Functions Test 40

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Functions Test 40

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Weekly Quiz Competition

Question 1
1 / -0

The domain of $$f(x)=\sqrt { x-4-2\sqrt { (x-5) } } -\sqrt { x-4+2\sqrt { (x-5) } } $$ is

A
$$[-5, \infty)$$

B
$$(-\infty , +2)$$

C
$$[5, \infty), ( -\infty, -2]$$

D
None of these

Solution
Question 2
1 / -0

The domain of the function $$f(x) = {}^{ 16-x }C_{ 2x-1 }+{}^{ 20-3x }C_{ 4x-5 }$$ , where the symbols have their usual meanings, is the set

A
$${2,3}$$

B
$${2,3, 4}$$

C
$${1, 2,3, 4, 5}$$

D
None of these

Solution
Question 3
1 / -0

The domain of $$f(x)=\dfrac { 1 }{ \sqrt { \left| \cos x \right| +\cos x} } $$ is

A
$$[-2n\pi, 2n\pi]$$

B
$$(2n\pi, 2n+1 \pi)$$

C
$$(\frac {(4n+1) \pi }{ 2 } , \frac {(4n+3) \pi }{ 2 })$$

D
$$(\frac { (4n-1)\pi }{ 2 } ,\frac { (4n+1)\pi }{ 2 } )$$

Solution
Question 4
1 / -0

The domain of the function $$\sqrt { (\log\ 5x) } $$ is

A
$$(1, \infty )$$

B
$$(0, \infty )$$

C
$$(0, 1 )$$

D
$$(0.5 , 1 )$$

Solution
Question 5
1 / -0

The domain of function $$f(x) = \log_{3 + x}(x^{2} -1)$$ is

A
$$(-3, -1) \cup (1, \infty )$$

B
$$\left [ -3, -1 \right) \cup \left [1, \infty \right )$$

C
$$\left (-3, -2 \right ) \cup \left (-2, -1 \right ) \cup \left (1, \infty \right )$$

D
$$\left [ -3, -2 \right) \cup \left (-2 -1 \right), \left (1, \infty \right )$$

Solution

The function $$f(x)$$ is to be defined when $$x^2 − 1 > 0$$. So,
$$x^2 > 1 \Rightarrow x< -1\ \text{or}\ x > 1$$
Also for $$f(x)$$ to be defined $$3 + x > 0$$ and $$3+x\neq 1$$
Therefore, $$x > −3$$ and $$x \neq −2$$
Hence, $$D_f = (− 3 − 2) \cup (− 2, -1) \cup (1,\infty)$$
Question 6
1 / -0

The domain of the function $$f(x) = \left [ log_{10} \left ( \frac{5x - x^{2}} {4} \right ) \right]^{1/2}$$ is

A
$$- \infty < x < \infty$$

B
$$1 \leqslant x \leqslant 4$$

C
$$ 4 \leqslant x \leqslant 16$$

D
$$ -1 \leqslant x \leqslant 1$$

Solution

We have $$f(x) = \left [ log_{10} \left ( \frac{5x - x^{2}} {4} \right ) \right]^{1/2}$$
From (1), clearly f(x) is defined for those values of x for
which $$log_{10} \left [\frac{5x - x^{2}} {4} \right] \geq 0 $$
$$ \Rightarrow \left (\frac {5x - x^{2}} {4} \right ) \geq 10^{0}$$
$$ \Rightarrow \left (\frac {5x - x^{2}} {4} \right ) \geq 1$$
$$\Rightarrow x^{2} - 5x + 4 \leq 0$$
$$\Rightarrow (x - 1) (x - 4) \leq 0$$
Hence the domain of the function is [1, 4]
Question 7
1 / -0

If $$\displaystyle {f}'(x) = g\,(x) $$ and $$\displaystyle {g}'(x) = - f\,(x) $$ for all $$ x $$ and $$ f\,(2) = 4 = {f}'(2) $$ then $$\displaystyle f^{2}\,(19) + g^{2} \,(19) $$ is

A
$$ 16 $$

B
$$ 32 $$

C
$$ 64 $$

D
None of these

Solution

Question 8

1 / -0

Let $$f(n)$$ denote the number of different ways in which the positive integer $$n$$ can be expressed as the sum of $$1s$$ and $$2s$$. For example, $$f(4) = 5$$, since $$4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1$$. Note that order of $$1s$$ and $$2s$$ is important.

