Tangents and its Equations Test 48

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Tangents and its Equations Test 48

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Weekly Quiz Competition

Question 1
1 / -0

At the point $P(a,a'')$ on the graph of $y=x^n$ , $(n \epsilon N)$ , in the first quadrant , a normal is drawn. The normal intersects the $y$ -axis at the point $(0,b)$ . If $\lim_{a\rightarrow 0}b=\displaystyle \frac{1}{2}$ , then n equals

A
$1$

B
$3$

C
$2$

D
$4$

Solution

$y=x^{n}$
$\displaystyle \dfrac{dy}{dx}=nx^{n-1}$
$\displaystyle (\dfrac{dy}{dx})_{(a,a^{n})}=na^{n-1}$
Slope of normal at P, $=-\dfrac{1}{na^{n-1}}$
Equation of normal at P is
$y-{ a }^{ n }=-\dfrac { 1 }{ n{ a }^{ n-1 } } (x-a)$
Since, it intersects y-axis at (0,b)
$\Rightarrow b-a^{n}=\dfrac{a}{na^{n-1}}$
$\Rightarrow \lim _{ a\rightarrow 0 }{ b- } \lim _{ a\rightarrow 0 }{ { a }^{ n } } =\lim _{ a\rightarrow 0 }{ (\dfrac { a }{ n{ a }^{ n-1 } } ) }$
$\Rightarrow \dfrac { 1 }{ 2 } =\lim _{ a\rightarrow 0 }{ (\dfrac { a }{ n{ a }^{ n-1 } } ) }$ ......(i)
Now, we will check by options.
Option C, satisfies (i)
Hence, $n=2$
Question 2
1 / -0

Let $f(x)$ = $\begin{cases} -x^2, {for \ x<0} \\x^2+8, {for \ x\geq 0} \end{cases}$ . Then $x$ -intercept of the line, that is, the tangent to the graph of $f(x)$ in both the intervals of its domain, is

A
zero

B
$-1$

C
$-2$

D
$-4$

Solution

At any point $(k, k^2+8)$ on $y=x^2+8$ , the slope of tangent is $\dfrac{dy}{dx}|_{x=k}=2k$ .....(1)
Similarly, at any point $(p, -p^2)$ on $y=-x^2$ , the slope of tangent is $\dfrac{dy}{dx}|_{x=p}=-2p$ ....(2)
For common tangent, $2k=-2p \Rightarrow k=-p$
Hence, the coordinates of point on common tangent and on $y=-x^2$ can be written as $(-k, -k^2)$ as shown in figure.
So, slope of common tangent is $\dfrac{8+k^2-(-k^2)}{k-(-k)}=\dfrac{8+2k^2}{2k}$
From (1), slope is also equal to $2k$ .
Hence, $\dfrac{8+2k^2}{2k}=2k$
Solving, we get $k= \pm 2$ .
Discarding negative value (since 'k' is positive as defined in the function), we get $k=2$
Hence, slope of common tangent is $2k=4$
The point common to tangent and $y=x^2+8$ is $(2,12)$
Hence, equation of common tangent is $y-12=4(x-2)$
To find $x$ intercept, substitute $y=0$ in the above equation.
We get $x$ intercept $=-1$
Question 3
1 / -0

A function $y=f(x)$ has a second-order derivative $f''(x)=6(x-1)$ . If its graph passes through the point $(2,1)$ and at the point tangent to the graph is $y=3x-5$ , then the value of $f(0)$ is

A
$1$

B
$-1$

C
$2$

D
$0$

Solution

$f''(x)=6x-6$
$\int { f''(x)dx } =\int { (6x-6)dx }$
$\Rightarrow f'(x)=3x^{2}-6x+C$
Since, it passes through (2,1) and tangent is $y=3x-5$
$\Rightarrow C=3$
So, $f'(x)=3x^{2}-6x+3$
$\int { f'(x)dx } =\int { (3{ x }^{ 2 }-6x+3)dx }$
$\Rightarrow f(x)=x^{3}-3x^{2}+3x+C_1$
Since, it passes through $(2,1)$
$\Rightarrow C_1=-1$
Hence, $f(x)=x^{3}-3x^{2}+3x-1$
$f(0)=-1$
Question 4
1 / -0

The tangent of the acute angle between the curves $y=|x^2-1|$ and $y=\sqrt {7-x^2}$ at their points of intersection is

