Differentiation Test 20

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Differentiation Test 20

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Weekly Quiz Competition

Question 1
1 / -0

If $$ \displaystyle {f}'\left ( x \right )=g\left ( x \right ) $$ and $$ \displaystyle {g}'\left ( x \right )=-f\left ( x \right ) $$ and $$ \displaystyle f\left ( 2 \right )=4={f}'\left ( 2 \right ) $$ then $$ \displaystyle f^{2}\left ( 16 \right )+g^{2}\left ( 16 \right ) $$ is

A
16

B
32

C
64

D
None of these

Solution

$$\displaystyle \frac { d }{ dx } \left( { f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right) \right) \\ =2\left[ f\left( x \right) f'\left( x \right) +g\left( x \right) g'\left( x \right) \right] \\ =2\left[ f\left( x \right) -g\left( x \right) f\left( x \right) \right] =0$$
Thus $${ f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right) $$ is constant.
Therefore $${ f }^{ 2 }\left( 16 \right) +{ g }^{ 2 }\left( 16 \right) ={ f }^{ 2 }\left( 2 \right) +{ g }^{ 2 }\left( 2 \right) \\ ={ f }^{ 2 }\left( 2 \right) +{ \left( f'\left( 2 \right) \right) }^{ 2 }=16+16=32$$
Question 2
1 / -0

If $$ \displaystyle y=\sec ^{ -1 }{ \left( \frac { x+1 }{ x-1 } \right) } +\sin ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) } $$ then $$ \displaystyle \frac{dy}{dx} $$ is equal to

A
$$0$$

B
$$ \displaystyle x+1 $$

C
$$1$$

D
$$-1$$

Solution

$$\displaystyle y=\sec ^{ -1 }{ \left( \frac { x+1 }{ x-1 } \right) } +\sin ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) } $$
$$\displaystyle =\cos ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) } +\sin ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) } =\frac { \pi }{ 2 } $$
However the above function is defined only for values of x, given by
$$\displaystyle -1\le \frac { x-1 }{ x+1 } \le 1$$
i.e. $$\displaystyle \frac { x-1 }{ x+1 } +1\ge 0$$ and $$\displaystyle \frac { x-1 }{ x+1 } -1\le 0$$
i.e. $$\displaystyle \frac { 2x }{ x+1 } \ge 0$$ and $$\displaystyle \frac { 2 }{ x+1 } \ge 0$$
i.e $$x<-1$$ or $$\ge 0$$ and $$x>-1$$
i.e $$x\ge 0$$
Hence we have
$$\displaystyle y=\frac { \pi }{ 2 } ,\quad x\ge 0\Rightarrow \frac { dy }{ dx } =0,\quad x>0$$
Question 3
1 / -0

Let $$f\left( x \right)=\sqrt { x-1 } +\sqrt { x+24-10\sqrt { x-1 } } ;1<x<26$$ be a real valued function. Then $$f'(x)$$ for $$1<x<26$$ is

A
$$0$$

B
$$\displaystyle \frac { 1 }{ \sqrt { x-1 } } $$

C
$$2\sqrt { x-1 } -5$$

D
none of these

Solution

Let, $$ f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}$$
$$ 1< x< 26$$   be a real valued   function ,
We   have,
$$ f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}$$
This can be written as
$$ f(x) = \sqrt{x -1} + \sqrt{\left ( 5 - \sqrt{x-1} \right )^{2} }$$
$$ \because 1< x< 26$$
$$ f(x) = \sqrt{x-1} + 5 - \sqrt{x-1} $$
$$ f(x) = 5 $$
On   differentiating   wrt   x , we get
$$ f{}'(x) = 0 $$
Hence ,Option A
Question 4
1 / -0

Find the area of the triangle whose vertices are $$(3,2), \ (-2, -3)$$ and $$(2,3)$$.

A
$$8$$ sq.unit

B
$$7$$ sq.unit

C
$$6$$ sq.unit

D
$$5$$ sq.unit

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 });$$ $$({ x }_{ 2 },y_2)$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$
$$ = \left |\dfrac{ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3 (y_1 - y_2) } {2} \right |$$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (3,2) $$ ; $$({ x }_{ 2 },{ y }_{ 2 }) = (-2,-3) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (2,3)$$
in the area formula, we get

Area = $$\left | \dfrac {3(-3-3) +(-2)(3-2) + 2 (2-(-3)) }{2}\right | = 5 $$ sq. unit
Question 5
1 / -0

