Application of Derivatives Test

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Application of Derivatives Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

The entire graph of the equation $$y=x^{2}+kx-x+9$$ is strictly above the $$x-$$axis if and only if :

A
$$k<7$$

B
$$-5< k <7$$

C
$$k>-5$$

D
$$none\ of\ these$$

Solution

$$y=x^2+kx-x+9$$
$$a>0$$
if $$b^2-4ac<0$$ & $$a>0$$
then the graph will always be above x-axis
$$ a = 1 \therefore a > 0 $$.
$$(K-1)^2-4\times 9<0$$
$$(K-1)^2-6^2<0$$
$$(K-1-6)(K-1+6)<0$$
$$(K-7)(K+5)<0$$
$$K>-5,K<7$$
$$-5<K<7$$
Option B.
Question 2
1 / -0

If $$y=e^{4x}+2e^{-x}$$ satisfied the equation $$\dfrac{d^{3}y}{dx^{3}}+A\dfrac{dy}{dx}+By=0$$ then value of $$|A+B|$$ is

A
$$36$$

B
$$25$$

C
$$24$$

D
$$15$$

Solution

$$\dfrac{d^3y}{dx^3}$$ = $$64e^{4x}$$ - $$2e^{-x}$$

$$\dfrac{dy}{dx}$$ = $$4e^{4x}$$ - $$2e^{-x}$$

$$\dfrac{d^3y}{dx^3}$$ + $$A$$$$\dfrac{dy}{dx}$$ + $$By$$ = 0

$$\Rightarrow$$ $$e^{4x} (64+4A+B)$$ + $$2e^{-x} (-1-A+B)$$ = 0

64 +4A + B = 0 -(I)
-1 - A + B = 0 -(II)

Solving (i) and (ii),
A = -13 ; B = -12

$$\therefore$$ $$|A + B|$$ = $$25$$

So, the correct option is 'B'.
Question 3
1 / -0

Let $$f(x)$$ be a differentiate function and $$f(\alpha)=f(\beta)=0(\alpha < \beta)$$, then in the interval $$(\alpha, \beta)$$.

A
$$f(x)+f'(x)=0$$ has at least one root

B
$$f(x)-f'(x)=0$$ has at least one real root

C
$$f(x)\cdot f'(x)=0$$ has at least one real root

D
None of these

Solution

If $$f(x)$$ is a differentiable function with
$$f(\alpha)=f(\beta)=0(\alpha< \beta)$$
then in interval $$(\alpha,\beta)$$ for some part
$$f'(x)<0$$
$$A+$$ are point
$$f(x)=0$$
$$f(x).f'(x)$$ has at least are real root.
Question 4
1 / -0

If $$f(x) = g(x) (x - \lambda)^2 \,$$ and $$\, g (\lambda)$$, where $$0 < x \le 1$$, then in this interval

A
Both $$f(x)$$ and $$g(x)$$ are increasing functions

B
Both $$f(x)$$ and $$g(x)$$ are decreasing function

C
$$f(x)$$ is an increasing function

D
$$g(x)$$ is an increasing function
Question 5
1 / -0

Let $$h(x) = f(x) - (f (x) )^2 + (f(x))^3$$ for every real number $$x$$, then

A
$$h$$ is increasing whenever $$f$$ is increasing and decreasing whenever $$f$$ is decreasing

B
$$h$$ is increasing whenever $$f$$ is decreasing

C
$$h$$ is decreasing whenever $$ f$$ is increasing

D
Nothing can be said in general

Solution

$$h(x)=f(x)-(f(x))^2+(f(x))^3$$

When $$x=2$$

$$h(x)=2-2^2+2^3=6$$

When $$x=-2$$

$$h(x)=-2-(-2)^2+(-2)^3=-14$$

Therefore, h is increasing whenever f is increasing and decreasing whenever f is decreasing.
Question 6
1 / -0

If $$f'\left( x \right) = \sqrt {2{x^2} - 1} $$ and $$y = f\left( {{x^2}} \right)$$, then $$\frac{{dy}}{{dx}}$$ at $$x = 1$$ is equal to

A
2

B
1

C
-2

D
-1

Solution

$$f'\left( x \right) = \sqrt {2{x^2} - 1} $$ $$y = f\left( {{x^2}} \right)$$

$$\begin{array}{l}\frac{{dy}}{{dx}} = f'\left( {{x^2}} \right) \times \left[ {2x} \right]\\now\,\,\,f'\left( x \right) = \sqrt {2{x^2} - 1} \\f'\left( {{x^2}} \right) = \sqrt {2{x^4} - 1} \\\therefore \,\frac{{dy}}{{dx}} = \sqrt {2{x^4} - 1} \times 2x = 2\end{array}$$
Question 7
1 / -0

Find the slope of the normal to the curve $$2x^{2} - xy + 3y^{2} = 18$$ at $$(3,1)$$.

