Differentiation Test 21

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Differentiation Test 21

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \frac{d}{dx}\left ( \tan ^{-1}\left ( \frac{\sqrt{x}-x}{1+x^{3/2}} \right ) \right )$$ equals $$\displaystyle ($$for $$x\geq 0)$$

A
$$\displaystyle \frac{1}{2\sqrt{x}(1+x)}-\frac{1}{1+x^{2}}$$

B
$$\displaystyle \frac{1}{2\sqrt{x}(1+x)}+\frac{1}{1+x^{2}}$$

C
$$\displaystyle \frac{1}{1+x}-\frac{1}{1+x^{2}}$$

D
None of these

Solution

Let, $$\displaystyle y=\tan ^{-1}\left ( \frac{\sqrt{x}-x}{1+x^{3/2}} \right )$$
$$\displaystyle y=\tan ^{-1}(\sqrt{x})-\tan ^{-1}x$$
$$\therefore \displaystyle \frac{dy}{dx}=\frac{1}{(1+x)}\times \frac{1}{2\sqrt{x}}-\frac{1}{1+x^{2}}$$
Question 2
1 / -0

The area of the triangle whose vertices are $$A(1,1), B(7, 3)$$ and $$C(12, 2)$$ is

A
$$25$$ square units

B
$$8$$ square units

C
$$16$$ square units

D
$$12$$ square units

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
$$\therefore$$
Area $$= \dfrac{1}{2} [1(1)+ 7(1)+ 12(-2)] = -8$$, but area can not be negative
Hence, area = 8 square units.
Question 3
1 / -0

$$\displaystyle \frac{d}{dx}\left ( \tan ^{-1}\left ( \frac{a-x}{1+ax} \right ) \right )$$ equals if ax > -1

A
$$\displaystyle \frac{a}{1+x^{2}}$$

B
$$\displaystyle \frac{1}{1+x^{2}}$$

C
$$\displaystyle- \frac{a}{1+x^{2}}$$

D
$$\displaystyle -\frac{1}{1+x^{2}}$$

Solution

Let, $$\displaystyle y=\tan ^{-1}\left ( \frac{a-x}{1+ax} \right )$$
$$\displaystyle y=\tan ^{-1}(a)-\tan ^{-1}(x)$$
$$\therefore \displaystyle \frac{dy}{dx}=-\frac{1}{1+x^2}$$
Question 4
1 / -0

If $$f(x) = \displaystyle \left | \cos x-\sin x \right |$$ then $$\displaystyle f'\left ( \dfrac{\pi}4 \right )$$ is equal to-

A
$$\displaystyle \sqrt{2}$$

B
$$\displaystyle -\sqrt{2}$$

C
$$0$$

D
$$Does\ not\ exist$$

Solution

$$f\left( x \right) =\left| \cos { x } -\sin { x } \right| $$
$$\displaystyle =\begin{cases} \cos { x-\sin { x } \quad \quad x<\dfrac { \pi }{ 4 } } \\ -\cos { x+\sin { x } } \quad x>\dfrac { \pi }{ 4 } \end{cases}$$
$$\displaystyle f'\left( x \right) =\begin{cases} -\sin { x } -\cos { x } \quad x<\dfrac { \pi }{ 4 } \\ +\sin { x+\cos { x } \quad x>\dfrac { \pi }{ 4 } } \end{cases}$$
Hence $$\displaystyle f'\left( \dfrac { \pi }{ 4 } \right) $$ does not exists.
Question 5
1 / -0

If $$\displaystyle y=\frac{1}{1+x^{\beta -\alpha}+x^{\gamma -\alpha}}+\frac{1}{1+x^{\alpha-\beta}+x^{\gamma -\beta }}+\frac{1}{1+x^{\alpha -\gamma }+x^{\beta-\gamma }}$$
then $$\displaystyle \frac{dy}{dx}$$ is equal to-

