Continuity and Differentiability Test

Result

Continuity and Differentiability Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

$$\frac { d }{ dx } (\sin ^{ -1 }{ \{ \frac { \sqrt { 1+x } +\sqrt { 1-x } }{ 2 } \} } )=$$

A
$$\frac { -1 }{ 2\sqrt { 1-{ x }^{ 2 } } } $$

B
$$\frac { 1 }{ 2\sqrt { 1-{ x }^{ 2 } } } $$

C
$$\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $$

D
$$\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } $$

Solution
Question 2
1 / -0

$$dx\left\{ sin{ }^{ -1 }(\frac { 5x+12\sqrt { 1-x{ }^{ 2 } } }{ 13 } ) \right\} =$$

A
$$\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $$

B
$$\frac { -1 }{ \sqrt { 1+{ x }^{ 2 } } } $$

C
$$\frac { 2 }{ \sqrt { 1-{ x }^{ 2 } } } $$

D
$$0$$
Question 3
1 / -0

If $$y=\cos ^{ -1 }{ (\frac { x-{ x }^{ -1 } }{ x+{ x }^{ -1 } } ) } \quad then\quad \frac { dy }{ dx } =$$

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

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

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

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

Solution
Question 4
1 / -0

The number of point at which the function F(x)= max$$\left\{ a-x,a+x,b \right\} -\infty <x<\infty ,0<a<b$$ cannot be differentiable is

A
1

B
2

C
3

D
none of these

Solution
Question 5
1 / -0

If $$x=\theta-\frac{1}{\theta},y=\theta+\frac{1}{\theta}$$ then $$\dfrac{dy}{dx}$$=

A
$$\dfrac{xy}{y^2+2}$$

B
y/x

C
-x/y

D
-y/x

Solution

$$x=\theta-\dfrac{1}{\theta}$$
$$\dfrac{dx}{d\theta}=1-\left(\dfrac{-1}{{\theta}^{2}}\right)$$
$$\therefore \dfrac{dx}{d\theta}=1+\dfrac{1}{{\theta}^{2}}$$
$$y=\theta+\dfrac{1}{\theta}$$
$$\dfrac{dy}{d\theta}=1+\left(\dfrac{-1}{{\theta}^{2}}\right)$$
$$\therefore \dfrac{dy}{d\theta}=1-\dfrac{1}{{\theta}^{2}}$$
$$\therefore \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{1-\dfrac{1}{{\theta}^{2}}}{1+\dfrac{1}{{\theta}^{2}}}$$
We have $$x=\theta-\dfrac{1}{\theta}$$ and $$y=\theta+\dfrac{1}{\theta}$$
$$\Rightarrow xy={\theta}^{2}-\dfrac{1}{{\theta}^{2}}$$
and $${y}^{2}={\left(\theta+\dfrac{1}{\theta}\right)}^{2}$$
$$\Rightarrow {y}^{2}={\theta}^{2}+\dfrac{1}{{\theta}^{2}}-2\theta\dfrac{1}{\theta}$$
$$\Rightarrow {y}^{2}={\theta}^{2}+\dfrac{1}{{\theta}^{2}}-2$$
$$\Rightarrow {y}^{2}+2={\theta}^{2}+\dfrac{1}{{\theta}^{2}}$$
$$\therefore \dfrac{dy}{dx}=\dfrac{1-\dfrac{1}{{\theta}^{2}}}{1+\dfrac{1}{{\theta}^{2}}}$$
$$=\dfrac{xy}{{y}^{2}+2}$$
Question 6
1 / -0

Let fbe twice differentiable function such that $${ g }^{ 1 }\left( x \right) =-f\left( x \right) and\quad \quad \quad \quad \quad { f }^{ 1 }\left( x \right) =g\left( x \right) ,\\ h(x)={ \left( f\left( x \right) \right) }^{ 2 }+{ \left( g\left( x \right) \right) }^{ 2 }.\quad Ifh(5)=11,\quad thenh(10)\quad is$$

A
22

B
11

C
0

D
8

Solution
Question 7
1 / -0

If $$y=\tan ^{ -1 }{ [x+\sqrt { 1+{ x }^{ 2 } } ] } $$ then $$\frac { dy }{ dx } =$$

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

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

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

D
None of these

Solution
Question 8
1 / -0

$$\frac { d[\sec ^{ -1 }{ (\sin { x } +{ x }^{ 2 }) } ] }{ dx } =$$

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

B
$$\frac { \cos { x } +2x }{ \sin { x } +{ x }^{ 2 }()\sqrt { { (\sin { x } +{ x }^{ 2 }) }^{ 2 }-1 } } $$

C
$$\frac { 2x }{ \sin { x } \sqrt { \sin { x } -1 } } $$

D
None of these
Question 9
1 / -0

If $$y={ tan }^{ -1 }\left( \dfrac { 1-{ cos }^{ 2 }x }{ 1+{ cos }^{ 2 } } \right) ,$$, then $$\dfrac { dy }{ xy } =$$

A
$$\dfrac { sinx }{ 1+{ cos }^{ 4 }x } $$

B
$$\dfrac { sinx }{ 1+{ cos }^{ 2 }x } $$

C
$$\dfrac { sin2x }{ 1+{ cos }^{ 4 }x } $$

D
$$\dfrac { sin2x }{ 1+{ cos }^{ 2 }x } $$

Solution
Question 10
1 / -0

Solve $$\frac{d\tan ^{-1}}{dx} (\frac{5x + 1}{3 - x - 6x^2}) = $$

A
0

B
1

C
$$\frac{1}{1 + (3x + 2)^2} + \frac{1}{1 + (2x - 1)^2}$$

D
$$\frac{3}{1 + (3x + 2)^2} + \frac{2}{1 + (2x - 1)^2}$$

Solution