Continuity and Differentiability Test

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Continuity and Differentiability Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The function given by y=$$ \left | x-1 \right | $$ is differentiable function and $$ f(1/n) $$ = 0 $$ \forall n\geq 1 $$ and n\epsilon I $$, then

A
f(x)=0, $$ x\epsilon (0,1] $$

B
f(0)=0, f'(0) =0

C
f(0)= 0=f'(0), $$ x\epsilon (0,1] $$

D
f(0)=0 and f'(0) need not to be zero

Solution

Given that f(x) is a continuous and differentiable function and$$f(\dfrac{1}{x})=0,x=n,n\epsilon l$$
$$\therefore f(0^{+})=f(\dfrac{1}{\infty })=0$$
since R.H.L=0,$$\therefore f(0)=0$$for f(x)to be continuous Alsio f'(0)
$$=\underset{h\rightarrow 0}{lim}\dfrac{f(h)-f(0)}{h-0}=\underset{h\rightarrow 0}{lim}\dfrac{f(h)}{h}=0$$
$$[Usingf(0)=0][\because f(0^{+})=0]$$Hence f(0)=0 f(0)
Question 2
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The largest value of c such that there exists a differential function $$ h(x) for -c < x < c $$ that is a solution of $$ y_1 = 1 +y^2 $$ with h(0) = 0 is

A
$$ 2\pi $$

B
$$ \pi $$

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

D
$$ \frac { \pi }{4} $$

Solution
Question 3
1 / -0

If $$y=\sin^{-1}\left\{\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}$$ then $$\dfrac{dy}{dx}=$$

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

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

C
$$\dfrac{1}{2(1+x^{2})}$$

D
$$none\ of\ these$$

Solution
Question 4
1 / -0

If $$y=\tan^{-1}\left\{\dfrac{\sqrt{1+x^{2}}-1}{x}\right\}$$ then $$\dfrac{dy}{dx}=?$$

A
$$\dfrac{1}{(1+x^{2})}$$

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

C
$$\dfrac{1}{2(1+x^{2})}$$

D
$$none\ of\ these$$

Solution

$$y=\tan^{-1}\left\{\dfrac{\sqrt{1+x^2}-1}{x}\right\}$$
Let $$x=\tan\theta$$
$$y=\tan^{-1}\left\{\dfrac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\right\}$$

$$y=\tan^{-1}\left(\dfrac{\sec\theta-1}{\tan\theta}\right)$$

$$y=\tan^{-1}\left(\dfrac{\dfrac{1}{\cos\theta}-1}{\dfrac{\sin\theta}{\cos\theta}}\right)$$

$$y=\tan^{-1}\left(\dfrac{1-\cos\theta}{\sin\theta}\right)$$

$$y=\tan^{-1}\left(\dfrac{2\sin^2\dfrac{\theta}{2}}{2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}}\right)$$

$$y=\tan^{-1}\left(\dfrac{\sin\dfrac{\theta}{2}}{\cos\dfrac{\theta}{2}}\right)$$

$$y=\tan^{-1}\left(\tan\dfrac{\theta}{2}\right)$$

$$y=\dfrac{\theta}{2}$$

Substituting $$\theta=\tan^{-1}x$$
$$y=\dfrac{\tan^{-1}x}{2}$$

$$\dfrac{dy}{dx}=\dfrac{d}{dx}\dfrac{\tan^{-1}x}{2}$$

$$\therefore$$ $$\dfrac{dy}{dx}=\dfrac{1}{2(1+x^2)}$$
Question 5
1 / -0

If $$y=\sec^{-1}\left(\dfrac{1}{2x^{2}-1}\right)$$ then $$\dfrac{dy}{dx}=?$$

A
$$\dfrac{-2}{(1+x^{2})}$$

B
$$\dfrac{-2}{(1-x^{2})}$$

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

D
$$none\ of\ these$$

Solution

$$y=\sec^{-1}\left(\dfrac{1}{2x^{2}-1}\right)$$

Put $$x=\cos\theta$$

As we know, $$\cos 2\theta=2\cos^2 \theta-1$$

Then $$y=\sec^{-1}\left(\dfrac{1}{2\cos^{2}\theta-1}\right)=\sec^{-1}(\sec 2\theta)=2\theta=2\cos^{-1}x$$.

$$\therefore \dfrac{dy}{dx}=\dfrac{-2}{\sqrt{1-x^{2}}}$$
Question 6
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Which of the following function is thrice differentiable at x=0?

A
$$f\left ( x \right )=\left | x^{3} \right |$$

B
$$f\left ( x \right )=x^{3}\left | x \right |$$

C
$$f\left ( x \right )=\left | x \right |\sin ^{3}x$$

D
$$f\left ( x \right )=x\left | \tan ^{3}x \right |$$
Question 7
1 / -0

If $$y=\sec^{-1}\left(\dfrac{x^{2}+1}{x^{2}-1}\right)$$ then $$\dfrac{dy}{dx}=?$$

A
$$\dfrac{-2}{(1+x^{2})}$$

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

C
$$\dfrac{-1}{(1-x^{2})}$$

D
$$none\ of\ these$$

Solution

$$y=\sec^{-1}\left(\dfrac{x^{2}+1}{x^{2}-1}\right)$$

Put $$x=\cot\theta$$.

