Total number of real values of x such that 12-x1-7-x-12-x1-7-12-64-3x1-7 is-are

(12+x)^1/7/x+(12+x)^1/7/12=64/3x^1/7 ⇒ (12+x)(12+x)^1/7=4.(64).x.x^1/7=2^8. x^8/7 ⇒ 12+x=x(2^8)^7/8 ∴ x=12/127 ...

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Byju's Answer

Standard XII

Mathematics

Roots of a Quadratic Equation

Total number ...

Question

Total number of real values of x such that $\frac{(12 + x)^{\frac{1}{7}}}{x} + \frac{(12 + x)^{\frac{1}{7}}}{12} = \frac{64}{3} x^{\frac{1}{7}}$ is/are

zero

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one

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seven

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two

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Solution

The correct option is B
one

$\frac{(12 + x)^{\frac{1}{7}}}{x} + \frac{(12 + x)^{\frac{1}{7}}}{12} = \frac{64}{3} x^{\frac{1}{7}} \Rightarrow (12 + x) (12 + x)^{\frac{1}{7}} = 4. (64) . x . x^{\frac{1}{7}} = 2^{8} . x^{\frac{8}{7}} \Rightarrow 12 + x = x (2^{8})^{\frac{7}{8}} ∴ x = \frac{12}{127}$

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Total number of real values of x such that $\frac{(12 + x)^{\frac{1}{7}}}{x} + \frac{(12 + x)^{\frac{1}{7}}}{12} = \frac{64}{3} x^{\frac{1}{7}}$ is/are

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Total number of real values of x such that (12+x)17x+(12+x)1712=643x17 is/are

The correct option is B one (12+x)17x+(12+x)1712=643x17⇒ (12+x)(12+x)17=4.(64).x.x17=28.x87⇒ 12+x=x(28)78∴ x=12127

Total number of real values of x such that $\frac{(12 + x)^{\frac{1}{7}}}{x} + \frac{(12 + x)^{\frac{1}{7}}}{12} = \frac{64}{3} x^{\frac{1}{7}}$ is/are