A
$$0$$

B
$$1$$

C
$$\displaystyle

(a+\beta +\gamma )X^{\alpha +\beta +\gamma -1}$$

D
None of these

Solution

$$\displaystyle y=\frac{1}{1+x^{\beta -\alpha}+x^{\gamma -\alpha}}+\frac{1}{1+x^{\alpha-\beta}+x^{\gamma -\beta }}+\frac{1}{1+x^{\alpha -\gamma }+x^{\beta-\gamma }}$$
$$\displaystyle y=\frac{1}{\displaystyle 1+\frac{x^{\beta }}{x^{\alpha}}+\frac{x^{\gamma }}{x^{\alpha} }}+\frac{1}{{\displaystyle 1+\frac{x^{\alpha }}{x^{\beta}}+\frac{x^{\gamma }}{x^{\beta} }}}+\frac{1}{\displaystyle 1+\frac{x^{\alpha}}{x^{\gamma} }+\frac{x^{\beta} }{x^{\gamma }}}$$
$$\displaystyle y=\frac{x^{\alpha }+x^{b}+x^{\gamma }}{x^{\alpha }+x^{\beta }+x^{\gamma }}=1$$
$$\therefore \cfrac{dy}{dx}=0$$
Question 6
1 / -0

If $$\displaystyle y=\left | \cos x \right |+\left | \sin x \right |$$ then $$\displaystyle \frac{dy}{dx}$$ at $$x=\dfrac{2\pi }{3}$$ is:

A
$$\displaystyle \frac{1-\sqrt{3}}{2}$$

B
$$0$$

C
$$\displaystyle \frac{\sqrt{3}-1}{2}$$

D
None of these

Solution

In the neighborhood of $$\:x=\dfrac{2\pi}3$$
$$\Rightarrow y=-\cos\:x+\sin\:x$$
$$\Rightarrow \left ( \cfrac{dy}{dx} \right )=\sin\:x+\cos\:x$$
Thus at$$\:x=\dfrac{2\pi}3$$
$$\cfrac{dy}{dx}=\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}=\dfrac{1}{2}\left ( \sqrt{3}-1 \right )$$
Question 7
1 / -0

$$\displaystyle \frac{d}{d\theta }\left ( \tan ^{-1}\left ( \frac{1-\cos \theta }{\sin \theta } \right ) \right )$$ equals if $$\displaystyle-\pi <\theta <\pi $$

A
$$\dfrac12$$

B
$$1$$

C
$$\sec \theta $$

D
$$cosec \theta $$

Solution

$$\dfrac{1-\cos \left(x\right)}{\sin \left(x\right)}$$
Let $$u=\dfrac{x}{2}$$
$$=\dfrac{-\left(1-2\sin ^2\left(u\right)\right)+1}{2\cos \left(u\right)\sin \left(u\right)}$$
$$=\dfrac{\sin \left(u\right)}{\cos \left(u\right)}$$
$$=\tan \left(u\right)=tan(\dfrac{x}{2})$$
So,
$$\frac{d}{dθ}\left(\arctan \left(\frac{1-\cos \left(θ\right)}{\sin \left(θ\right)}\right)\right)=\frac{d}{dθ}\left(\arctan \left(tan\frac{\theta}{2}\right)\right)=\dfrac{d}{d\theta}(\dfrac{\theta}{2})=\dfrac{1}{2}$$
Question 8
1 / -0

Find the area of the right-angled triangle whose vertices are $$(2, -2)$$ , $$(-2, 1)$$ and $$(5, 2).$$

A
$$\displaystyle 5\sqrt{2}$$ sq. units

B
$$\displaystyle \dfrac{25}{2}$$ sq. units

C
$$\displaystyle 15\sqrt{2}$$ sq. units

D
$$10$$ sq. units

Solution

Let the points be $$ A(2,-2), B(-2,1), C(5,2) $$.

Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1} \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Hence, Length of side AB $$ = \sqrt { \left(-2-2 \right) ^{ 2 }+\left(1 + 2\right) ^{ 2 } } = \sqrt { 16+ 9 } = \sqrt { 25 } = 5 $$
Length of side BC $$ = \sqrt {\left(5 + 2\right) ^{ 2 }+\left(2-1\right) ^{ 2 } } = \sqrt { 49 + 1 } = \sqrt { 50} $$
Length of side AC $$ = \sqrt { \left(5-2 \right) ^{ 2 }+\left(2+ 2\right) ^{ 2 } } = \sqrt { 9+16 } = \sqrt { 25 } = 5 $$
Since, $$ {(\sqrt { 50 }) }^{2} = { 5 }^{2} + { 5 }^{2} $$,
$$ \Rightarrow {BC}^{2} ={AB}^{2} + {AC}^{2} $$
Hence, the triangle has a right angle at $$A$$, with $$AB$$ and $$AC$$ as base and height.
So, area of the triangle $$ABC = \cfrac {1}{2} \times base \times height = \cfrac {1}{2} \times 5 \times 5= \cfrac {25}{2}$$ sq units.
Question 9
1 / -0

Find the area of the triangle whose vertices are $$(a, b + c), (a, b - c)$$ and $$(-a, c)$$.

