Differentiation Test 17

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Differentiation Test 17

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Weekly Quiz Competition

Question 1
1 / -0

If $$y=logx^3+3 sin^{-1}x+kx^2$$, then find $$\displaystyle \frac {dy}{dx}$$

A
$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

B
$$3\cdot \displaystyle \frac {1}{x^3}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

C
$$3\cdot \displaystyle \frac {1}{x}-3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

D
$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+2x$$

Solution

Here, $$y=\log x^3+3 \sin^{-1} x+kx^2$$

On differentiating we get

$$\displaystyle \dfrac {dy}{dx}=\dfrac {d}{dx}[\log x^3]+\dfrac {d}{dx}[3 \sin^{-1}x]+\dfrac {d}{dx}[kx^2]$$

$$=3\displaystyle \dfrac {d}{dx}[\log x]+3\dfrac {d}{dx}(\sin^{-1}x)+k\dfrac {d}{dx}(x^2)$$

$$=3\cdot \displaystyle \dfrac {1}{x}+3\cdot \dfrac {1}{\sqrt {1-x^2}}+k(2x)$$
Question 2
1 / -0

If $$\displaystyle y=\frac { x }{ a+\displaystyle\frac { x }{ b+\displaystyle\frac { x }{ a+\displaystyle\frac { x }{ b+.....\infty } } } } $$, then $$\cfrac{dy}{dx} =$$

A
$$\displaystyle\frac{a}{ab+2ay}$$

B
$$\displaystyle\frac{b}{ab+2by}$$

C
$$\displaystyle\frac{a}{ab+2by}$$

D
$$\displaystyle\frac{b}{ab+2ay}$$

Solution

$$\displaystyle y = \frac{x}{a+\frac{x}{b+y}}$$
$$\Rightarrow \displaystyle y =\frac{x(b+y)}{ab+ay+x}$$
$$\Rightarrow aby+ay^2+xy=bx+xy\Rightarrow aby+ay^2=bx$$
Differentiating both sides w.r.t $$x$$
$$\Rightarrow\displaystyle (ab+2ay)\frac{dy}{dx}=b\therefore \frac{dy}{dx}=\frac{b}{ab+2ay}$$
Question 3
1 / -0

Given : $$f(x)=4x^3-6x^2\cos2a+3x \sin 2a.\sin 6a+\sqrt{\ln (2a-a^2)}$$ then

A
$$f(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$

B
$${f}'(\displaystyle \frac{1}{2})<0$$

C
$$f'(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$

D
$${f}'(\displaystyle \frac{1}{2})>0$$

Solution

$$f'(x)=12x^{2}-12xcos2a+3sin2a.sin6a+0$$
$$f'(1/2)=\frac{12}{4}-\frac{12}{2}cos2a+3sin2a.sin6a$$
$$=3-6cos2a+1.5(cos4a-cos8a)$$
So, it will alwyas be greater than $$0$$
Question 4
1 / -0

If $$y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) + \sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)$$, then $$\displaystyle\frac{dy}{dx}$$ equals

A
$$1$$

B
$$0$$

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

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

Solution

$$y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) +

\sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)$$
$$y=\cos ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) +\sin ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) } } =\displaystyle \frac { \pi }{ 2 } $$ ............. $$\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } } $$
$$\therefore \displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left(\displaystyle \frac { \pi }{ 2 } \right) =0$$
Question 5
1 / -0

The solution set of $${f}'(x)>{g}'(x)$$ where $$f(x)=\displaystyle \frac{1}{2}(5^{2x+1})$$ & $$g(x)= 5^x+4x(\ln 5)$$ is

A
$$x>1$$

B
$$0< x< 1$$

C
$$x \leq 0$$

D
$$x>0$$

Solution

Given, $$f(x)= \dfrac {1}{2}(5^{2x+1})$$
$$f'(x)= 5^{2x+1}\times \log 5$$
$$g'(x)=5^{x}l\log 5 +4\ln 5=\ln 5(5^{x}+4)$$
$$f'(x)>g'(x)$$
$$=>5\times 5^{2x}>(5^{x}+4)$$
Let $$t=5^{x}$$
$$=> 5t^{2}-t-4>0$$
$$=>(t-1)(t+\frac{4}{5})>0$$$$=>t>1$$ or $$ t>-\frac{4}{5}$$
$$=>5^{x}>5^{0} =>x>0$$
so, $$x>0$$
Question 6
1 / -0

