }{ dx } =\cfrac { 9x }{ 16y } =m_{1}, m_{2} =\pm \cfrac { 3\sqrt { 2 } }{ 2 }$
Therefore

$tan\left( \theta \right) =\dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\left| -\cfrac { 6\sqrt { 2 } }{ 7 } \right| \Rightarrow \theta ={ tan }^{ -1 }\left( \cfrac { 6\sqrt { 2 } }{ 7 } \right)$

The triangle formed by the tangent to the curve

$\displaystyle f(x)=x^{2}+bx-b$ at the point

$(1, 1)$ and the coordinates axes, lies in the first quadrant. If its area is

$2$ then the value of b is

0%
-1
0%
3
0%
-3
0%
1

Explanation

$f'(x) = 2x+b$

$\Rightarrow m = 2+b$
So equation of tangent at

$(1,1)$ is

$(y-1) = (b+2)(x-1)$
intercept on the axes are

$(b+1)/(b+2)$ and

$-(b+1)$

$\Rightarrow \mid -\dfrac{1}{2}\dfrac{(b+1)}{(b+2)} (b+1)\mid = 2$
'

$\Rightarrow \mid \dfrac{(b+1)^2}{b+2}\mid = 4$

$\Rightarrow b = -3$ is only solution

Let

$y=f(x)$ be the equation of a parabola which is touches by the line

$y=x$ at the point where

$x=1$ . Then

0%
$f^{'}\left ( 0 \right )=f^{'}\left ( 1 \right )$
0%
$f^{'}\left ( 1 \right )=1$
0%
$f\left ( 0 \right )+f^{'}\left ( 0 \right )+f^{''}\left ( 0 \right )=1$
0%
$2f\left ( 0 \right )=1-f^{'}\left ( 0 \right )$

Explanation

Solving

$y=f\left( x \right)$ and

$y=x$ we get

$f\left( x \right) =x$
Differentiating this w.r.t x we get

${ f }^{ ' }\left( x \right) =1$ ...(1)
At

$x=1$ we get

${ f }^{ ' }\left( 1 \right) =1$
Differentiating (1) w.r.t x we get

${ f }^{ " }\left( x \right) =0$
For

$x=0$

$f\left( 0 \right) =0$ ,

${ f }^{ ' }\left( 0 \right) =1$ and

${ f }^{ " }\left( 0 \right) =0$
Thus,

$f\left( 0 \right) +{ f }^{ ' }\left( 0 \right) +{ f }^{ " }\left( 0 \right) =0+1+0=1$

$f'(0)=f'(1)=1$ and

$2f'(0) = 1-f'(0)=0$
Hence, options 'A', 'B', 'C' and 'D' are correct.

Let the parabolas

$y=x^{2}+ax+b$ and

$y=x(c-x)$ touch each other at the point

$(1, 0)$ . Then

0%
$a=-3$
0%
$b=1$
0%
$c=2$
0%
$b+c=3$

Explanation

As point

$\left( 1,0 \right)$ lies on

$y={ x }^{ 2 }+ax+b\Rightarrow a+b=-1$ ...(1)
And also lies on

$y=x\left( c-x \right) \Rightarrow c=1$ ...(2)
They touch each other, so slope of tangents drawn from

$(1, 0)$ will be same, then

$\cfrac { dy }{ dx } =2x+a=c-2x\Rightarrow a=c-4\Rightarrow a=-3$ ...(3)
Using (1) and (3) we get

$-3+b=1\Rightarrow b=2$ ...(4)
Now using (2) and (4) we get

$b+c=2+1=3$
Hence, options 'A' and 'D' are correct.