As we know, $$\cos 2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$$

Then $$y=\sec^{-1}\left(\dfrac{1+\tan^{2}\theta}{1-\tan^{2}\theta}\right)=\sec^{-1}\left(\dfrac{1}{\cos 2\theta}\right)=\sec^{-1}(\sec 2\theta)=2\theta=2\cot^{-1}x$$.

$$\therefore \dfrac{dy}{dx}=\dfrac{-2}{1+x^2}$$
Question 8
1 / -0

If $$y=\tan ^{-1} \sqrt{\dfrac{x+1}{x-1}}, $$ then $$ \dfrac{d y}{d x} $$ is

A
$$\dfrac{-1}{2|x| \sqrt{x^{2}-1}} $$

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

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

D
$$ None \ of \ these$$

Solution

Let $$ x=\sec \theta $$

$$\text { Then } y=\tan ^{-1} \sqrt{\dfrac{\sec \theta+1}{\sec \theta-1}} $$

$$\quad=\tan ^{-1} \sqrt{\dfrac{1+\cos \theta}{1-\cos \theta}}=\tan ^{-1}\left(\cot \dfrac{\theta}{2}\right)$$

$$\Rightarrow y=\tan ^{-1}\left\{\tan \left(\dfrac{\pi}{2}-\frac{\theta}{2}\right)\right\}=\dfrac{\pi}{2}-\dfrac{1}{2} \sec ^{-1} x$$

$$\Rightarrow \dfrac{d y}{d x}=-\dfrac{1}{2} \times \dfrac{1}{|x| \sqrt{x^{2}-1}}$$
Question 9
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Directions For Questions

If $$\phi (x)$$ is a differentialble real valued function satisfying $$\phi (x)+2\phi (x) \le 1$$, then it can be adjusted as $$e^{2x}\phi' (x) +2e${2x}\phi (x) \ge e^{2x}$$ or $$\dfrac {d}{dx}\left(e^{2x}\phi (x)-\dfrac {e^{2x}}{2}\right)\ge 0$$ or $$\dfrac {d}{dx}e^{2x} \left(\phi (x)-\dfrac {1}{2}\right)\ge 0$$.
Here $$e^{2x}$$ is called integration factor which helps in creating single differential coefficient as shown above. Answer the following questions:

...view full instructions

If $$P(1)=0$$ and $$\dfrac {dP(x)}{dx} > P(x)$$ for all $$x \ge 1$$, then

A
$$P(x) > 0\forall x > 1$$

B
$$P(x)$$ is a constant function

C
$$P(x) < 0\forall x > 1$$

D
None of these

Solution

$$\dfrac {dP(x)}{dx} > P(x)$$
$$\Rightarrow \ e^{-x}\dfrac {dP(x)}{dx}-e^{-x}P(x) > 0$$
$$\Rightarrow \dfrac {d}{dx}(P(x)e^{-x}) > 0$$
$$\Rightarrow P(x)e^{-x}$$ is an increasing function.
$$\Rightarrow P(x)e^{-x} > P(1)e^{-1} \forall a \ge 1$$
$$\Rightarrow P(x) e^{-x} > 0 \forall x > 1\Rightarrow P(x) > 0 \forall x > 1$$.
Question 10
1 / -0

The function $$ f(x)=\dfrac{x}{1+\left | x \right |} $$ is differentiable

A
only at non-integer points

B
everywhere

C
only at $$x=0$$

D
none of these

Solution

Thus $$f(x)$$ is differentiable at $$x=0$$
$$f(x)=\dfrac{x}{1+\left | x \right |}$$is everywhere differentiable

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 34

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Continuity and Differentiability Test - 34

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SHARING IS CARING

Question 1

f(x)=0, $$ x\epsilon (0,1] $$

f(0)=0, f'(0) =0

f(0)= 0=f'(0), $$ x\epsilon (0,1] $$

f(0)=0 and f'(0) need not to be zero

Solution

Question 2

$$ 2\pi $$

$$ \pi $$

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

$$ \frac { \pi }{4} $$

Solution

Question 3

$$\dfrac{-1}{2\sqrt{1-x^{2}}}$$

$$\dfrac{1}{2\sqrt{1-x^{2}}}$$

$$\dfrac{1}{2(1+x^{2})}$$

$$none\ of\ these$$

Solution

Question 4

$$\dfrac{1}{(1+x^{2})}$$

$$\dfrac{2}{(1+x^{2})}$$

$$\dfrac{1}{2(1+x^{2})}$$

$$none\ of\ these$$

Solution

Question 5

$$\dfrac{-2}{(1+x^{2})}$$

$$\dfrac{-2}{(1-x^{2})}$$

$$\dfrac{-2}{\sqrt{1-x^{2}}}$$

$$none\ of\ these$$

Solution

Question 6

$$f\left ( x \right )=\left | x^{3} \right |$$

$$f\left ( x \right )=x^{3}\left | x \right |$$

$$f\left ( x \right )=\left | x \right |\sin ^{3}x$$

$$f\left ( x \right )=x\left | \tan ^{3}x \right |$$

Question 7

$$\dfrac{-2}{(1+x^{2})}$$

$$\dfrac{2}{(1+x^{2})}$$

$$\dfrac{-1}{(1-x^{2})}$$

$$none\ of\ these$$

Solution

Question 8

$$\dfrac{-1}{2|x| \sqrt{x^{2}-1}} $$

$$ \dfrac{-1}{2 x \sqrt{x^{2}-1}} $$

$$ \dfrac{1}{2 x \sqrt{x^{2}-1}} $$

$$ None \ of \ these$$

Solution

Question 9

$$P(x) > 0\forall x > 1$$

$$P(x)$$ is a constant function

$$P(x) < 0\forall x > 1$$

None of these

Solution

Question 10

only at non-integer points

everywhere

only at $$x=0$$

none of these

Solution

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