Continuity and Differentiability Test

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Continuity and Differentiability Test - 35

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Question 1
1 / -0

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 $$H(x_0)=0$$ for some $$x=x_0$$ and $$\dfrac {d}{dx}H(x) > 2cxH(x)$$ for all $$x \le x_0$$, where $$c > 0$$, then

A
$$H(x)=0$$ has root for $$x > x_0$$

B
$$H(x)=0$$ has no root for $$x > x_0$$

C
$$H(x)=0$$ is constant function

D
None of these

Solution

Given that $$\dfrac {d}{dx}H(x) > 2cxH(x)$$
$$\Rightarrow \ e^{-cx^2}\dfrac {d}{dx}H(x)-e^{-cx^2}2cxH(x) > 0$$
$$\Rightarrow \ \dfrac {d}{dx}(H(x)e^{-cx^2}) > 0$$
$$\Rightarrow \ H(x)e^{-cx^2} > 0$$ is an increasing function.
But $$H(x_0)=0$$ and $$e^{-cx^2}$$ is always positive.
$$\Rightarrow \ H(x) > 0$$ for all $$x > x_0$$
$$\Rightarrow \ H(x)$$ cannot be zero for any $$x > x_0$$.
Question 2
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Given a function $$f:[0,4] \rightarrow \mathrm{R}$$ is differentiable, then for $$\displaystyle \operatorname{some} \alpha, \beta \in(0,2), \int_{0}^{4} f(t) d t$$ equals to

A
$$f\left(\alpha^{2}\right)+f\left(\beta^{2}\right)$$

B
$$2 \alpha f\left(\alpha^{2}\right)+2 \beta f\left(\beta^{2}\right)$$

C
$$\alpha f\left(b^{2}\right)+\beta f\left(\alpha^{2}\right)$$

D
$$f(\alpha) f(\beta)[f(\alpha)+f(\beta)]$$

Solution

$$\displaystyle I=\int_{0}^{4} f(t) dt$$
Put $$t=x^2$$

$$\Rightarrow d t=2 x dx$$

$$\displaystyle \therefore I=2 \int_{0}^{2} x f\left(x^{2}\right) dx$$

From Lagrange's Mean Value Theorem
$$\dfrac{\displaystyle \int_{0}^{2} 2 x f\left(x^{2}\right) d x-\int_{0}^{0} 2 x f\left(x^{2}\right) d x}{2-0}=2 y f\left(y^{2}\right)$$

$$\Rightarrow \displaystyle \int_{0}^{2} 2 x f\left(x^{2}\right) d x=2 \times 2 y f\left(y^{2}\right) \\$$

$$=2\left\{\dfrac{2 \alpha f\left(\alpha^{2}\right)+2 \beta f\left(\beta^{2}\right)}{2}\right\}\\$$
(where $$0<\beta<y<\alpha<2$$, and using intermediate Mean Value Theorem.)
Question 3
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$$ f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2) $$ and $$ g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x) $$
The value of $$ f(3) $$ is

A
1

B
0

C
-1

D
-2

Solution

$$ f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2) $$ and $$ g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x) $$

$$\text { Here put } g^{\prime}(1)=a, g^{\prime \prime}(2)=b \quad\quad\quad(1)$$

$$\text { Then } f(x)=x^{2}+a x+b, f(1)=1+a+b \Rightarrow f^{\prime}(x)=2 x+a$$

$$\quad f^{\prime \prime}(x)=2 $$

$$\therefore g(x)=(1+a+b) x^{2}+(2 x+a) x+2=x^{2}(3+a+b)+a x\quad+2$$

$$\Rightarrow g^{\prime}(x)=2 x(3+a+b)+a \text { and } g^{\prime \prime}(x)=2(3+a+b)$$

$$\text { Hence, } g^{\prime}(1)=2(3+a+b)+a \quad\quad\quad(2)$$

$$g^{\prime \prime} (2)=2(3+a+b)\quad\quad\quad(3)$$

From (1),(2) and (3) we have

$$ a=2(3+a+b)+a $$

$$\text { and } b=2(3+a+b) $$

$$\Rightarrow 3+a+b=0 \text { and } b+2 a+6=0$$

Hence, $$ b=0 $$ and $$ a=-3 . $$ So, $$ f(x)=x^{2}-3 x $$ and $$ g(x)=-3 x+2 $$

$$ f(x)=x^{2}-3 x \Rightarrow f(3)=9-9=0 $$
Question 4
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If f(x) = $$\displaystyle \int_{0}^{x}(f(t))^2 dt f:R→R$$ be differentiable function and f(g(x)) is differentiable function at x=a, then

