Limits and Derivatives Test

Result

Limits and Derivatives Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\displaystyle \lim_{x\rightarrow 0}\dfrac {1-\cos^{3}x}{x\sin x\cos x}$$ is

A
$$\dfrac {2}{5}$$

B
$$\dfrac {3}{5}$$

C
$$\dfrac {3}{2}$$

D
$$\dfrac {3}{4}$$

Solution
Question 2
1 / -0

Integrate:
$$lim_{x\rightarrow 0}\dfrac{(1-\cos{2x})^{2}}{2x\tan{x}-x\tan{2x}}$$

A
$$2$$

B
$$\dfrac{-1}{2}$$

C
$$-2$$

D
$$\dfrac{1}{2}$$

Solution
Question 3
1 / -0

The value of $$\lim_{x \rightarrow 0} \left(\dfrac{\tan x}{x}\right)^{1/x^{3}}$$ is-

A
$$0$$

B
$$\infty$$

C
$$e^{1/4}$$

D
$$Does\ not\ exist$$

Solution
Question 4
1 / -0

$$ \underset { x\rightarrow 0 }{ lim } \cfrac { { \left( 25 \right) }^{ x }-2\left( 15 \right)^ x+{ 9 }^{ x } }{ cos6x-cos2x } $$ is equal to :

A
$$ log \left ( \cfrac {5} {3} \right ) $$

B
$$ \cfrac {1} {4} log 15 $$

C
$$ - \cfrac {1} {16} \left( \cfrac {5} {3} \right) ^2 $$

D
$$log \left ( \cfrac {3} {5} \right ) $$
Question 5
1 / -0

$$Lt_{x\to 0} \dfrac{sin x - x+\dfrac{x^3}{6}}{x^5}=$$_________

A
$$\dfrac{1}{120}$$

B
$$\dfrac{-1}{120}$$

C
$$0$$

D
$$\dfrac{1}{6}$$

Solution
Question 6
1 / -0

If x + y = sin (x + y) then $$\dfrac{dy}{dx}$$ =

A
$$\dfrac{1}{2}$$

B
0

C
-1

D
$$\dfrac{1}{3}$$

Solution
Question 7
1 / -0

$${d}{dx}(sin^{5} x. sin5x) =$$

A
$$sin^{4}x. sin 5x $$

B
$$5 sin^{4}x. sin 6x$$

C
$$5 sin^{4}x . sin 5x$$

D
$$-5 sin^{4}x. sin 6x$$

Solution
Question 8
1 / -0

The value of $$\lim _{ x\rightarrow 0 }{ \dfrac { 1-\cos { ^{ 3 }x } }{ x\sin { x\cos { x } } } }$$ is

A
$$\dfrac{2}{5}$$

B
$$\dfrac{3}{5}$$

C
$$\dfrac{3}{2}$$

D
$$\dfrac{3}{4}$$

Solution
Question 9
1 / -0

If the anti derivative of f(x) is $$e^x$$ and that of g(x) of cos x respectively, then $$\int f(x) cos x dx + \int g(x) e^x dx=$$_____________.

A
$$f(x) g(x) +c$$

B
$$f(x) + g(x) +c$$

C
$$-e^xcos x +c$$

D
$$e^xcos x +c$$

Solution
Question 10
1 / -0

The ant derivative of $$\dfrac{1}{\left(21+x\tan x\right)^{2}}$$ with respect to $$x$$ is equal to

A
$$\dfrac{\tan x}{x\tan x+1}+C$$

B
$$\dfrac{\cot x}{x\tan x+1}+C$$

C
$$\dfrac{-\tan x}{x\tan x+1}+C$$

D
$$\dfrac{-1}{x\tan x+1}+C$$

Solution