If for a non-zero $$x,$$ the function $$f(x)$$ satisfies the equation $$\displaystyle af\left( x \right)+bf\left( \frac { 1 }{ x } \right) =\frac { 1 }{ x } -5\left( a\neq b \right) $$ then $$f'(x)$$ is equal to

A
$$\displaystyle \frac { 1 }{ { b }^{ 2 }-{ a }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } +b \right) $$

B
$$\displaystyle \frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } +b \right) $$

C
$$\displaystyle \frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } -b \right) $$

D
none of these

Solution

We have, $$\displaystyle af\left( x \right)+bf\left( \frac { 1 }{ x } \right) =\frac { 1 }{ x } -5$$
Substituting $$\displaystyle\frac{1}{x}$$ in place of $$x,$$ we have
$$\displaystyle af\left( \frac { 1 }{ x } \right) +bf\left( x \right)=x-5$$ ...(2)
Eliminating $$\displaystyle f\left( \frac { 1 }{ x } \right) $$ from equation (1) and (2), we have
$$\displaystyle \left( { a }^{ 2 }-{ b }^{ 2 } \right) f\left( x \right) =a\left( \frac { 1 }{ x } -5 \right) -b\left( x-5 \right) $$
$$\displaystyle \Rightarrow f\left( x \right) =\frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left[ \frac { a }{ x } -bx+5\left( b-a \right) \right] $$
$$\displaystyle \therefore f'\left( x \right) =\frac { 1 }{ { b }^{ 2 }-{ a }^{ 2 } } \left[ \frac { a }{ { x }^{ 2 } } +b \right] $$
Question 6
1 / -0

If $${ S }_{ n }$$ denotes the sum of $$n$$ terms of $$g.p$$. whose common ratio is $$r$$, then $$\displaystyle \left( r-1 \right) \frac { d{ S }_{ n } }{ dr } $$ is equal to

A
$$\left( n-1 \right) { S }_{ n }+n{ S }_{ n-1 }$$

B
$$\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }$$

C
$$\left( n-1 \right) { S }_{ n }$$

D
None of these

Solution

We have, $$\displaystyle {S}_{n}=\frac { a\left( { r }^{ n }+1 \right) }{ r-1 } $$
$$\Rightarrow \left( r-1 \right) { S }_{ n }={ ar }^{ n }-a$$
Differentiating both sides with respect to $$r,$$ we get
$$\displaystyle \left( r-1 \right) \frac { { dS }_{ n } }{ dr } +{ S }_{ n }=na{ r }^{ n-1 }-0$$
$$\displaystyle \Rightarrow \left( r-1 \right) \frac { { dS }_{ n } }{ dr } =na{ r }^{ n-1 }-{ S }_{ n }$$
$$=n[n$$th term from of $$G.P.]$$ $$-{S}_{n}$$
$$=n\left( { S }_{ n }-{ S }_{ n-1 } \right) -{ S }_{ n }$$
$$=\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }.$$
Question 7
1 / -0

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $$(2,2)$$, $$(4,4)$$ and $$(2,6)$$.

A

5

B
3

C
1

D
0

Solution

Let $$ A(2,2), B(4,4) $$ and $$ C(2,6) $$ be the vertices of a given triangle ABC.

Let D, E, and F be the midpoints of AB, BC and CA respectively.

Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y }_{

2 }) $$ is calculated by the formula $$ \left( \dfrac { { x }_{ 1 }+{ x

}_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right) $$

Using this formula,the coordinates of D, E, and F are given as $$\displaystyle D \left (\frac{2+4}{2},\frac {2+4}{2} \right ), E\left (\frac {4+2}{2},\frac {4+6}{2}\right ) $$ and $$ F\left (\dfrac {2+2}{2},\dfrac {2+6}{2} \right ) $$

i.e., $$ D(3,3), E(3,5) and F(2,4) $$

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \dfrac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (3,3) $$ ; $$({ x

}_{ 2 },{ y }_{ 2 }) = (3,5) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (2,4)$$

in the area formula, we get

Area of triangle DEF $$ = \left| \dfrac { 3(5-4)+(3)(4-3)+2(3-5) }{ 2

} \right| = \left| \dfrac { 3 +3 -4 }{ 2 } \right| =

\dfrac {2}{2} = 1 \ sq \ units $$
Question 8
1 / -0

Find $$ \displaystyle \frac{dy}{dx}$$ if $$ y=x^{x} $$

A
$$ \displaystyle x^{x}\left ( lnx+1 \right ) $$

B
$$ \displaystyle x^{x}\left ( lnx-1 \right ) $$

C
$$ \displaystyle x .x^{x-1}$$

D
$$ \displaystyle x^{x-1}\left ( lnx+1 \right ) $$

Solution

$$y = x^x$$
Take log both sides,
$$\log y = x\log x$$
Now differentiate both side w.r.t $$x$$
$$\Rightarrow \cfrac{1}{y}\cfrac{dy}{dx} = 1+\log x$$
$$\Rightarrow \cfrac{dy}{dx} =x^x(1+\log x)$$
Question 9
1 / -0

