Application of Derivatives Test

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Application of Derivatives Test - 64

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Question 1
1 / -0

The function $$f(x)=x-ln|2x+1|,x\epsilon \left(-100,\dfrac{-1}{2}\right)\cup \left(\dfrac{-1}{2},\dfrac{1}{2}\right)$$ is decreasing in interval

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

B
$$\left(-100,\dfrac{-1}{2}\right)$$

C
$$\left(\dfrac{-1}{2},0\right)$$ only

D
$$\left(0,\dfrac{1}{2}\right)$$ only

Solution

Given $$f\left( x \right) =x-log\left| 2x+1 \right| $$

Differentiate function w.r.t. x, we get,

$${ f }^{ \prime }\left( x \right) =\frac { d }{ dx } \left[ x-log\left| 2x+1 \right| \right] $$

$$\therefore { f }^{ \prime }\left( x \right) =1-\frac { 1 }{ 2x+1 } \times \frac { d }{ dx } \left( 2x+1 \right) $$

$$\therefore { f }^{ \prime }\left( x \right) =1-\frac { 1 }{ 2x+1 } \times 2$$

$$\therefore { f }^{ \prime }\left( x \right) =\frac { 2x+1-2 }{ 2x+1 } $$

$$\therefore { f }^{ \prime }\left( x \right) =\frac { 2x-1 }{ 2x+1 } $$

For function to be decreasing, $${ f }^{ \prime }\left( x \right) <0$$

$$\therefore \frac { 2x-1 }{ 2x+1 } <0$$

Case 1) $$2x-1<0$$ and $$2x+1>0$$

$$\therefore 2x<1$$ and $$2x>-1$$

$$\therefore x<\frac { 1 }{ 2 } $$ and $$x>-\frac { 1 }{ 2 } $$

Thus, function is decreasing in the interval $$\left( \frac { -1 }{ 2 } ,\frac { 1 }{ 2 } \right) $$

Case 2) $$2x-1>0$$ and $$2x+1<0$$

$$\therefore 2x>1$$ and $$2x<-1$$

$$\therefore x>\frac { 1 }{ 2 } $$ and $$x<-\frac { 1 }{ 2 } $$
Satisfying both the conditions simultaneously is not possible.
Question 2
1 / -0

For what values of a , $$f(x) = -x^3 +4ax^2 +2x-5$$ is decreasing $$\forall$$ x

A
$$(1, 2)$$

B
$$(3, 4)$$

C
R

D
no value of a

Solution

Given $$y=f\left( x \right) =-{ x }^{ 3 }+4a{ x }^{ 2 }+2x-5$$

For continuously decreasing function, $$\frac { dy }{ dx } <0$$

$$\therefore \frac { d }{ dx } \left[ -{ x }^{ 3 }+4a{ x }^{ 2 }+2x-5 \right] <0$$

$$\therefore -3{ x }^{ 2 }+4a\left( 2x \right) +2<0$$

$$\therefore -3{ x }^{ 2 }+8ax+2<0$$

$$\therefore 3{ x }^{ 2 }-8ax-2>0$$

$$\therefore \frac { -\left( -8a \right) \pm \sqrt { { \left( -8a \right) }^{ 2 }-4\times 3\times \left( -2 \right) } }{ 2\times 3 } >0$$

$$\therefore \frac { 8a\pm \sqrt { 64{ a }^{ 2 }+24 } }{ 6 } >0$$

$$\therefore 8a\pm \sqrt { 64{ a }^{ 2 }+24 } >0$$

Consider first equation $$8a+\sqrt { 64{ a }^{ 2 }+24 } >0$$

$$\therefore \sqrt { 64{ a }^{ 2 }+24 } >-8a$$
Squaring both sides, we get,

$$\therefore 64{ a }^{ 2 }+24>64{ a }^{ 2 }$$

$$\therefore 24>0$$
We don't get value of a in this condition.

Consider second equation $$8a-\sqrt { 64{ a }^{ 2 }+24 } >0$$
$$\therefore 8a>\sqrt { 64{ a }^{ 2 }+24 } $$
Squaring both sides, we get,

$$\therefore 64{ a }^{ 2 }>64{ a }^{ 2 }+24$$
$$\therefore 0>24$$
This condition is not possible.

Thus, option (D) is correct.
Question 3
1 / -0

Equation of the tangent line at $$y=\dfrac{a}{4}$$ to the curve $$y\left( { x }^{ 2 }+{ a }^{ 2 } \right) ={ ax }^{ 2 }$$.

A
$$8y=3\sqrt { 3 }x -a$$

B
$$8y=3\sqrt { 3 }x -5a$$

C
$$8y=3\sqrt { 3 }x +a$$

D
None

Solution
Question 4
1 / -0

Which of the following is not always correct for the function $$f(x)$$ and $$g(x)$$ these are inverse to each other.

A
If $$f(x)$$ is an increasing function, then $$g(x)$$ will also be increasing.

B
If $$f(x)$$ is an odd function, then $$g(x)$$ will also be an odd function.

C
Tangent at $$(\alpha,\ \beta )$$ to $$f(x)$$ is parallel to tangent at $$(\beta,\ \alpha)$$ to $$g(x)$$

D
Tangent at $$(\alpha,\ \beta )$$ to $$f(x)$$ and tangent at is parallel to $$(\beta,\ \alpha)$$ to $$g(x)$$ forms complementary angles with x-axis

Solution
Question 5
1 / -0

Solve it:-
$$y = x + \dfrac{1}{x},$$

A
$$x=1$$ is a point of local maximum

B
$$x=-1$$ is a point of local minimum

C
Local maximum value > Local minimum value

D
Local maximum value < Local minimum value

Solution
Question 6
1 / -0

The increasing function in $$(0,\ \pi /4)$$ is

A
$$\cos x+\sin x$$

B
$$\cos x-\sin x$$

C
$$\dfrac {\sin x}{x}$$

D
$$\dfrac {x}{\sin x}$$

Solution
Question 7
1 / -0

In the interval $$\left( {7,\infty } \right),f(x) = \left| {x - 5} \right| + 2\left| {x - 7} \right|$$ is

A
Increasing

B
Decreasing

C
Constant

D
Cannot be estimated

Solution
Question 8
1 / -0

Let $$f :\left[ 2,4 \right] \rightarrow \left[ 3,5 \right] $$ be a bijective decreasing function, then find $$\int _{ 2 }^{ 4 }{ f(t)dt- } \int _{ 3 }^{ 5 }{ { f }^{ -1 }(t)dt. } $$

A
2

B
14

C
0

D
10/3

Solution
Question 9
1 / -0

The increasing function in $$\left(0,\pi/4\right)$$ is

A
$$\cos x+\sin x$$

B
$$\cos x-\sin x$$

C
$$\dfrac{\sin x}{x}$$

D
$$\dfrac{x}{\sin x}$$

Solution
Question 10
1 / -0

If $$f(x)=\cos x+a^{2}x+b$$ is an increasing function for all values of $$x$$, then the value which $$'a'$$ can take.

A
$$a\in [-1,1]$$

B
$$a\in(-\infty,-1]\cup [1,\infty)$$

C
$$a\in [-1,\infty)$$

D
$$a\in(-\infty,1]$$

Solution