A
g(x) must be differentiable at x=a

B
g(x) may be non-differentiable at x=a

C
g(x) must be discontinuous at x=a

D
None of the above

Solution

Here, $$ f'(x)=(f(x))^2>0;\dfrac{d}{dx} f(g(x)) |_{x=a} $$
$$=f'(g(x))\displaystyle \lim_{x→a}\dfrac{g(x)-g(a)}{x-a}$$
It implies that g(x) must be differentiable at x=a.
Question 5
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Directions For Questions

Let $$f: R \rightarrow R$$ be a differentiable function such that $$f(x)=x^{2}+\int_{0}^{x} e^{-t} f(x-t) d t$$

...view full instructions

$$y=f(x)$$ is

A
injective but not surjective

B
surjective but not injective

C
bijective

D
neither injective nor surjective

Solution

$$\begin{aligned}f(x)=& x^{2}+\int_{0}^{x} e^{-t} f(x-t) d t \\=& x^{2}+\int_{0}^{x} e^{-(x-t)} f(x-(x-t)) d t \\&=x^{2}+e^{-x} \int_{0}^{x} e^{t} f(t) d t\end{aligned}$$
Differentiating w.r.t. $$x,$$ we get
$$\begin{aligned}\Rightarrow f^{\prime}(x)=& 2 x-e^{-x} \int_{0}^{x} e^{t} f(t) d t+e^{-x} \cdot e^{x} f(x) \\&=2 x-e^{-x} \int_{0}^{x} e^{t} f(t) d t+f(x)\end{aligned}$$
$$\Rightarrow f^{\prime}(x)=2 x+x^{2} \\$$
$$\Rightarrow f(x)=\dfrac{x^{3}}{3}+x^{2}+c\\$$
Also $$f(0)=0\\$$ [using equation from equation]
$$\Rightarrow f(x)=\dfrac{x^{3}}{3}+x^{2} \\$$
$$\Rightarrow f^{\prime}(x)=x^{2}+2 x\\$$
$$\Rightarrow f^{\prime}(x)=0$$ has real roots, hence $$f^{\prime}(x)$$ is non-monotonic
Hence, $$f(x)$$ is many-one, but range is $$R$$, hence surjective.
Question 6
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If $$f:R \rightarrow (0, \infty)$$ be a differentiable function $$f(x)$$ satisfying $$f(x+ y) - f(x - y) = f(x) \{ f(y) - f(y) - y \}, \forall x, y \epsilon R, (f(y) \neq f(-y)$$ for all $$y \epsilon R)$$ and $$f' (0) = 2010$$.
Now answer the following questions

Which of the following is true for $$f(x)$$

A
$$f(x)$$ is one-one and into

B
$$\{ f(x) \}$$ is non-periodic, where {.} denotes fractional part of x.

C
$$f(x) = 4$$ has only two solutions.

D
$$f(x) = f^{-1} x$$ has only one solution

Solution

Here, $$2 f' (x) = \displaystyle \lim_{h \rightarrow 0} \left ( \frac{f(x + h) - f(x)}{h} + \frac{f(x x - h) - f(x)}{-h} \right )$$
$$ = \displaystyle \lim_{h \rightarrow 0} \left ( \frac{f(x + h) - f(x - h)}{h} \right )$$ (i)
$$\therefore 2f'(0) \displaystyle \lim_{h \rightarrow 0} \left ( \frac{f(h) - f (-h)}{h} + \frac{f(-h) - f(0)}{-h} \right )$$
$$ = \displaystyle \lim_{h \rightarrow 0}\frac{f(h) - f(-h)}{h}$$ ... ..(ii)
Now by given relation, we have
$$f(h) - f(-h) = \frac{f(x + h) - f(x - h)}{-h}$$ and $$f(0) = 1$$
From Eqs. (i) and (ii), we have $$\frac{f'(x)}{f(x)} = 2010$$
$$\Rightarrow$$ $$f(x) = e^{2010e}, f(0) = 1$$
$$\therefore \{ f(x) \}$$ is non-periodic.
Question 7
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If $$u=\sin^{-1}\Bigg(\dfrac{2x}{1+x^2}\Bigg)$$ and $$v=\tan^{-1}\Bigg(\dfrac{2x}{1-x^2}\Bigg)$$, then $$\dfrac{du}{dv}$$ is

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

B
$$x$$

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

D
$$1$$

Solution

d
Question 8
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The function $$f(x)=\Bigg\{\dfrac{\sin x}{x}+\cos x,if\,x\neq 0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k,if\,x=0$$ is continuous at $$x=0$$,then the value of k is

A
$$3$$

B
$$2$$

C
$$1$$

D
$$1.5$$

Solution

B
Question 9
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The function $$f(x)=|x|+|x-1|$$ is

A
continuous at $$x = 0$$ as well as at $$x = 1$$.

