Tangents and its Equations Test 43

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Tangents and its Equations Test 43

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Question 1
1 / -0

A curve $$y=me^{mx}$$ where $$m > 0$$ intersects y-axis at a point $$P$$.
What is the slope of the curve at the point of intersection $$P$$ ?

A
$$m$$

B
$$m^2$$

C
$$2m$$

D
$$2m^2$$

Solution

$$y = m {e}^{mx}, \; m > 0$$

Therefore,
Slope $$= \cfrac{dy}{dx} = {m}^{2} {e}^{mx}$$

Substituting $$x = 0$$, we have

$$Slope={m}^{2} {e}^{m \cdot 0} = {m}^{2}$$
Question 2
1 / -0

A curve $$y=me^{mx}$$ where $$m > 0$$ intersects y-axis at a point $$P$$.
How much angle does the tangent at $$P$$ make with y-axis ?

A
$$\tan^{-1}m^2$$

B
$$\cot^{-1}(a+m^2)$$

C
$$\sin^{-1}(\dfrac{1}{\sqrt{1+m^4}})$$

D
$$\sec^{-1}\sqrt{1+m^4}$$
Question 3
1 / -0

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

A
-1

B
1

C
0

D
None of these

Solution

Putting $$y=x$$ in $$y = \sqrt{4-x^{2}}$$ , we get $$x = \sqrt{2}, -\sqrt{2}$$.

So, the point is $$(\sqrt{2}, \sqrt{2})$$.

Differentiating $$y^{2}+x^{2} = 4$$ w.r.t. x,$$2y \dfrac{dy}{dx}+ 2x = 0$$ or $$\dfrac{dy}{dx}= -\dfrac{x}{y}$$
$$\Rightarrow at(\sqrt{2}, \sqrt{2}), \dfrac{dy}{dx} = -1$$
Question 4
1 / -0

The number of tangents to the cure $$x^{3/2}+y^{3/2}=2a^{3/2}, a> 0$$, which are equally inclined to the axes, is

A
2

B
1

C
0

D
4

Solution
Question 5
1 / -0

If m is the slope of a tangent to the curve $$e^{y}=1+x^{2},$$ then

A
$$\left | m \right |> 1$$

B
$$m> 1$$

C
$$m> -1$$

D
$$\left | m \right |\leq 1$$

Solution

Differentiating w.r.t.x, we get $$e^{y} \dfrac{dy}{dx} = 2x$$

$$\Rightarrow \dfrac{dy}{dx} = \dfrac{2x}{1+x^{2}} (\because e^{y} = 1 +x^{2})$$

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

But $$1 + \left | x \right |^{2} - 2\left | x \right |=(1-\left | x \right |)^{2}\geq 0$$
$$\Rightarrow 1 + \left | x \right |^{2} \geq 2\left | x \right | $$

$$\therefore \left | m \right |\leq 1$$
Question 6
1 / -0

At the point $$P(a, a^{n})$$ 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 $$\underset{a\rightarrow b}{lim}b=\dfrac{1}{2}$$, then n equals

A
1

B
3

C
2

D
4

Solution
Question 7
1 / -0

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

A
$$(0,1)$$

B
$$( 1, 0 )$$

C
$$(1, 1)$$

D
None of these

Solution

Differentiating w.r.t.x, we get
$$1 + \dfrac{dy}{dx} = e^{xy}\left ( y + x\dfrac{dy}{dx} \right )$$ or$$\dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$$

As the tangent is parallel along $$y-axis$$

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

$$ \Rightarrow 1-xe^{xy}=0$$This holds for x = 1, y = 0
Question 8
1 / -0

The abscissa of points P and Q in the curve $$y = e^{x}+e^{-x}$$ such that tangents at P and Q make $$60^{o}$$ with the x-axis

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

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

C
ln $$\left ( \dfrac{\sqrt{7}+\sqrt{3}}{2} \right )$$

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

Solution
Question 9
1 / -0

If x=4 y = 14 is a normal to the curve $$y^{2}=ax^{3}-\beta $$ at (2,3) then the value of $$\alpha +\beta $$ is

A
9

B
-5

C
7

D
-7

Solution
Question 10
1 / -0

At what points of curve $$y = \dfrac{2}{3}x^{3}+\dfrac{1}{2}x^{2}$$, the tangent makes the equal with the axis?

A
$$(\dfrac{1}{2},\dfrac{5}{24})$$ and $$\left ( -1,\dfrac{-1}{6} \right )$$

B
$$(\dfrac{1}{2},\dfrac{4}{9})$$ and $$ ( -1,0)$$

C
$$\left ( \dfrac{1}{3},\dfrac{1}{7} \right )$$ and$$ \left ( -3, \dfrac{1}{2} \right )$$

D
$$\left ( \dfrac{1}{3},\dfrac{4}{47} \right )$$ and $$\left ( -1, \dfrac{1}{2} \right )$$

Solution