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Sets, Relation and Function

117357 The range of the function $y = 3$ sin $(\sqrt{\frac{π^{2}}{16} - x^{2}})$ is

1

[0, \sqrt{3 / 2}]

2

[0, 1]

3

[0, 3 / \sqrt{2}]

4

[0, \infty]

Explanation:

C Given,
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - x^{2}})$
$\frac{π^{2}}{16} - x^{2} > 0$
$x^{2} < \frac{π^{2}}{16}$
$x < \frac{π}{4}$
Hence, domain is $[0, \frac{π}{4}]$
When, $x = 0$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - 0})$
$y = 3 \sin \frac{π}{4}$
$y = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$
When, $x = \frac{π}{4}$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - {(\frac{π}{4})}^{2}})$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - \frac{π^{2}}{16}})$
$y = 3 \sin 0$
$y = 0$ Hence, range of the function is $[0, \frac{3}{\sqrt{2}}]$

WB JEE-2014

Sets, Relation and Function

117358 For the function $f (x) = [\frac{1}{[x]}]$ , where $[x]$ denotes the greatest integer less than or equal to $x$ , which of the following statements are true?

1 The domain is

(- \infty, \infty)

2 The range is

{0} \cup {- 1} \cup {1}

3 The domain is

(- \infty, 0) \cup [1, \infty)

4 The range is

{0} \cup {1}

Explanation:

\&c) : Given, the function,
$f (x) = [\frac{1}{[x]}]$
Domain $= R - {f (x) = 0}$
When, $[x] = 0, f (x) = 0$
$x \in [0, 1]$ $∴$ Domain $= (- \infty, 0) \cup [1, \infty)$
and, range $= {0} \cup {- 1} \cup {1}$

WB JEE-2015

Sets, Relation and Function

117359 ${X \in R; | \cos x | \geq \sin x} \cap [0, \frac{3 π}{2}]$ is equal to

1

[0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]

2

[0, \frac{π}{4}] \cup [\frac{π}{2}, \frac{3 π}{2}]

3

[0, \frac{π}{4}] \cup [\frac{5 π}{4}, \frac{3 π}{2}]

4

[0, \frac{3 π}{2}]

Explanation:

A Given,
${x \in R : | \cos x | \geq \sin x} \cap [0, \frac{3 π}{2}]$
When we draw the graphs of the function $| \cos x |$ and $\sin$ $x$ , it is clear that -
$| \cos x | \geq \sin x when,$
$x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]$
$∴ x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}] \cap [0, \frac{3 π}{2}]$
$\Rightarrow x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]$
original image

WB JEE-2015

Sets, Relation and Function

117360 Given that, $x$ is a real number satisfying $\frac{5 x^{2} - 26 x + 5}{3 x^{2} - 10 x + 3} < 0$ , then

1

x < \frac{1}{5}

2

\frac{1}{5} < x < 3

3

x > 5

4

\frac{1}{5} < x < \frac{1}{3}

or

3 < x < 5

Explanation:

D Given,
$\frac{5 x^{2} - 26 x + 5}{3 x^{2} - 10 x + 3} < 0$
$\Rightarrow \frac{5 x^{2} - 25 x - x + 5}{3 x^{2} - 9 x - x + 3} < 0$
$\Rightarrow \frac{5 x (x - 5) - 1 (x - 5)}{3 x (x - 3) - 1 (x - 3)} < 0$
$\frac{(5 x - 1) (x - 5)}{(3 x - 1) (x - 3)} < 0$
$(5 x - 1) (x - 5) < 0 or (3 x - 1) (x - 3) < 0$
$x \in (5 \cdot \frac{1}{5}) or x \in (3, \frac{1}{3})$ Hence, $\frac{1}{5} < x < \frac{1}{3}$ or $3 < x < 5$

WB JEE-2015

Sets, Relation and Function

117357 The range of the function $y = 3$ sin $(\sqrt{\frac{π^{2}}{16} - x^{2}})$ is

1

[0, \sqrt{3 / 2}]

2

[0, 1]

3

[0, 3 / \sqrt{2}]

4

[0, \infty]

Explanation:

C Given,
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - x^{2}})$
$\frac{π^{2}}{16} - x^{2} > 0$
$x^{2} < \frac{π^{2}}{16}$
$x < \frac{π}{4}$
Hence, domain is $[0, \frac{π}{4}]$
When, $x = 0$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - 0})$
$y = 3 \sin \frac{π}{4}$
$y = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$
When, $x = \frac{π}{4}$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - {(\frac{π}{4})}^{2}})$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - \frac{π^{2}}{16}})$
$y = 3 \sin 0$
$y = 0$ Hence, range of the function is $[0, \frac{3}{\sqrt{2}}]$

WB JEE-2014

Sets, Relation and Function

117358 For the function $f (x) = [\frac{1}{[x]}]$ , where $[x]$ denotes the greatest integer less than or equal to $x$ , which of the following statements are true?

