Application of Derivatives Test

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Application of Derivatives Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

If the function $$f(x)=2\tan x+(2a+1)\log _{ e }{ \left| \sec { x } \right| } +(a-2)x$$ is increasing on $$R$$, then

A
$$a\in \left (\dfrac {1}{2},\infty \right)$$

B
$$a\in \left (-\dfrac {1}{2},\dfrac {1}{2}\right)$$

C
$$a=\dfrac {1}{2}$$

D
$$a\in R$$

Solution

$$f(x)=2tanx+(2a+1)log_e \mid secx \mid +(a-2)x$$
$$f'(x)=2sec^2x+(2a+1)\dfrac{secxtanx}{secx}+(a-2)1$$
$$=2sec^2x+(2a+1)tanx+(a-2)$$
Let $$tanx=t$$
$$=2(t^2+1)+(2a+1)t+(a-2)>0$$
$$=2t^2+2at+t+a>0$$
$$(2a+1)^2-4.2.a<0$$
$$4a^2+1+4a-8a<0$$
$$4a^2-4a+1<0$$
$$4(a-\dfrac{1}{2})^2)<0$$
$$a=\dfrac{1}{2}$$
Question 2
1 / -0

Consider the equation $$x^y=e^{x-y}$$
What is $$\dfrac{d^2y}{dx^2}$$ at $$x=1$$ equal to ?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

$${x}^{y} = {e}^{x - y}$$
Taking $$\log$$ both sides, we have

$$y \log{x} = \left( x - y \right) \log{e}$$

$$y \log{x} = x - y ..... \left( 1 \right)$$

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

At $$x = 1 \Rightarrow y = 1$$

Differentiating equation $$\left( 1 \right)$$ w.r.t. $$x$$, we have

$$\cfrac{dy}{dx} \log{x} + \cfrac{y}{x} = 1 - \cfrac{dy}{dx}$$

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

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

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{y \log{x}}{x \left( 1 + \log{x} \right)}$$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{\log{x}}{{\left( 1 + \log{x} \right)}^{2}}$$

Again differentiating above equation w.r.t. $$x$$, we have

$$\cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{{\left( 1 + \log{x} \right)}^{2} \cdot \frac{1}{x} - \log{x} \cdot 2 \left( 1 + \log{x} \right) \frac{1}{x}}{{\left( 1 + \log{x} \right)}^{4}}$$

$$\Rightarrow \cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{\frac{1}{x} \left( 1 - \log{x} \right)}{{\left( 1 + \log{x} \right)}^{3}}$$

At $$x = 1$$,
$$\cfrac{{d}^{2}y}{d{x}^{2}} = 1$$
Question 3
1 / -0

Consider the equation $$x^y=e^{x-y}$$
What is $$\dfrac{dy}{dx}$$ at $$x=1$$ equal to ?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

$${x}^{y} = {e}^{x - y}$$

Taking $$\log$$ both sides, we have

$$y \log{x} = \left( x - y \right) \log{e}$$

$$y \log{x} = x - y ..... \left( 1 \right)$$

At $$x = 1 \Rightarrow y = 1$$

Differentiating equation $$\left( 1 \right)$$ w.r.t. $$x$$, we have

$$\cfrac{dy}{dx} \log{x} + \cfrac{y}{x} = 1 - \cfrac{dy}{dx}$$

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

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

At $$x = 1, \; y = 1$$

$$\cfrac{dy}{dx} = 0$$
Question 4
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 5
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 6
1 / -0

The function $$f(x)=4-3x+3x^2-x^3$$ is

A
decreasing on $$R$$

B
increasing on $$R$$

C
strictly decreasing on $$R$$

D
strictly increasing on $$R$$

Solution

$${f}' \left(x\right) = -3+6x-3x^{2} = -3\left(x^{2}-2x+1\right) = -3\left(x-1\right)^{2} \le 0.$$

$$\Rightarrow {f}' \left(x\right) \le 0$$ for all $$x \in R \Rightarrow f\left(x\right)$$ is decreasing on $$R$$.
Question 7
1 / -0

The real value of $$k$$ for which $$f(x)=x^2+kx+1$$ is increasing on $$(1, 2)$$, is

A
$$-2$$

B
$$-1$$

C
$$1$$

D
$$2$$

Solution

$$f'(x)=(2x+k)$$.

$$1 < x < 2\Rightarrow 2 < 2x , 4 \Rightarrow 2+k < 2x+k < 4+k \Rightarrow 2+k < f'(x) < 4+k$$

$$f(x)$$ is increasing $$\Leftrightarrow (2x+k) \ge 0\Leftrightarrow 2+k\ge 0\Leftrightarrow k \ge -2$$

$$\therefore \ $$ least value of $$k$$ is $$-2$$
Question 8
1 / -0

$$f(x)=\sin x-kx $$ is decreasing for all $$x \in R$$, when

A
$$k < 1$$

B
$$k \le 1$$

C
$$k > 1$$

D
$$k \ge 1$$

Solution
Question 9
1 / -0

The function $$f(x)=3x+\cos 3x$$ is

A
increasing on $$R$$

B
decreasing on $$R$$

C
strictly increasing on $$R$$

D
strictly decreasing on $$R$$

Solution
Question 10
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 $$\dfrac{dy}{dx}= -\dfrac{x}{y}$$
$$\Rightarrow at(\sqrt{2}, \sqrt{2}), \dfrac{dy}{dx} = -1$$

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 50

Result

Application of Derivatives Test - 50

Score

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Rank

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SHARING IS CARING

Question 1

$$a\in \left (\dfrac {1}{2},\infty \right)$$

$$a\in \left (-\dfrac {1}{2},\dfrac {1}{2}\right)$$

$$a=\dfrac {1}{2}$$

$$a\in R$$

Solution

Question 2

$$0$$

$$1$$

$$2$$

$$4$$

Solution

Question 3

$$0$$

$$1$$

$$2$$

$$4$$

Solution

Question 4

$$m$$

$$m^2$$

$$2m$$

$$2m^2$$

Solution

Question 5

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

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

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

$$\sec^{-1}\sqrt{1+m^4}$$

Question 6

decreasing on $$R$$

increasing on $$R$$

strictly decreasing on $$R$$

strictly increasing on $$R$$

Solution

Question 7

$$-2$$

$$-1$$

$$1$$

$$2$$

Solution

Question 8

$$k < 1$$

$$k \le 1$$

$$k > 1$$

$$k \ge 1$$

Solution

Question 9

increasing on $$R$$

decreasing on $$R$$

strictly increasing on $$R$$

strictly decreasing on $$R$$

Solution

Question 10

-1

1

0

None of these

Solution

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