$$f : N\rightarrow N$$ is

One-one and onto

One-one and into

Many-one and onto

Many-one and into

Solution

$$6 = 0(2) + 6(1) = 1(2) + 4(1) = 2(2) + 2(1) = 3(2) + 0(1)$$

Number of $$2s$$	Number of $$1s$$	Number of permutations
$$0$$	$$6$$	$$1$$
$$1$$	$$4$$	$$\dfrac {5!}{4!} = 5$$
$$2$$	$$2$$	$$\dfrac {4!}{2!2!} = 6$$
$$3$$	$$0$$	$$\dfrac {3!}{3!} = 1$$
		$$Total = 13$$

$$\therefore f(6) = 13$$

Now, $$f(f(6)) = f(13)$$

Number of $$1s$$	Number of $$2s$$	Number of permutations
$$13$$	$$0$$	$$1$$
$$11$$	$$1$$	$$\dfrac {12!}{11!} = 12$$
$$9$$	$$2$$	$$\dfrac {11!}{9!2!} = 55$$
$$7$$	$$3$$	$$\dfrac {10!}{7!3!} = 120$$
$$5$$	$$4$$	$$\dfrac {9!}{5!4!} = 126$$
$$3$$	$$5$$	$$\dfrac {8!}{3!5!} = 56$$
$$1$$	$$6$$	$$\dfrac {7!}{6!} = 7$$
		$$Total = 377$$

$$\therefore f(f(6)) = f(13) = 377$$

$$f(1) = 1(1)$$

$$f(2) = 2 (1, 1$$ or $$2)$$

$$f(3) = 3(1, 1, 1\ or\ 2, 1\ or\ 1, 2)$$

$$f(4) = 5$$ (explained in the paragraph)

By taking higher value of $$n$$ in $$f(n)$$, we always get more value of $$f(n)$$. Hence, $$f(x)$$ is one-one. Clearly, $$f(x)$$ is into.

Question 9
1 / -0

let $$f(x) = sin^2 x/2 + cos ^2 x/2 $$ and $$g(x) = sec^2 x - tan ^2 x.$$ The two functions are equal over the set

A
$$\phi$$

B
$$R$$

C
$$R-{ x:x (2n+1) \frac{\pi}{2}, n\in1}$$

D
None of these

Solution
Question 10
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) = $$ \underset{x\rightarrow 0}{lim}f(x) $$
$$ =\underset{x\rightarrow 0}{lim}\dfrac{2x-sin^{-1}x}{2x+tan^{-1}x} $$

$$ = \underset{x\rightarrow 0}{lim}\dfrac{2-\frac{(sin^{-1}x)}x}{2+\frac{(tan^{-1}x)}x}$$
$$\dfrac{2-1}{2+1}=\dfrac13$$

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Re-Attempt Weekly Quiz Competition

Functions Test 40

Result

Functions Test 40

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SHARING IS CARING

Question 1

$$[-5, \infty)$$

$$(-\infty , +2)$$

$$[5, \infty), ( -\infty, -2]$$

None of these

Solution

Question 2

$${2,3}$$

$${2,3, 4}$$

$${1, 2,3, 4, 5}$$

None of these

Solution

Question 3

$$[-2n\pi, 2n\pi]$$

$$(2n\pi, 2n+1 \pi)$$

$$(\frac {(4n+1) \pi }{ 2 } , \frac {(4n+3) \pi }{ 2 })$$

$$(\frac { (4n-1)\pi }{ 2 } ,\frac { (4n+1)\pi }{ 2 } )$$

Solution

Question 4

$$(1, \infty )$$

$$(0, \infty )$$

$$(0, 1 )$$

$$(0.5 , 1 )$$

Solution

Question 5

$$(-3, -1) \cup (1, \infty )$$

$$\left [ -3, -1 \right) \cup \left [1, \infty \right )$$

$$\left (-3, -2 \right ) \cup \left (-2, -1 \right ) \cup \left (1, \infty \right )$$

$$\left [ -3, -2 \right) \cup \left (-2 -1 \right), \left (1, \infty \right )$$

Solution

Question 6

$$- \infty < x < \infty$$

$$1 \leqslant x \leqslant 4$$

$$ 4 \leqslant x \leqslant 16$$

$$ -1 \leqslant x \leqslant 1$$

Solution

Question 7

$$ 16 $$

$$ 32 $$

$$ 64 $$

None of these

Solution

Question 8

One-one and onto

One-one and into

Many-one and onto

Many-one and into

Solution

Question 9

$$\phi$$

$$R$$

$$R-{ x:x (2n+1) \frac{\pi}{2}, n\in1}$$

None of these

Solution

Question 10

2

1/3

2/3

-1/3

Solution

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