1 The domain is

(- \infty, \infty)

2 The range is

{0} \cup {- 1} \cup {1}

3 The domain is

(- \infty, 0) \cup [1, \infty)

4 The range is

{0} \cup {1}

Explanation:

\&c) : Given, the function,
$f (x) = [\frac{1}{[x]}]$
Domain $= R - {f (x) = 0}$
When, $[x] = 0, f (x) = 0$
$x \in [0, 1]$ $∴$ Domain $= (- \infty, 0) \cup [1, \infty)$
and, range $= {0} \cup {- 1} \cup {1}$

WB JEE-2015

Sets, Relation and Function

117359 ${X \in R; | \cos x | \geq \sin x} \cap [0, \frac{3 π}{2}]$ is equal to

1

[0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]

2

[0, \frac{π}{4}] \cup [\frac{π}{2}, \frac{3 π}{2}]

3

[0, \frac{π}{4}] \cup [\frac{5 π}{4}, \frac{3 π}{2}]

4

[0, \frac{3 π}{2}]

Explanation:

A Given,
${x \in R : | \cos x | \geq \sin x} \cap [0, \frac{3 π}{2}]$
When we draw the graphs of the function $| \cos x |$ and $\sin$ $x$ , it is clear that -
$| \cos x | \geq \sin x when,$
$x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]$
$∴ x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}] \cap [0, \frac{3 π}{2}]$
$\Rightarrow x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]$
original image

WB JEE-2015

Sets, Relation and Function

117360 Given that, $x$ is a real number satisfying $\frac{5 x^{2} - 26 x + 5}{3 x^{2} - 10 x + 3} < 0$ , then

1

x < \frac{1}{5}

2

\frac{1}{5} < x < 3

3

x > 5

4

\frac{1}{5} < x < \frac{1}{3}

or

3 < x < 5

Explanation:

D Given,
$\frac{5 x^{2} - 26 x + 5}{3 x^{2} - 10 x + 3} < 0$
$\Rightarrow \frac{5 x^{2} - 25 x - x + 5}{3 x^{2} - 9 x - x + 3} < 0$
$\Rightarrow \frac{5 x (x - 5) - 1 (x - 5)}{3 x (x - 3) - 1 (x - 3)} < 0$
$\frac{(5 x - 1) (x - 5)}{(3 x - 1) (x - 3)} < 0$
$(5 x - 1) (x - 5) < 0 or (3 x - 1) (x - 3) < 0$
$x \in (5 \cdot \frac{1}{5}) or x \in (3, \frac{1}{3})$ Hence, $\frac{1}{5} < x < \frac{1}{3}$ or $3 < x < 5$

WB JEE-2015

Sets, Relation and Function

117357 The range of the function $y = 3$ sin $(\sqrt{\frac{π^{2}}{16} - x^{2}})$ is

1

[0, \sqrt{3 / 2}]

2

[0, 1]

3

[0, 3 / \sqrt{2}]

4

[0, \infty]

Explanation:

C Given,
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - x^{2}})$
$\frac{π^{2}}{16} - x^{2} > 0$
$x^{2} < \frac{π^{2}}{16}$
$x < \frac{π}{4}$
Hence, domain is $[0, \frac{π}{4}]$
When, $x = 0$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - 0})$
$y = 3 \sin \frac{π}{4}$
$y = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$
When, $x = \frac{π}{4}$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - {(\frac{π}{4})}^{2}})$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - \frac{π^{2}}{16}})$
$y = 3 \sin 0$
$y = 0$ Hence, range of the function is $[0, \frac{3}{\sqrt{2}}]$

WB JEE-2014

Sets, Relation and Function

117358 For the function $f (x) = [\frac{1}{[x]}]$ , where $[x]$ denotes the greatest integer less than or equal to $x$ , which of the following statements are true?

1 The domain is

(- \infty, \infty)

2 The range is

{0} \cup {- 1} \cup {1}

3 The domain is

(- \infty, 0) \cup [1, \infty)

4 The range is

{0} \cup {1}

Explanation:

\&c) : Given, the function,
$f (x) = [\frac{1}{[x]}]$
Domain $= R - {f (x) = 0}$
When, $[x] = 0, f (x) = 0$
$x \in [0, 1]$ $∴$ Domain $= (- \infty, 0) \cup [1, \infty)$
and, range $= {0} \cup {- 1} \cup {1}$

WB JEE-2015

Sets, Relation and Function

117359 ${X \in R; | \cos x | \geq \sin x} \cap [0, \frac{3 π}{2}]$ is equal to

1

[0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]

2

[0, \frac{π}{4}] \cup [\frac{π}{2}, \frac{3 π}{2}]

3

[0, \frac{π}{4}] \cup [\frac{5 π}{4}, \frac{3 π}{2}]

4

[0, \frac{3 π}{2}]

Explanation:

A Given,
${x \in R : | \cos x | \geq \sin x} \cap [0, \frac{3 π}{2}]$
When we draw the graphs of the function $| \cos x |$ and $\sin$ $x$ , it is clear that -
$| \cos x | \geq \sin x when,$
$x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]$
$∴ x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}] \cap [0, \frac{3 π}{2}]$
$\Rightarrow x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]$
original image

