Application of Derivatives Test

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Application of Derivatives Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

The slope of the tangent to the curve $$x=t^2+3t-8, y=2t^2-2t-5$$ at the point $$(2, -1)$$ is

A
$$\dfrac {22}{7}$$

B
$$\dfrac {6}{7}$$

C
$$\dfrac {7}{6}$$

D
$$-\dfrac {6}{7}$$

Solution

Given curve

$$x = t^2 + 3t - 8$$ and $$y = 2t^2 - 2t - 5$$

slope of tangent to the curve $$= \dfrac{dy}{dx} = \dfrac{dy/dt}{dx / dt}$$

$$\therefore \dfrac{dx}{dt} = 2t + 3 , \, \dfrac{dy}{dt} = 4t - 2$$

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{4t - 2}{2t + 3}$$

$$\left(\dfrac{dy}{dx} \right)_{t = 2} $$ $$\dfrac{4(2) - 2}{2(2) + 3} = \dfrac{6}{7}$$
Question 2
1 / -0

The normal to the curve $$x^2=4y$$ passing through $$(1, 2)$$ is

A
$$x+y=3$$

B
$$x-y=3$$

C
$$x+y=1$$

D
$$x-y=1$$

Solution

Curve is $$x^2=4y$$

Diff wrt to x
$$2x=\dfrac{4dy}{dx}\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{2}$$

Slope of normal $$=\dfrac{-1}{dy/dx}=\dfrac{-2}{x}$$
Let (h, k) be the point where normal and curve intersects
$$\therefore$$ Slope of normal at (h, k)$$=-2/h$$

Equation of normal passes through (h, k)
$$y-y_1=m(x-x_1)$$
$$y-k=\dfrac{-2}{h}(x-h)$$

Since normal passes through $$(1, 2)$$
$$2-k=\dfrac{-2}{h}(1-h)$$
$$k=2+\dfrac{2}{h}(1-h)$$ ..........$$(1)$$

since (h, k) lies on the curve $$x^2=4y$$
$$h^2=4k$$
$$k=\dfrac{h^2}{4}$$ ....... $$(2)$$

using $$(1)$$ and $$(2)$$
$$2+\dfrac{2}{h}(1-h)=\dfrac{h^2}{4}$$
$$\dfrac{2}{h}=\dfrac{h^2}{4}$$
$$h=2$$

Putting $$h=2$$ in $$(2)$$
$$k=\dfrac{h^2}{4}, k=1$$
$$h=2$$ and $$k=1$$ putting in equation of normal

$$\Rightarrow y-k=\dfrac{-2(k-h)}{h}$$

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

$$\Rightarrow y-1=-x+2\\$$
$$\Rightarrow x+y=2+1\\$$
$$\Rightarrow x+y=3$$.
Question 3
1 / -0

Let $$f(x)={x}^{3}+a{x}^{2}+bx+5{\sin}^{2}x$$ be an increasing function on the set $$R$$. Then, $$a$$ and $$b$$ satisfy

A
$${a}^{2}-3b-15> 0$$

B
$${a}^{2}-3b+15> 0$$

C
$${a}^{2}-3b+15< 0$$

D
$$a> 0$$ and $$b> 0$$

Solution

$$f(x)=x^3+ax^2+bx+5\sin^2x$$

$$f^{\prime}(x)=3x^2+2ax+b+5.2\sin x\cos x>0$$

$$f^{\prime}(x)=3x^2+2ax+b+5\sin 2x >0$$

$$-1\le \sin 2x\le 1$$

$$\sin 2x=-1$$

$$f^{\prime}(x)=3x^2+2ax+b+5(-1)=3x^2+2ax+b-5 \ge 0$$

$$f^{\prime \prime}=6x+2a$$

$$6x+2a=0$$ or, $$x=\dfrac{-2a}{6}=\dfrac{-a}{3}$$

so now,

$$3\left(\dfrac{-a}{3}\right)^2+2a.\left(\dfrac{-a}{3}\right)+b-5\ge 0$$

$$\dfrac{a^2}{3}-\dfrac{2a^2}{3}+b-5\ge 0$$

$$\dfrac{-a^2}{3}+b-5\ge 0$$

$$a^2-3b+15<0$$
Question 4
1 / -0

If the function $$f(x)=\cos\left| x \right| -2ax+b$$ increases along the entire number scale, then

A
$$a=b$$

B
$$a=\cfrac{1}{2}b$$

C
$$a\le -\cfrac{1}{2}$$

D
$$a> -\cfrac{3}{2}$$

Solution

$$\because f(x)=\cos |x|-2ax+b$$ increase in R
$$=\cos x-2ax+b$$ (as $$\cos (-x)=\cos x$$)
$$f'(x)=-\sin x-2a$$
for increasing $$f(x)$$, $$f'(x)\geq 0$$
$$-\sin x-2a\geq 0$$
$$\sin x+2a\leq 0$$
$$2a\leq -\sin x$$ ($$\because$$ maximum value of $$\sin x=1$$)
$$2a\leq -1$$
$$a\leq -\dfrac{1}{2}$$.
Question 5
1 / -0

Function $$f(x)={a}^{x}$$ is increasing on $$R$$, if

A
$$a> 0$$

B
$$a< 0$$

C
$$0< a< 1$$

D
$$a> 1$$

Solution

$$f(x) = a^x$$

$$f'(x) = a^x \log a$$

function is increasing on R.

$$a^x \log a > 0$$

$$a^x > 0 \, \& \, \log a > 0$$

or

$$a^x < 0 \, \& \, \log a < 0$$

But log function is always positive

$$\log a > 0$$

$$\Rightarrow a > 1$$
Question 6
1 / -0

The function $$f(x)=\cfrac{\lambda \sin x+2\cos x}{\sin x+\cos x}$$ is increasing, if

