If the function f x -0-254-3 x-42 x is defined- then the possible of x is - areA- 2-5B- 3-8C- 4-7D- 6

The correct options are C 4.7 D 6f(x)=√(0.25^4 3x 4^2x) For f(x) to be defined nbsp; 0.25^4 3x 4^2x≥ 0 ⇒ 0.25^4 3x≥4^2x ⇒(14)^4 3x≥ 4^2x ⇒ 4^3x 4≥4^2x ⇒ 3x 4≥ ...

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Byju's Answer

Standard XII

Mathematics

Real Valued Functions

If the functi...

Question

If the function $f (x) = \sqrt{{0.25}^{4 - 3 x} - 4^{2 x}}$ is defined, then the possible of $x$ is/are

2.5

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3.8

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4.7

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6

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Solution

The correct options are
C $4.7$
D $6$
$f (x) = \sqrt{{0.25}^{4 - 3 x} - 4^{2 x}}$
For $f (x)$ to be defined
${0.25}^{4 - 3 x} - 4^{2 x} \geq 0 \Rightarrow {0.25}^{4 - 3 x} \geq 4^{2 x} \Rightarrow {(\frac{1}{4})}^{4 - 3 x} \geq 4^{2 x} \Rightarrow 4^{3 x - 4} \geq 4^{2 x} \Rightarrow 3 x - 4 \geq 2 x \Rightarrow x \geq 4$

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If the function f(x)=√0.254−3x−42x is defined, then the possible of x is/are

The correct options are C 4.7 D 6f(x)=√0.254−3x−42x For f(x) to be defined 0.254−3x−42x≥0⇒0.254−3x≥42x⇒(14)4−3x≥42x⇒43x−4≥42x⇒3x−4≥2x⇒x≥4

If the function $f (x) = \sqrt{{0.25}^{4 - 3 x} - 4^{2 x}}$ is defined, then the possible of $x$ is/are