Relations and Functions Test

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Relations and Functions Test - 70

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Question 1
1 / -0

The value of f(0), so that the function
f(x) = $$ \dfrac{2x-sin^{-1}x}{2x+tan^{-1}x} $$ is continuous at each point in its domain, is equal to

A
2

B
1/3

C
2/3

D
-1/3

Solution

The function f is clearly continuous at each point in its domain except possibly at x=0 Given that f(x) is continuous at x=0
Therfore,f (0) = $$ \underset{x\rightarrow 0}{lim}f(x) $$
$$ =\underset{x\rightarrow 0}{lim}\dfrac{2x-sin^{-1}x}{2x+tan^{-1}x} $$

$$ = \underset{x\rightarrow 0}{lim}\dfrac{2-\frac{(sin^{-1}x)}x}{2+\frac{(tan^{-1}x)}x}$$
$$\dfrac{2-1}{2+1}=\dfrac13$$
Question 2
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The function $$f(x)= \dfrac{(3^{x}-1^{})^2}{\sin x. \ln(1+x)}, x\neq 0 $$ , is continuous at $$x=0$$. Then the value of $$f(0)$$ is

A
2log 3

B
$$ (\log_{e}3)^{2} $$

C
$$ \log_{e} 6 $$

D
None of these

Solution

Given f(x) is continuous at $$x=0$$
$$ \Rightarrow \underset{x\rightarrow 0}{\lim}f(x)=f(0) $$
$$ \Rightarrow \underset{x\rightarrow 0}{\lim}\dfrac{(3^{x}-1)^{2}}{\sin x\ln(1+x)}=f(0)$$
$$ \Rightarrow f(0)=\underset{x\rightarrow 0}{\lim} \dfrac{\bigg({\dfrac{3^x-1}{x}}\bigg)^2}{\dfrac{\sin x}{x}\dfrac{\ln(1+x)}{x}}=$$ $$ (\log_e3)^{2} $$
Question 3
1 / -0

Directions For Questions

If $$a_{o} = x, a_{n+1} = f(a_n)$$, where n = 0, 1, 2,.....then answer the following question

...view full instructions

If f: $$R\rightarrow R$$ be given by $$f(x) = 3 + 4x$$ and $$a_n = A + Bx$$, then which of the following is not true?

A
A + B + 1 = $$2^{2n + 1}$$

B
| A - B| = 1`

C
$$lim_{n \to \infty} \dfrac{A}{B} = -1$$

D
None of these

Solution
Question 4
1 / -0

Let $$g(x) = f(x) - 1$$. If $$f(x) + f(1 - x) = 2 \space \forall \space x \space \epsilon \space R$$, then $$g(x)$$ is symmetrical about

A
Origin

B
The line $$x = \frac{1} {2}$$

C
The point $$\left (1, 0 \right )$$

D
The point $$\left (\frac {1} {2}, 0 \right )$$

Solution
Question 5
1 / -0

Directions For Questions

$$f(x) = \begin{cases} x-1, -1 \leq x \leq 0\\x^2, 0\leq x\leq 1 \end{cases}$$ and g(x) = sin x, consider the functions.
$$h_1(x) = f(|g(x)|) \space and \space h_2(x) = |f(g(x))|$$.

...view full instructions

Which of the following is not true about $$h_1(x)$$?

A
It is periodic function with period $$\pi$$

B
Range is [0,1]

C
Domain if R

D
None of these

Solution
Question 6
1 / -0

Directions For Questions

If $$a_{o} = x, a_{n+1} = f(a_n)$$, where n = 0, 1, 2,.....then answer the following question

...view full instructions

If f: $$R\rightarrow R$$ be given by $$f(x) = 3 + 4x$$ and $$a_n = A + Bx$$, then which of the following is not true?

A
A + B + 1 = $$2^{2n + 1}$$

B
| A - B| = 1`

C
$$\displaystyle \lim_{n \to \infty} \dfrac{A}{B} = -1$$

D
None of these

Solution

Since $$a_1 = g(x) = 3 + 4x$$
$$\therefore a_2 = g\{g^2(x)\} = g(3+4x) = 3 + 4(3+4x) = (4^2 - 1) + 4^x$$

$$a_3 = g\{g^(x)\} = g(15 + 4^2x) = 3 + 4 (15 + 4^2x) = 63 + 4^3x = (4^3 - 1) + 4^3 x$$

Similarly, we get $$a_n = (4^n - 1) + 4^n x$$
$$\Rightarrow A = 4^n - 1 \space and B = 4^n$$

$$\Rightarrow A + B + 1 = 2^{2n + 1}, |a - b| = 1\space and\space lim_{n \to \infty}\dfrac{4^n - 1}{4^n}$$

$$= lim_{n \to \infty}\left(1 - \dfrac{1}{4^n}\right) = 1$$
Question 7
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Directions For Questions

$$f(x) = \begin{cases} 2x + a, x \geq -1\\ bx^2 + 3, x < -1 \end{cases}$$

and $$g(x) = \begin{cases} x + 4, 0 \leq x \leq 4\\ -3x -2, -2 < x < 0 \end{cases}$$

...view full instructions

g(f(x)) is not defined if

A
$$a\space \epsilon ( 10, \infty) b\space \epsilon (5, \infty)$$

B
$$a\space \epsilon ( 4, 10) b\space \epsilon (5, \infty)$$

C
$$a\space \epsilon ( 10, \infty) b\space \epsilon (0, 1)$$

D
$$a\space \epsilon ( 4, 10) b\space \epsilon (1, 5)$$

Solution
Question 8
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Let $$f(x)$$ and $$g(x)$$ be differentiable for $$0\times < 1$$ such that $$f(0)=0, g(0), f(1)=6$$. Let there exist a real number $$c$$ in $$(0,1)$$ such that $$f'(c)=2g'(c)$$, then the value of $$g(1)$$ must be

A
$$1$$

B
$$3$$

C
$$2.5$$

D
$$-1$$

Solution
Question 9
1 / -0

If g is the inverse of function $$f$$ and $$f'(x) = \frac{1}{1 + x}$$, then the value of g'(x) is equal to:

A
$$1 + x^7$$

B
$$\frac{1}{1 + [g(x)]^7}$$

C
$$1 + [g(x)]^7$$

D
$$7x^6$$

Solution

Since g is the inverse of $$f, f^{-1}(x) = g(x)$$
$$\therefore f[f^{-1}(x)] = f [g(x)] = x$$
$$\therefore f'[g(x)] \cdot \frac{d}{dx} [g(x)] = 1$$
$$\therefore f'[g(x)] \times g'(x) = 1$$
$$\therefore g'(x) = \frac{1}{f'[g(x)]}$$, where $$f'(x) = \frac{1}{1 + x^7}$$
$$\therefore g(x) = 1 + [g(x)]^7$$
Question 10
1 / -0

Computers use

A
decimal system

B
binary system

C
base 5 system

D
base 6 system

Solution

binary system