Directions For Questions

If f: $$R\rightarrow R$$ and $$f(x)=g(x)+h(x)$$ where $$g(x)$$ is a polynomial and $$h(x)$$ is a continuous and differentiable bounded function on both sides, then $$f(x)$$ is one-one, we need to differentiate $$f(x)$$. If $$f'(x)$$ changes sign in domain of $$f$$, then $$f $$ is many-one else one-one.

...view full instructions

$$f:R\rightarrow R$$ and $$\displaystyle f(x)=\frac {x(x^4+1)(x+1)+x^4+2}{x^2+x+1}$$, then $$f(x)$$ is

A
one-one ito

B
many-one onto

C
one-one onto

D
many-one into

Solution

$$\displaystyle f(x)=\frac {x(x^4+1)(x+1)+x^4+2}{x^2+x+1}$$
$$\displaystyle \quad \quad =\frac {x(x^4+1)(x+1)+(x^4+1)+1}{x^2+x+1}$$
$$\displaystyle \quad \quad =\frac {(x^4+1)(x^2+x+1)+1}{x^2+x+1}$$
$$\displaystyle \quad \quad =(x^4+1)+\frac {1}{x^2+x+1}$$
$$\displaystyle f'(x) =4x^3-\frac {2x+1}{(x^2+x+1)^2}=$$ not always positive or negative
Thus, $$f$$ is many one.
Also range and co-domain of $$f$$ are not same,
Hence is many-one into function
Question 7
1 / -0

Let $$\displaystyle f\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n } }{ n } \right) =\frac { f\left( { x }_{ 1 } \right) +f\left( { x }_{ 2 } \right) +...+f\left( { x }_{ n } \right) }{ n } $$ where all $${ x }_{ i }\in R$$ are independent to each other and $$n\in N$$. if $$f(x)$$ is differentiable and $$f'\left( 0 \right) =a,f\left( 0 \right) =b$$ and $$f'\left( x \right) $$ is equal to

A
$$a$$

B
$$0$$

C
$$b$$

D
None of these

Solution

Differentiating the given equation w.r.t. $${ x }_{ 1 },$$ we get
$$\displaystyle \dfrac { 1 }{ n } f'\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n } }{ n } \right) =\dfrac { f'\left( { x }_{ 1 } \right) }{ n } $$
$$[$$Since all $${ x }_{ i }'s$$ are independent to each other, $$\displaystyle \therefore \dfrac { d{ x }_{ i } }{ { dx }_{ j } } =0$$ if $$i\neq j$$ and $$\displaystyle \dfrac { { dx }_{ i } }{ { dx }_{ j } } =1$$ if $$(i=j)]$$
On putting $${ x }_{ 1 }={ x }_{ 2 }=...={ x }_{ n-1 }=0$$ and $${ x }_{ n }=x,$$ we get $$\displaystyle f'\left( \dfrac { x }{ n } \right) =f'\left( 0 \right) =a.$$
On integrating, we get $$\displaystyle nf'\left( \dfrac { x }{ n } \right) =ax+c$$
Since $$f(0)=b,$$ we have $$c=nb$$
$$\displaystyle \therefore nf'\left( \dfrac { x }{ n } \right) =ax+nb\Rightarrow nf'\left( x \right) =nax+nb\Rightarrow f\left( x \right) =ax+b.$$
$$\displaystyle \therefore f'\left( x \right) =a,\forall x\in R$$
Question 8
1 / -0

Suppose the function $$f(x)-f(2x)$$ has the derivative $$5$$ at $$x=1$$ and derivative $$7$$ at $$x=2$$.The derivative of the function $$f(x)-f(4x)$$ at $$x=1$$, has the value equal to