A
$$\dfrac {-11}{3}$$

B
$$\dfrac {3}{11}$$

C
$$\dfrac {11}{3}$$

D
$$\dfrac {-3}{11}$$

Solution

Equation of curve $$2{ x }^{ 2 }-xy+3{ y }^{ 2 }=18$$

Differentiating above equation w.r.t $$x,$$

$$4x-y-x\dfrac { dy }{ dx }+6y\dfrac { dy }{ dx } =0$$

$$4x-y+\dfrac { dy }{ dx } (6y-x)=0$$

$$\dfrac { dy }{ dx } (6y-x)=y-4x$$

$$\dfrac { dy }{ dx } =\dfrac { y-4x }{ 6y-x } $$

$$\dfrac { dy }{ dx| } _{ (3,1) }=\dfrac { 1-4\left( 3 \right) }{ 6-3 } $$

$$\dfrac { dy }{ dx| } _{ (3,1) }=\dfrac { -11 }{ 3 } $$

Slope of normal to the curve
$$=\dfrac { -1 }{ \dfrac { dy }{ dx| } _{ (3,1) } } $$

$$=\dfrac { -1 }{ \dfrac { -11 }{ 3 } } $$

$$=\dfrac { 3 }{ 11 } $$

$$\boxed { \therefore Ans =\dfrac { 3 }{ 11 } } $$
Question 8
1 / -0

Area of the triangle formed by the tangent at $$x=2$$ on the curve $$y= \dfrac{8}{4+x^2}$$ with the coordinate axes is (in sq. units)

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

$$y=\dfrac {8}{4+x^2}$$

$$At \,\,x=2\ y=1$$

$$y' =\dfrac {-8(2x)}{(4+x^2)^2}=\dfrac {-16\times 2}{8^2}=\dfrac {-1}{2}$$

Tangent is $$y=mx+c$$

$$y=\dfrac {-1}{2}x+c$$

passes through $$(2, 1)\ C=2$$

$$2y+x=4$$
Base of triangle $$=4$$

Height of triangle $$=\dfrac {4}{2}=2$$

Area of $$\Delta =\dfrac {1}{2}\times 4\times 2=4\ sq.unit$$
Question 9
1 / -0

The curve $$y = ax^3 + bx^2 + cx + 8$$ touches $$x-$$ axis at $$P(-2, 0)$$ and cuts $$y-$$ axis at a point $$Q$$ where its gradient is $$3$$. The values of $$a, b, c$$ are respectively ?

A
$$-\dfrac{5}{4}, -3, 3$$

B
$$0, \dfrac{1}{4},3$$

C
$$\dfrac{1}{4}, 0, 3$$

D
$$\dfrac{1}{4}, - \dfrac{1}{4}, 3$$

Solution

$$y=ax^{3}+bx^{2}+cx+8$$
$$\Rightarrow 0=-8a+4b+c(-2)+8$$
$$0=-8a+4b-2c+8 ----(1)$$

It touches $$y-$$ axis at $$Q\Rightarrow x=0$$

$$y=8$$

$$Q.(0,8)$$

$$\dfrac{dy}{dx}=3ax^{2}+2bx+c$$

Given gradient is $$3$$ at $$G(0,8)$$

$$3=c$$
$$\therefore c=3$$

$$\because$$ the curve touches $$x-$$ axis

$$\therefore$$ slope of the tangent at $$P$$ is $$0$$

$$\dfrac{dy}{dx}=3ax^{2}+2bx+3$$

$$0=12a-4b+3$$

$$12a-4b=-3 ---- (1)$$

eqn $$(1)$$ $$-8a+4b-6+8=0$$
$$-8a+4b=-2 --- (ii)$$

Adding $$(10$$ & $$(2)$$, we get

$$12a-4b=-3$$
$$-8a+4b=-2$$
____________
$$4a=-5$$
$$a=-5/4$$
$$b=-3$$
$$\boxed{\therefore a=-5/4, b=-3, c=3}$$
Question 10
1 / -0

The curve satisfying D.E $$y dx - (x+3{y}^{2})dy=0$$ and passing through the point $$(1,1)$$ also passes through the point:

A
$$\left( \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right) $$

B
$$\left( -\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } \right) $$

C
$$\left( \cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } \right) $$

D
$$\left(- \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right) $$

Solution

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Re-Attempt Weekly Quiz Competition

Application of Derivatives Test - 42

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Application of Derivatives Test - 42

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SHARING IS CARING

Question 1

$$k<7$$

$$-5< k <7$$

$$k>-5$$

$$none\ of\ these$$

Solution

Question 2

$$36$$

$$25$$

$$24$$

$$15$$

Solution

Question 3

$$f(x)+f'(x)=0$$ has at least one root

$$f(x)-f'(x)=0$$ has at least one real root

$$f(x)\cdot f'(x)=0$$ has at least one real root

None of these

Solution

Question 4

Both $$f(x)$$ and $$g(x)$$ are increasing functions

Both $$f(x)$$ and $$g(x)$$ are decreasing function

$$f(x)$$ is an increasing function

$$g(x)$$ is an increasing function

Question 5

$$h$$ is increasing whenever $$f$$ is increasing and decreasing whenever $$f$$ is decreasing

$$h$$ is increasing whenever $$f$$ is decreasing

$$h$$ is decreasing whenever $$ f$$ is increasing

Nothing can be said in general

Solution

Question 6

2

1

-2

-1

Solution

Question 7

$$\dfrac {-11}{3}$$

$$\dfrac {3}{11}$$

$$\dfrac {11}{3}$$

$$\dfrac {-3}{11}$$

Solution

Question 8

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 9

$$-\dfrac{5}{4}, -3, 3$$

$$0, \dfrac{1}{4},3$$

$$\dfrac{1}{4}, 0, 3$$

$$\dfrac{1}{4}, - \dfrac{1}{4}, 3$$

Solution

Question 10

$$\left( \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right) $$

$$\left( -\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } \right) $$

$$\left( \cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } \right) $$

$$\left(- \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right) $$

Solution

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