A
$$2ac$$

B
$$2bc$$

C
$$b(a + c)$$

D
$$c (a - b)$$

Solution

Area of a triangle $$ (A) =\left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (a,b+c) $$ , $$({ x }_{ 2

},{ y }_{ 2 }) = (a,b-c) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (-a,c) $$

$$ A=\left| \cfrac { a(b-c-c)+a(c-b-c)-a(b+c-b+c) }{ 2 } \right| $$
$$=\left| \cfrac { a(b-2c)+a(-b)-a(2c) }{ 2 } \right|$$
$$ =\left| \cfrac { ab-2ac-ab-2ac }{ 2 } \right| $$
$$=\left| \cfrac { -4ac }{ 2 } \right| $$
$$=2ac$$ square units
Question 10
1 / -0

Let $$\displaystyle y=(1+x^{2})\tan^{-1}(x-x)$$ and $$\displaystyle f(x)=\frac1{2x}\frac {dy}{dx},$$ then $$f(x)+\cot^{-1}x$$ is equal to

A
$$0$$

B
$$\displaystyle \frac {\pi}{2}$$

C
$$\displaystyle -\frac {\pi}{2}$$

D
$$\pi$$

Solution

$$\displaystyle y=(1+x^{2})\tan^{-1}(x-x)$$
$$\Rightarrow \displaystyle \dfrac {dy}{dx}=2x\tan^{-1}x+(1+x^2).\dfrac{1}{1+x^2}-1=2x\tan^{-1}x$$
$$\displaystyle\Rightarrow f(x)= \dfrac {\dfrac {dy}{dx}}{2x}=\tan^{-1}x$$
$$\therefore \displaystyle f(x)+\cot^{-1}x= \tan^{-1}+\cot^{-1}x=\dfrac {\pi}{2}$$

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Differentiation Test 21

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Differentiation Test 21

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Question 1

$$\displaystyle \frac{1}{2\sqrt{x}(1+x)}-\frac{1}{1+x^{2}}$$

$$\displaystyle \frac{1}{2\sqrt{x}(1+x)}+\frac{1}{1+x^{2}}$$

$$\displaystyle \frac{1}{1+x}-\frac{1}{1+x^{2}}$$

None of these

Solution

Question 2

$$25$$ square units

$$8$$ square units

$$16$$ square units

$$12$$ square units

Solution

Question 3

$$\displaystyle \frac{a}{1+x^{2}}$$

$$\displaystyle \frac{1}{1+x^{2}}$$

$$\displaystyle- \frac{a}{1+x^{2}}$$

$$\displaystyle -\frac{1}{1+x^{2}}$$

Solution

Question 4

$$\displaystyle \sqrt{2}$$

$$\displaystyle -\sqrt{2}$$

$$0$$

$$Does\ not\ exist$$

Solution

Question 5

$$0$$

$$1$$

$$\displaystyle (a+\beta +\gamma )X^{\alpha +\beta +\gamma -1}$$

None of these

Solution

Question 6

$$\displaystyle \frac{1-\sqrt{3}}{2}$$

$$0$$

$$\displaystyle \frac{\sqrt{3}-1}{2}$$

None of these

Solution

Question 7

$$\dfrac12$$

$$1$$

$$\sec \theta $$

$$cosec \theta $$

Solution

Question 8

$$\displaystyle 5\sqrt{2}$$ sq. units

$$\displaystyle \dfrac{25}{2}$$ sq. units

$$\displaystyle 15\sqrt{2}$$ sq. units

$$10$$ sq. units

Solution

Question 9

$$2ac$$

$$2bc$$

$$b(a + c)$$

$$c (a - b)$$

Solution

Question 10

$$0$$

$$\displaystyle \frac {\pi}{2}$$

$$\displaystyle -\frac {\pi}{2}$$

$$\pi$$

Solution

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$$\displaystyle

(a+\beta +\gamma )X^{\alpha +\beta +\gamma -1}$$