If x -5-3- then the correct options is are A- x3-125-27-B- x4-0-625-1- x -1-5- 1-3-D- 1-x2-1-25- -

Here, nbsp;x∈ [ 5,3] Splitting into intervals x∈[ 5,0)∪[0,3] x^3∈[ 125,0)∪[0,27] ⇒ x^3∈[ 125,27] x^4∈(0,625]∪[0,81] ⇒ x^4∈[0,625] 1x∈( ∞, 15]∪[13,∞) x^2∈[0 ...

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Byju's Answer

Standard XII

Mathematics

Properties of Inequalities

If x ∈[-5,3],...

Question

If $x \in [- 5, 3],$ then the correct option(s) is (are)

x^{3} \in [- 125, 27]

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x^{4} \in [0, 625]

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\frac{1}{x} \in [- \frac{1}{5}, \frac{1}{3}]

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\frac{1}{x^{2}} \in [\frac{1}{25}, \infty)

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Solution

The correct options are
A $x^{3} \in [- 125, 27]$
B $x^{4} \in [0, 625]$
D $\frac{1}{x^{2}} \in [\frac{1}{25}, \infty)$
Here, $x \in [- 5, 3]$
Splitting into intervals
$x \in [- 5, 0) \cup [0, 3]$
$x^{3} \in [- 125, 0) \cup [0, 27]$
$\Rightarrow x^{3} \in [- 125, 27]$

$x^{4} \in (0, 625] \cup [0, 81]$
$\Rightarrow x^{4} \in [0, 625]$

$\frac{1}{x} \in (- \infty, - \frac{1}{5}] \cup [\frac{1}{3}, \infty)$

$x^{2} \in [0, 25]$
$\Rightarrow \frac{1}{x^{2}} \in [\frac{1}{25}, \infty)$

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Similar questions

If $x + \frac{1}{x} = 3$ , calculate $x^{2} + \frac{1}{x^{2}}, x^{3} + \frac{1}{x^{3}} a n d x^{4} + \frac{1}{x^{4}}$ .

Q. If

x

is a rational number satisfying

(1 - x) (1 + x + x^{2} + x^{3} + x^{4})

= \frac{31}{32} .

Then

1 + x + x^{2} + x^{3} + x^{4} + x^{5}

x^{3} - x - 4

added to

x^{4} - x^{3} - x^{2} + x + 3

to obtain

x^{4} + x^{2} - 1

?
If true then enter

1

and if false then enter

0

Q. If

x \neq - 1

, then find the equation of

\frac{x^{5} + x^{4} + x^{3} + x^{2}}{x^{3} + x^{2} + x + 1} = 1

Q. If

(1 + x + x^{2}) (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + . . .

then,

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If x∈[−5,3], then the correct option(s) is (are)

The correct options are A x3∈[−125,27] B x4∈[0,625] D 1x2∈[125,∞)Here, x∈[−5,3] Splitting into intervals x∈[−5,0)∪[0,3] x3∈[−125,0)∪[0,27] ⇒x3∈[−125,27] x4∈(0,625]∪[0,81] ⇒x4∈[0,625] 1x∈(−∞,−15]∪[13,∞) x2∈[0,25] ⇒1x2∈[125,∞)

If $x \in [- 5, 3],$ then the correct option(s) is (are)