WB JEE-2015

Sets, Relation and Function

117360 Given that, $x$ is a real number satisfying $\frac{5 x^{2} - 26 x + 5}{3 x^{2} - 10 x + 3} < 0$ , then

1

x < \frac{1}{5}

2

\frac{1}{5} < x < 3

3

x > 5

4

\frac{1}{5} < x < \frac{1}{3}

or

3 < x < 5

Explanation:

D Given,
$\frac{5 x^{2} - 26 x + 5}{3 x^{2} - 10 x + 3} < 0$
$\Rightarrow \frac{5 x^{2} - 25 x - x + 5}{3 x^{2} - 9 x - x + 3} < 0$
$\Rightarrow \frac{5 x (x - 5) - 1 (x - 5)}{3 x (x - 3) - 1 (x - 3)} < 0$
$\frac{(5 x - 1) (x - 5)}{(3 x - 1) (x - 3)} < 0$
$(5 x - 1) (x - 5) < 0 or (3 x - 1) (x - 3) < 0$
$x \in (5 \cdot \frac{1}{5}) or x \in (3, \frac{1}{3})$ Hence, $\frac{1}{5} < x < \frac{1}{3}$ or $3 < x < 5$

WB JEE-2015

Sets, Relation and Function

117357 The range of the function $y = 3$ sin $(\sqrt{\frac{π^{2}}{16} - x^{2}})$ is

1

[0, \sqrt{3 / 2}]

2

[0, 1]

3

[0, 3 / \sqrt{2}]

4

[0, \infty]

Explanation:

C Given,
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - x^{2}})$
$\frac{π^{2}}{16} - x^{2} > 0$
$x^{2} < \frac{π^{2}}{16}$
$x < \frac{π}{4}$
Hence, domain is $[0, \frac{π}{4}]$
When, $x = 0$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - 0})$
$y = 3 \sin \frac{π}{4}$
$y = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$
When, $x = \frac{π}{4}$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - {(\frac{π}{4})}^{2}})$
$y = 3 \sin (\sqrt{\frac{π^{2}}{16} - \frac{π^{2}}{16}})$
$y = 3 \sin 0$
$y = 0$ Hence, range of the function is $[0, \frac{3}{\sqrt{2}}]$

WB JEE-2014

Sets, Relation and Function

117358 For the function $f (x) = [\frac{1}{[x]}]$ , where $[x]$ denotes the greatest integer less than or equal to $x$ , which of the following statements are true?

1 The domain is

(- \infty, \infty)

2 The range is

{0} \cup {- 1} \cup {1}

3 The domain is

(- \infty, 0) \cup [1, \infty)

4 The range is

{0} \cup {1}

Explanation:

\&c) : Given, the function,
$f (x) = [\frac{1}{[x]}]$
Domain $= R - {f (x) = 0}$
When, $[x] = 0, f (x) = 0$
$x \in [0, 1]$ $∴$ Domain $= (- \infty, 0) \cup [1, \infty)$
and, range $= {0} \cup {- 1} \cup {1}$

WB JEE-2015

Sets, Relation and Function

117359 ${X \in R; | \cos x | \geq \sin x} \cap [0, \frac{3 π}{2}]$ is equal to

1

[0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]

2

[0, \frac{π}{4}] \cup [\frac{π}{2}, \frac{3 π}{2}]

3

[0, \frac{π}{4}] \cup [\frac{5 π}{4}, \frac{3 π}{2}]

4

[0, \frac{3 π}{2}]

Explanation:

A Given,
${x \in R : | \cos x | \geq \sin x} \cap [0, \frac{3 π}{2}]$
When we draw the graphs of the function $| \cos x |$ and $\sin$ $x$ , it is clear that -
$| \cos x | \geq \sin x when,$
$x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]$
$∴ x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}] \cap [0, \frac{3 π}{2}]$
$\Rightarrow x \in [0, \frac{π}{4}] \cup [\frac{3 π}{4}, \frac{3 π}{2}]$
original image

WB JEE-2015

Sets, Relation and Function

117360 Given that, $x$ is a real number satisfying $\frac{5 x^{2} - 26 x + 5}{3 x^{2} - 10 x + 3} < 0$ , then

1

x < \frac{1}{5}

2

\frac{1}{5} < x < 3

3

x > 5

4

\frac{1}{5} < x < \frac{1}{3}

or

3 < x < 5

Explanation:

D Given,
$\frac{5 x^{2} - 26 x + 5}{3 x^{2} - 10 x + 3} < 0$
$\Rightarrow \frac{5 x^{2} - 25 x - x + 5}{3 x^{2} - 9 x - x + 3} < 0$
$\Rightarrow \frac{5 x (x - 5) - 1 (x - 5)}{3 x (x - 3) - 1 (x - 3)} < 0$
$\frac{(5 x - 1) (x - 5)}{(3 x - 1) (x - 3)} < 0$
$(5 x - 1) (x - 5) < 0 or (3 x - 1) (x - 3) < 0$
$x \in (5 \cdot \frac{1}{5}) or x \in (3, \frac{1}{3})$ Hence, $\frac{1}{5} < x < \frac{1}{3}$ or $3 < x < 5$

WB JEE-2015