A
$$\lambda< 1$$

B
$$\lambda> 1$$

C
$$\lambda< 2$$

D
$$\lambda> 2$$

Solution

$$f(x) = \dfrac{\lambda \sin x + 2 \cos x}{\sin x + \cos x}$$

for increasing

$$f'(x) > 0$$

$$\Rightarrow \dfrac{(\sin x + \cos x) (\lambda \cos x - 2 \sin x) - (\lambda \sin x + 2 \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} > 0$$

$$\Rightarrow \dfrac{\lambda - 2}{(\sin x + \cos x)^2} > 0$$

but $$(\sin x + \cos x)^2 > 0$$

$$\therefore \lambda - 2 > 0$$

$$\lambda > 2$$
Question 7
1 / -0

If the function $$f(x)={x}^{3}-9k{x}^{2}+27x+30$$ is increasing on $$R$$, then

A
$$-1< k< 1$$

B
$$k< -1$$ or $$k> 1$$

C
$$0< k< 1$$

D
$$-1< k< 0$$

Solution

$$f(x)=x^3-9kx^2+27x+30$$ is increasing on R.
$$f(x)=x^3-9kx^2+27x+30$$
$$f'(x)=3x^2-18kx+27$$
$$=3(x^2-6kx+9)$$
Given $$f(x)$$ is increasing on R
$$\Rightarrow f'(x) > 0$$ for all $$x\in R$$
$$\Rightarrow 3(x^2-6kx+9) > 0$$
$$(x^2-6kx+9) > 0$$ all $$x\in R$$
$$ax^2+bx+c > 0$$
so, $$(-6k)^2-4(1)(9) < 0$$
$$\Rightarrow 36k^2-36 < 0$$
$$(k+1)(k-1) < 0$$
It can be possible when $$(k+1) < 0$$ and $$d(k-1) > 0$$
$$\Rightarrow k < -1$$ and $$k > 1$$ (not possible)
or, $$(k+1) > 0$$ and $$(k-1) < 0$$
$$k > -1$$ and $$k < 1$$
$$-1 < k < 1$$ so option A is correct.
Question 8
1 / -0

If the function $$f(x)={x}^{2}-kx+5$$ is increasing on $$[2,4]$$, then

A
$$k\in (2,\infty)$$

B
$$k\in (-\infty,2)$$

C
$$k\in (4,\infty)$$

D
$$k\in (-\infty,4)$$

Solution

$$f(x)=x^2-kx+5$$ is increasing in $$x\in [2, 4]$$
$$f'(x)=2x-k > 0$$
$$2x > k$$ $$2\leq x\leq 4$$
$$k < 2x$$ $$4\leq 2x \leq 8$$
k should less than the minimum value of $$2x$$
$$k < 4$$
$$k\in (-\infty, 4)$$
Final Answer.
Question 9
1 / -0

Function $$f(x)=\log _{ a }{ { x }_{ } } $$ is increasing on $$R$$, if

A
$$0< a< 1$$

B
$$a> 1$$

C
$$a< 1$$

D
$$a> 0$$

Solution

We have
$$f(x)=log_ax$$
Differentiate with respect x, we get
$$f'(x)=\dfrac{1}{x log a}$$
$$\because$$ function is increasing on R
$$\dfrac{1}{x log a} > 0$$
$$a > 1$$.
Question 10
1 / -0

The function $$f(x)={x}^{9}+3{x}^{7}+64$$ is increasing on

A
$$R$$

B
$$(-\infty,0)$$

C
$$(0,\infty)$$

D
$${R}_{0}$$

Solution

$$f(x)=x^9+3x^7+64$$
$$f'(x)=9x^8+21x^6$$
$$=3x^6(3x^2+7)$$
$$\because$$ function is increasing
$$3x^6(3x^2+7) > 0$$
$$\Rightarrow$$ function is increasing on R.

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 49

Result

Application of Derivatives Test - 49

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SHARING IS CARING

Question 1

$$\dfrac {22}{7}$$

$$\dfrac {6}{7}$$

$$\dfrac {7}{6}$$

$$-\dfrac {6}{7}$$

Solution

Question 2

$$x+y=3$$

$$x-y=3$$

$$x+y=1$$

$$x-y=1$$

Solution

Question 3

$${a}^{2}-3b-15> 0$$

$${a}^{2}-3b+15> 0$$

$${a}^{2}-3b+15< 0$$

$$a> 0$$ and $$b> 0$$

Solution

Question 4

$$a=b$$

$$a=\cfrac{1}{2}b$$

$$a\le -\cfrac{1}{2}$$

$$a> -\cfrac{3}{2}$$

Solution

Question 5

$$a> 0$$

$$a< 0$$

$$0< a< 1$$

$$a> 1$$

Solution

Question 6

$$\lambda< 1$$

$$\lambda> 1$$

$$\lambda< 2$$

$$\lambda> 2$$

Solution

Question 7

$$-1< k< 1$$

$$k< -1$$ or $$k> 1$$

$$0< k< 1$$

$$-1< k< 0$$

Solution

Question 8

$$k\in (2,\infty)$$

$$k\in (-\infty,2)$$

$$k\in (4,\infty)$$

$$k\in (-\infty,4)$$

Solution

Question 9

$$0< a< 1$$

$$a> 1$$

$$a< 1$$

$$a> 0$$

Solution

Question 10

$$R$$

$$(-\infty,0)$$

$$(0,\infty)$$

$${R}_{0}$$

Solution

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