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

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

C
$\displaystyle \frac {5\sqrt 3}{4}$

D
$\displaystyle \frac {3\sqrt 5}{4}$

Solution

The given curves are $y=|x^{2}-1|$ and $y=\sqrt{7-x^2}$
To find the point of intersection, let us equate both the curves.
$|x^{2}-1|=\sqrt{7-x^2}$
Squaring on both sides gives us $x^{4}-x^{2}-6=0$
$(x^2-3)(x^2+2) =0$
On solving above quadratic equation, we get $x^{2}=3$
The point of intersection is $\left(\pm \sqrt{3},2\right)$
The slope of the tangent to $y=|x^{2}-1|$ at the point $\left( \sqrt{3},2\right)$ is
$m_{1}=\displaystyle\dfrac{dy}{dx}=2x=2\sqrt{3}$
The slope of tangent to $y=\sqrt{7-x^2}$ at the point $\left(\sqrt{3},2\right)$ is
$m_{2}=\displaystyle\dfrac{dy}{dx}=\dfrac{-x}{\sqrt{7-x^2}}=\dfrac{-\sqrt{3}}{2}$
Let $\theta$ be the acute angle between the two tangents.
$\displaystyle\tan\theta = \dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}$
$\Rightarrow \displaystyle\tan\theta = \dfrac{-5\sqrt{3}}{4}$
$\therefore$ The tangent of the acute angle between the curves is $\displaystyle\dfrac{5\sqrt{3}}{4}$ .
Hence, option C is correct.
Question 5
1 / -0

The angle made by the tangent of the curve $x=a (t+\sin t \cos t)$ , $y=a(1+sint)^2$ with the $x- axis$ at any point on it is

A
$\displaystyle \frac {1}{4}(\pi +2t)$

B
$\displaystyle \frac {1-\sin t}{\cos t}$

C
$\displaystyle \frac {1}{4}(2t-\pi)$

D
$\displaystyle \frac {1+\sin t}{\cos 2t}$

Solution

$x=a (t+sin t cos t)$
$\dfrac{dx}{dt}=2acos^{2}t$
$y=a(1+sint)^2$
$\dfrac{dy}{dt}=2a(cost+costsint)$
$\Rightarrow \displaystyle \dfrac{dy}{dx}=\dfrac{1+sint}{cost}$
$\displaystyle = \dfrac{(cos\dfrac{t}{2}+sin\dfrac{t}{2})^{2}}{cos^{2}\dfrac{t}{2}-sin^{2}\dfrac{t}{2}}$
$\displaystyle = \dfrac{1+tan{\dfrac{t}{2}}}{1-tan{\dfrac{t}{2}}}$
$\displaystyle =tan(\dfrac{\pi}{4}+\dfrac{t}{2})$
Slope of tangent to the curve $= tan(\dfrac{\pi}{4}+\dfrac{t}{2})$
So, the angle made by the tangent to the curve with the x-axis $=\dfrac{\pi}{4}+\dfrac{t}{2}$ or $\dfrac{1}{4}(\pi+2t)$
Question 6
1 / -0

The abscissas of points $P$ and $Q$ on the curve $y=e^x+e^{-x}$ such that tangents at $P$ and $Q$ make $60^{\circ}$ with the $x$ -axis are

A
$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {7}\right )$ and $\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 5} {2}\right )$

B
$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$

C
$\ln \left (\displaystyle \frac {\sqrt 7-\sqrt 3} {7}\right )$

D
$\pm\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$

Solution

$y=e^{x}+e^{-x}$
$\displaystyle\frac{dy}{dx}=e^{x}-e^{-x}=\frac{e^{2x}-1}{e^{x}}$
Let P $(x_{1},y_{1})$ and Q $(x_2,y_2)$ be the points on the given curve.
Slope of tangent at P is $\displaystyle m_1=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$
$\Rightarrow \displaystyle \sqrt{3}=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$
$\displaystyle\Rightarrow e^{2x_1}-\sqrt{3}e^{x}-1=0$
$\displaystyle\Rightarrow (e^{x_1}-\frac{\sqrt{3}}{2})^{2}=\frac{7}{4}$
$\displaystyle\Rightarrow e^{x_1}=\pm\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}$
$\displaystyle\Rightarrow {x_1}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$ as logarithm of negative numbers is not defined.
Similarly, $\displaystyle\Rightarrow {x_2}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$ as it makes same angle, $60^{0}$ with x-axis.
Question 7
1 / -0

The point on the curve $y^{2} = x ,$ the tangent at which makes an angle of $45^{0}$ with positive direction of $x -$ axis will be given by

A
$\left (\displaystyle \frac{1}{2},\displaystyle \frac{1}{4} \right )$

B
$\left ( \displaystyle \frac{1}{2}, \displaystyle \frac{1}{2} \right )$

C
$(2,4)$

D
$\left ( \displaystyle \frac{1}{4}, \displaystyle \frac{1}{2} \right )$

Solution

Given equation of curve is $y^{2} = x$
$\displaystyle \frac{dy}{dx}=\frac{1}{2y}$
Slope of tangent $=\dfrac{1}{2y}$
Also, given slope of tangent $=tan 45^{\circ}$
$\displaystyle\frac{1}{2y}=1\Rightarrow y=\frac{1}{2}$
$\displaystyle \Rightarrow x=\frac{1}{4}$
Hence, the required point is $\left ( \dfrac{1}{4},\dfrac{1}{2} \right )$
Question 8
1 / -0