$$ \displaystyle f^{ ' }\left( x \right) =g\left( x \right) $$ and $$ \displaystyle g^{ ' }\left( x \right) =-f\left( x \right)$$ for all real x and $$ \displaystyle f\left( 5 \right) =2=f^{ ' }\left( 5 \right) $$ then $$ \displaystyle f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right) $$ is -

A
$$2$$

B
$$4$$

C
$$8$$

D
None of these

Solution

Given, $$ \displaystyle f^{ ' }\left( x \right) =g\left( x \right)$$ and $$g^{ ' }\left( x \right) =-f\left( x \right)$$
Now $$ \displaystyle \frac { d }{ dx } \left[ f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right) \right] =2f\left( x \right) f^{ ' }\left( x \right) +2g\left( x \right) g^{ ' }\left( x \right) $$
$$ \displaystyle =2f\left( x \right) g\left( x \right) -2g\left( x \right) f\left( x \right) =0$$
$$ \displaystyle \therefore \quad f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right)$$ =constant
$$ \displaystyle f^{ 2 }\left( 5 \right) +g^{ 2 }\left( 5 \right) = 4 + 4 = 8$$
$$ \displaystyle \therefore \quad f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right) $$ = 8
Question 10
1 / -0

If $$y = \displaystyle \tan^{-1}\left (\cot x \right ) +\cot^{-1}(\tan x),$$ then $$\displaystyle \frac{dy}{dx}$$ is equal to-

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$-2$$

Solution

$$\dfrac{d}{dx}\left(\tan^{-1} \left(\cot \left(x\right)\right)+\cot^{-1} \left(\tan \left(x\right)\right)\right)$$
$$=\dfrac{d}{dx}\left(\tan^{-1} \left(\cot \left(x\right)\right)\right)+\dfrac{d}{dx}\left(\cot^{-1} \left(\tan \left(x\right)\right)\right)$$
$$=-\dfrac{\csc ^2\left(x\right)}{\cot ^2\left(x\right)+1}-\dfrac{\sec ^2\left(x\right)}{\tan ^2\left(x\right)+1}$$
$$=-\dfrac{2\csc ^2\left(x\right)\sec ^2\left(x\right)}{\left(\tan ^2\left(x\right)+1\right)\left(\cot ^2\left(x\right)+1\right)}=-2\dfrac{csc^2(x)sec^2{(x)}}{csc^2{(x)}sec^2{(x)}}=-2$$

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Re-Attempt Weekly Quiz Competition

Differentiation Test 20

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Differentiation Test 20

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Rank

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SHARING IS CARING

Question 1

16

32

64

None of these

Solution

Question 2

$$0$$

$$ \displaystyle x+1 $$

$$1$$

$$-1$$

Solution

Question 3

$$0$$

$$\displaystyle \frac { 1 }{ \sqrt { x-1 } } $$

$$2\sqrt { x-1 } -5$$

none of these

Solution

Question 4

$$8$$ sq.unit

$$7$$ sq.unit

$$6$$ sq.unit

$$5$$ sq.unit

Solution

Question 5

$$\displaystyle \frac { 1 }{ { b }^{ 2 }-{ a }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } +b \right) $$

$$\displaystyle \frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } +b \right) $$

$$\displaystyle \frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } -b \right) $$

none of these

Solution

Question 6

$$\left( n-1 \right) { S }_{ n }+n{ S }_{ n-1 }$$

$$\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }$$

$$\left( n-1 \right) { S }_{ n }$$

None of these

Solution

Question 7

5

3

1

0

Solution

Question 8

$$ \displaystyle x^{x}\left ( lnx+1 \right ) $$

$$ \displaystyle x^{x}\left ( lnx-1 \right ) $$

$$ \displaystyle x .x^{x-1}$$

$$ \displaystyle x^{x-1}\left ( lnx+1 \right ) $$

Solution

Question 9

$$2$$

$$4$$

$$8$$

None of these

Solution

Question 10

$$1$$

$$0$$

$$-1$$

$$-2$$

Solution

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