X- aLt-a-2x-3x-3a-x-2-x-

The correct option is A 2/3√(3) Lt_ x→ a √( a +2 x ) √( 3 x ) #160;√( 3 a + x ) 2√( x ) #160; #160;=Lt_ x→ a √( a +2 x ) √( 3 x ) #160; ...

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Byju's Answer

Standard XII

Mathematics

AM,GM,HM Inequality

x→ aLt√a+2x-√...

Question

$L t x \to a \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$

\frac{2}{\sqrt{3}}

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- \frac{1}{\sqrt{3}}

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\frac{2}{3 \sqrt{3}}

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\frac{1}{\sqrt{3}}

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Solution

The correct option is A $\frac{2}{3 \sqrt{3}}$
$L t_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}}$
$= L t_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} \times \frac{\sqrt{a + 2 x} + \sqrt{3 x}}{\sqrt{a + 2 x} + \sqrt{3 x}} \times \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{3 a + x} + 2 \sqrt{x}}$
$= L t_{x \to a} \frac{a - x}{3 (a - x)} \times \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= L t_{x \to a} \frac{1}{3} \times \frac{2 \sqrt{a} + 2 \sqrt{a}}{\sqrt{3 a} + \sqrt{3 a}}$
$= L t_{x \to a} \frac{1}{3} \times \frac{4 \sqrt{a}}{2 \sqrt{3 a}}$
$= \frac{2}{3 \sqrt{3}}$
Hence, $L t_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} = \frac{2}{3 \sqrt{3}}$

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Similar questions

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

3 [\frac{3 x - 1}{2 x + 3}] - 2 [\frac{2 x + 3}{3 x - 1}] = 5; x \neq \frac{1}{3}, - \frac{3}{2}

Q. Solve the equations:

\frac{2 x^{3} - 3 x^{2} + x + 1}{2 x^{3} - 3 x^{2} - x - 1} = \frac{3 x^{3} - x^{2} + 5 x - 13}{3 x^{3} - x^{2} - 5 x + 13}

Q. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 {2 x}^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a x - 12,

then value of

a^{2}

Q. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a x - 12

, then

a

is equal to

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Ltx→a√a+2x−√3x√3a+x−2√x=

$L t x \to a \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$