A
$$19$$

B
$$9$$

C
$$17$$

D
$$14$$

Solution

The derivative will be,$$f'(x)-2f'(2x)$$
$$f'(x)-2f'(2x)|_{1}=5=f'(1)-2f'(2)...........(1)$$
$$f'(x)-2f'(2x)|_{2}=7=f'(2)-2f'(4)$$
Hence, $$ 14=2f'(2)-4f'(4)$$ ........................(2)
Adding (1) and (2),
$$19=2f'(2)-4f'(4)+f'(1)-2f'(2)$$
$$\Rightarrow f'(1)-4f'(4)=19$$
Question 9
1 / -0

$$y=\sqrt{\sin x+\sqrt{\sin x +\sqrt{\sin x+-\infty }}}$$ then $$\displaystyle \frac{dy}{dx}$$ equals:$$(\sin x> 0)$$

A
$$\displaystyle \frac{\cos x}{2y-1}$$

B
$$\displaystyle \frac{y}{2\tan x+y\sec x}$$

C
$$\displaystyle \frac{1}{\sqrt{1+4\sin x}}$$

D
$$\displaystyle \frac{2\cos x}{\sin x+2y}$$

Solution

$$y=\sqrt{\sin x\sqrt{\sin x \sqrt{\sin x+-\infty }}}$$
Taking square on both the sides, we get
$${ y }^{ 2 }=\sin { x } +\sqrt { \sin x\sqrt { \sin x\sqrt { \sin x+-\infty } } } $$
$${ y }^{ 2 }=\sin { x } +y$$ ............ Since $$y=\sqrt{\sin x\sqrt{\sin x \sqrt{\sin x+-\infty }}}$$
$$2y\displaystyle \frac { dy }{ dx } =\cos { x } +\displaystyle \frac { dy }{ dx } $$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { \cos { x } }{ 2y-1 } $$
Question 10
1 / -0

If P(x) is a polynomial such that $$P\left ( x^{2}+1 \right )=\left \{ P\left ( x^{2} \right ) \right \}^{2}+1$$ and $$P(0)=0$$ then $$P^{'}(0)$$ is equal to

A
$$1$$

B
$$0$$

C
$$-$$1

D
none of these

Solution

Given, $$P\left ( x^{2}+1 \right )=\left \{ P\left ( x^{2} \right ) \right \}^{2}+1$$ ...(1)
Given $$P(0)=0$$
Put $$x=0$$ in (1)
$$\Rightarrow P(1)={P(0)}^{2}+1$$
$$\Rightarrow P(1)=1$$
Also, put $$x=1$$ in (1)
$$\Rightarrow P(2)=2$$
Also, since, P(x) is a polynomial so we have P(x)=x
$$\Rightarrow P'(x)=1$$
$$P'(0)=1$$

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Re-Attempt Weekly Quiz Competition

Differentiation Test 17

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Differentiation Test 17

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SHARING IS CARING

Question 1

$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

$$3\cdot \displaystyle \frac {1}{x^3}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

$$3\cdot \displaystyle \frac {1}{x}-3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+2x$$

Solution

Question 2

$$\displaystyle\frac{a}{ab+2ay}$$

$$\displaystyle\frac{b}{ab+2by}$$

$$\displaystyle\frac{a}{ab+2by}$$

$$\displaystyle\frac{b}{ab+2ay}$$

Solution

Question 3

$$f(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$

$${f}'(\displaystyle \frac{1}{2})<0$$

$$f'(x)$$ is not defined at $$x=\displaystyle \frac{1}{2}$$

$${f}'(\displaystyle \frac{1}{2})>0$$

Solution

Question 4

$$1$$

$$0$$

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

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

Solution

Question 5

$$x>1$$

$$0< x< 1$$

$$x \leq 0$$

$$x>0$$

Solution

Question 6

one-one ito

many-one onto

one-one onto

many-one into

Solution

Question 7

$$a$$

$$0$$

$$b$$

None of these

Solution

Question 8

$$19$$

$$9$$

$$17$$

$$14$$

Solution

Question 9

$$\displaystyle \frac{\cos x}{2y-1}$$

$$\displaystyle \frac{y}{2\tan x+y\sec x}$$

$$\displaystyle \frac{1}{\sqrt{1+4\sin x}}$$

$$\displaystyle \frac{2\cos x}{\sin x+2y}$$

Solution

Question 10

$$1$$

$$0$$

$$-$$1

none of these

Solution

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