If the curve represented parametrically by the equations $x=2 \ln\cot t+1$ and $y=\tan t+ \cot t$

A
tangent and normal intersect at the point $(2,1)$

B
normal at $t=\displaystyle \frac{\pi}{4}$ is parallel to the $y$ axis

C
tangent at $t=\displaystyle \frac{\pi}{4}$ is parallel to the line $y=x$

D
tangent at $t=\displaystyle \frac{\pi}{4}$ is parallel to the $x$ axis

Solution

$x=2 \ln\cot t+1$
$\displaystyle \frac{dx}{dt}=-\frac{2cosec^{2}t}{\cot t}=-\frac{2}{\sin t\cos t}$
$y=\tan t+ \cot t$
$\displaystyle \frac{dy}{dt}=sec^{2}t-cosec^{2}t=\frac{sin^{2}t-cos^{2}t}{sin^{2}t cos^{2}t}$
$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{sin^{2}t-cos^{2}t}{sin2t}$
$\Rightarrow \displaystyle \frac{dy}{dx}=\cot 2t$
Slope of tangent at $t=\dfrac{\pi}{4}$ is 0.
Hence, tangent is parallel to x-axis.
Question 9
1 / -0

A curve is represented by the equations, $\displaystyle x = \sec^2t$ and $y =\cot t$ , where $t$ is a parameter. If the tangent at the point $P$ on the curve where $\displaystyle t = \dfrac{\pi}{4}$ meets the curve again at the point $Q$ then $|PQ|$ is equal to

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

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

C
$\displaystyle \frac {2\sqrt 5}{3}$

D
$\displaystyle \frac {3\sqrt 5}{2}$

Solution

$x=\sec^{2}t$
$y=\cot t$
$\dfrac{dy}{dx}=-\dfrac{\csc^{2}t}{2\sec^{2}t\cdot\tan t}=-\dfrac{\cot^{3}t}{2}$

Now,
$\left.\dfrac{dy}{dx}\right|_{t=45^{0}}=-\dfrac{1}{2}$
$\left.x\right|_{t=45^{0}}=2$
$\left.y\right|_{t=45^{0}}=1$

Hence the equation of the tangent will be
$y-1=-\dfrac{1}{2}(x-2)\Rightarrow2y-2=-x+2\Rightarrow x+2y=4$ ...(i)

Now,
$\sec^{2}t=1+\tan^{2}t$
$\sec^{2}t=1+\dfrac{1}{\cot^{2}t}$
Or
$x=1+\dfrac{1}{y^{2}}\Rightarrow xy^{2}=y^{2}+1$ is the equation of curve.

Solving the equation of tangent and equation of curve we get
$4-2y=1+\dfrac{1}{y^{2}}$
Or
$4y^{2}-2y^{3}=y^{2}+1\Rightarrow2y^{3}-3y^{2}+1=0$
$y=-\dfrac{1}{2}$ and $y=1$
For $y=1; x= 2$
For $y=-\dfrac{1}{2}; x = 5$

Hence,
$Q=\left(5,-\dfrac{1}{2}\right)$ and $P=(2,1)$
$\therefore PQ=\dfrac{3\sqrt{5}}{2}$ .
Question 10
1 / -0

The x-intercept of the tangent at any arbitrary point of the curve $\displaystyle \frac {a}{x^2} + \frac {b}{y^2} = 1$ is proportional to:

A
square of the abscissa of the point of tangency

B
square root of the abscissa of the point of tangency

C
cube of the abscissa of the point of tangency

D
cube root of the abscissa of the point of tangency

Solution

let $P \left( \dfrac { \sqrt { a } }{ \cos { \theta } } ,\dfrac { \sqrt { b } }{ \sin { \theta } } \right)$ be any point of curve $\dfrac { a }{ x^{ 2 } } +\dfrac { b }{ y^{ 2 } } =1$
$\dfrac { -2a }{ x^{ 3 } } -\dfrac { 2b }{ y^{ 3 } } \dfrac { dy }{ dx } =0$
$\dfrac { dy }{ dx }$ at point P $=-\sqrt { \dfrac { b }{ a } } \cot ^{ 3 }{ \theta }$
Equation of tangent at point P: $\displaystyle y-\frac { \sqrt { b } }{ \sin { \theta } } =-\sqrt { \frac { b }{ a } } \cot ^{ 3 }{ \theta } \left( x-\frac { \sqrt { a } }{ \cos { \theta } } \right)$
for X-intercept put $x=0$
$\displaystyle 0-\frac { \sqrt { b } }{ \sin { \theta } } =-\sqrt { \frac { b }{ a } } \cot ^{ 3 }{ \theta } \left( x-\frac { \sqrt { a } }{ \cos { \theta } } \right)$
$\displaystyle \Rightarrow x=\sqrt { a } \left( \frac { \sin ^{ 2 }{ \theta } }{ \cos ^{ 3 }{ \theta } } +\frac { 1 }{ \cos { \theta } } \right) =\frac { \sqrt { a } }{ \cos ^{ 3 }{ \theta } }$
Therefore, X-intercept is proportional to cube of abcissa.