B
continuous at $$x = 1$$ but not at $$x = 0$$.

C
discontinuous at $$x = 0$$ as well as at $$x = 1$$.

D
continuous at $$x = 0$$ but not at $$x = 1$$.

Solution

a
Question 10
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If $$f(x) = \sin^{-1} \left ( \frac{4^{x + \frac{1}{2}}}{1 + 2^{4x}} \right )$$, which of the following is not the derivative of f(x)?

A
$$\frac{2.4^{x} \cdot \log 4}{1 + 4^{2x}}$$

B
$$\frac{4^{x + 1} \cdot \log 2}{1 + 4^{2x}}$$

C
$$\frac{4^{x + 1} \cdot \log 4}{1 + 4^{4x}}$$

D
$$\frac{2^{2^(x + 1)} \cdot \log 2}{1 + 2^{4x}}$$

Solution

Put $$4^x = \tan \theta$$. Then $$\theta = \tan^{-1}(4^x)$$
$$\therefore f(x) = \sin^{-1} \left ( \frac{2 \tan \theta}{1 + \tan \theta} \right )$$
$$= \sin^{-1} (\sin 2 \theta)$$
$$= 2 \theta$$
$$= 2 \tan^{-1} (4^x)$$
$$\therefore f'(x) = 2 \times \frac{1}{1 + (4^x)^2} \times 4^x \log 4$$
$$= \frac{2.4^x \cdot \log 4}{1 + 4^{2x}}$$ ....(a)
$$= \frac{2.4^x \cdot 2 \log 2}{1 + 4^{2x}}$$
$$= \frac{4^{x + 1} \cdot \log 2}{1 + 4^{2x}}$$ ...(b)
$$= \frac{(2^{2})^{x + 1} \cdot \log 2}{1 + 2^{4x}}$$
$$ = \frac{2^{2^(x + 1)} \cdot \log 2}{1 + 2^{4x}}$$ ....(d)

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 35

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Continuity and Differentiability Test - 35

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SHARING IS CARING

Question 1

$$H(x)=0$$ has root for $$x > x_0$$

$$H(x)=0$$ has no root for $$x > x_0$$

$$H(x)=0$$ is constant function

None of these

Solution

Question 2

$$f\left(\alpha^{2}\right)+f\left(\beta^{2}\right)$$

$$2 \alpha f\left(\alpha^{2}\right)+2 \beta f\left(\beta^{2}\right)$$

$$\alpha f\left(b^{2}\right)+\beta f\left(\alpha^{2}\right)$$

$$f(\alpha) f(\beta)[f(\alpha)+f(\beta)]$$

Solution

Question 3

1

0

-1

-2

Solution

Question 4

g(x) must be differentiable at x=a

g(x) may be non-differentiable at x=a

g(x) must be discontinuous at x=a

None of the above

Solution

Question 5

injective but not surjective

surjective but not injective

bijective

neither injective nor surjective

Solution

Question 6

$$f(x)$$ is one-one and into

$$\{ f(x) \}$$ is non-periodic, where {.} denotes fractional part of x.

$$f(x) = 4$$ has only two solutions.

$$f(x) = f^{-1} x$$ has only one solution

Solution

Question 7

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

$$x$$

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

$$1$$

Solution

Question 8

$$3$$

$$2$$

$$1$$

$$1.5$$

Solution

Question 9

continuous at $$x = 0$$ as well as at $$x = 1$$.

continuous at $$x = 1$$ but not at $$x = 0$$.

discontinuous at $$x = 0$$ as well as at $$x = 1$$.

continuous at $$x = 0$$ but not at $$x = 1$$.

Solution

Question 10

$$\frac{2.4^{x} \cdot \log 4}{1 + 4^{2x}}$$

$$\frac{4^{x + 1} \cdot \log 2}{1 + 4^{2x}}$$

$$\frac{4^{x + 1} \cdot \log 4}{1 + 4^{4x}}$$

$$\frac{2^{2^(x + 1)} \cdot \log 2}{1 + 2^{4x}}$$

Solution

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