1 - ces 4x USO a x V1ovx 4 xao x >0 -

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Let

f (x) = \{\begin{matrix} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{matrix} .

Then, f (x) is continuous at x = 4 when
(a) a = 0, b = 0
(b) a = 1, b = 1
(c) a = −1, b = 1
(d) a = 1, b = −1.

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20. If f( 2x + 1/x) = x + 1/4x + 1 (x is not equal to 0) , the value of f(x) is (1) 4x (2) 1/4(2x + 1/x) (3) 1/4x (4) 4(2x + 1/x)

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(a) x

^{4}

\frac{8}{9}

^{2}

+1 = 3x

^{3}

+3x.
(b) (x

^{4}

+ 1) - 4x (x

^{2}

+1) - 3x

^{2}

= 0.

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\frac{x^{4} + x^{2} + 1}{x^{2} - 4 x - 5} > 0

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x \neq 0

∣ ∣ ∣ ∣ \begin{matrix} x + 1 & 2 x + 1 & 3 x + 1 2 x & 4 x + 3 & 6 x + 3 4 x + 4 & 6 x + 4 & 8 x + 4 \end{matrix} ∣ ∣ ∣ ∣ = 0

, then

x + 1

is equal to

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