The curve

$\displaystyle \frac{x^{n}}{a^{n}}+\frac{y^{n}}{b^{n}}=2$ touches the line

$\displaystyle \frac{x}{a}+\frac{y}{b}=2$ at the point

0%
$(b, a)$
0%
$(a, b)$
0%
$(1, 1)$
0%
$\displaystyle \left ( \frac{1}{a}, \frac{1}{b} \right )$

Explanation

Slope of curve

$\cfrac { { x }^{ n } }{ { a }^{ n } } +\cfrac { { y }^{ n } }{ { b }^{ n } } =2$ is

$\cfrac { dy }{ dx } =-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } }$
Slope of tangent

$\cfrac { { x } }{ { a } } +\cfrac { { y } }{ { b } } =2$ is

$\cfrac { dy }{ dx } =-\cfrac { { b } }{ { a } }$
Equating both slopes, we get

$-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } } =-\cfrac { { b } }{ { a } }$
Only

$x=a\quad y=b$ satisfy the above equation.
Hence, option 'B' is correct.

The normal to the curve

$2x^{2}+y^{2}=12$ at the point

$(2, 2)$ cuts the curve again at

0%
$\displaystyle \left ( -\frac{22}{9}, -\frac{2}{9} \right )$
0%
$\displaystyle \left ( \frac{22}{9}, \frac{2}{9} \right )$
0%
$(-2, -2)$
0%
none of these

Explanation

For curve

$2{ x }^{ 2 }+{ y }^{ 2 }=12$
Slope of normal is

$-\cfrac { dx }{ dy } =\cfrac { 2y }{ 4x } =\cfrac { 1 }{ 2 }$
And equation of normal is

$y-2=\cfrac { 1 }{ 2 } (x-2)\\ 2y-x=2$
Solving it with equation of curve, we get

$y=2$ and

$y=-\cfrac { 2 }{ 9 }$
Which gives

$x=2$ and

$x=-\cfrac { 22 }{ 9 }$
Hence coordinate of other point is

$\left( -\cfrac { 22 }{ 9 } ,-\cfrac { 2 }{ 9 } \right)$
Hence, option 'A' is correct.

The area bounded by the axes of reference and the normal to

$y=\log_{e}x$ at the point

$(1, 0)$ is

0%
1 unit $^{2}$
0%
2 unit $^{2}$
0%
$\displaystyle \frac{1}{2}$ unit $^{2}$
0%
none of these

Explanation

For curve

$y=\log _{ e }{ x }$
Slope of normal at point

$(1,0)$ is

$-\cfrac { dx }{ dy } =-x=-1$
Equation of normal is

$y-0=-1(x-1)\\ x+y=1$
Therefore X intercept is

$\\ X=1$
And Y intercept is

$Y=1$
Hence area

$=\cfrac { 1 }{ 2 } \times 1\times 1=\cfrac { 1 }{ 2 }$

A point on the ellipse

$4x^{2}+9y^{2}=36$ where the tangent is equally inclided to the axes is

0%
$\displaystyle \left ( \frac{9}{\sqrt{13}}, \frac{4}{\sqrt{13}} \right )$
0%
$\displaystyle \left ( -\frac{9}{\sqrt{13}}, \frac{4}{\sqrt{13}} \right )$
0%
$\displaystyle \left ( \frac{9}{\sqrt{13}}, -\frac{4}{\sqrt{13}} \right )$
0%
$\displaystyle \left ( \frac{4}{\sqrt{13}}, -\frac{9}{\sqrt{13}} \right )$

Explanation

Slope of tangent to the curve

$4{ x }^{ 2 }+9{ y }^{ 2 }=36$ is

$1$ . since it is equally inclined to the axes.

$\cfrac { dy }{ dx } =\cfrac { 4x }{ 9y } =1\\ \Rightarrow x=\cfrac { 9y }{ 4 }$
On solving obtained equation with the equation of curve, we get

$y=\pm \cfrac { 4 }{ \sqrt { 13 } }$
Substituting value of y we get

$(x,y)=\left( \cfrac { 9 }{ \sqrt { 13 } } ,\cfrac { 4 }{ \sqrt { 13 } } \right) ,\left( -\cfrac { 9 }{ \sqrt { 13 } } ,\cfrac { 4 }{ \sqrt { 13 } } \right) ,\left( \cfrac { 9 }{ \sqrt { 13 } } ,-\cfrac { 4 }{ \sqrt { 13 } } \right)$
Hence, options 'A', 'B' and 'C' are correct.

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.