Ans: C

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 48

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Tangents and its Equations Test 48

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Question 1

111

333

222

444

Solution

Question 2

zero

−1-1−1

−2-2−2

−4-4−4

Solution

Question 3

111

−1-1−1

222

000

Solution

Question 4

532\displaystyle \frac {5\sqrt 3}{2}253​​

352\displaystyle \frac {3\sqrt 5}{2}235​​

534\displaystyle \frac {5\sqrt 3}{4}453​​

354\displaystyle \frac {3\sqrt 5}{4}435​​

Solution

Question 5

14(π+2t)\displaystyle \frac {1}{4}(\pi +2t)41​(π+2t)

1−sin⁡tcos⁡t\displaystyle \frac {1-\sin t}{\cos t}cost1−sint​

14(2t−π)\displaystyle \frac {1}{4}(2t-\pi)41​(2t−π)

1+sin⁡tcos⁡2t\displaystyle \frac {1+\sin t}{\cos 2t}cos2t1+sint​

Solution

Question 6

ln⁡(3+77)\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {7}\right )ln(73​+7​​) and ln⁡(3+52)\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 5} {2}\right )ln(23​+5​​)

ln⁡(3+72)\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )ln(23​+7​​)

ln⁡(7−37)\ln \left (\displaystyle \frac {\sqrt 7-\sqrt 3} {7}\right )ln(77​−3​​)

±ln⁡(3+72)\pm\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )±ln(23​+7​​)

Solution

Question 7

(12,14)\left (\displaystyle \frac{1}{2},\displaystyle \frac{1}{4} \right )(21​,41​)

(12,12)\left ( \displaystyle \frac{1}{2}, \displaystyle \frac{1}{2} \right )(21​,21​)

(2,4)(2,4)(2,4)

(14,12)\left ( \displaystyle \frac{1}{4}, \displaystyle \frac{1}{2} \right )(41​,21​)

Solution

Question 8

tangent and normal intersect at the point (2,1)(2,1)(2,1)

normal at t=π4t=\displaystyle \frac{\pi}{4}t=4π​ is parallel to the yyy axis

tangent at t=π4t=\displaystyle \frac{\pi}{4}t=4π​ is parallel to the line y=xy=xy=x

tangent at t=π4t=\displaystyle \frac{\pi}{4}t=4π​ is parallel to the xxx axis

Solution

Question 9

532\displaystyle \frac {5\sqrt 3}{2}253​​

552\displaystyle \frac {5\sqrt 5}{2}255​​

253\displaystyle \frac {2\sqrt 5}{3}325​​

352\displaystyle \frac {3\sqrt 5}{2}235​​

Solution

Question 10

square of the abscissa of the point of tangency

square root of the abscissa of the point of tangency

cube of the abscissa of the point of tangency

cube root of the abscissa of the point of tangency

Solution

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$1$

$3$

$2$

$4$

$-1$

$-2$

$-4$

$1$

$-1$

$2$

$0$

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

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

$\displaystyle \frac {5\sqrt 3}{4}$

$\displaystyle \frac {3\sqrt 5}{4}$

$\displaystyle \frac {1}{4}(\pi +2t)$

$\displaystyle \frac {1-\sin t}{\cos t}$

$\displaystyle \frac {1}{4}(2t-\pi)$

$\displaystyle \frac {1+\sin t}{\cos 2t}$

$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {7}\right )$ and $\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 5} {2}\right )$

$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$

$\ln \left (\displaystyle \frac {\sqrt 7-\sqrt 3} {7}\right )$

$\pm\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$

$\left (\displaystyle \frac{1}{2},\displaystyle \frac{1}{4} \right )$

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

$(2,4)$

$\left ( \displaystyle \frac{1}{4}, \displaystyle \frac{1}{2} \right )$

tangent and normal intersect at the point $(2,1)$

normal at $t=\displaystyle \frac{\pi}{4}$ is parallel to the $y$ axis

tangent at $t=\displaystyle \frac{\pi}{4}$ is parallel to the line $y=x$

tangent at $t=\displaystyle \frac{\pi}{4}$ is parallel to the $x$ axis

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

$\displaystyle \frac {5\sqrt 5}{2}$

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

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