Let a-b- R-a0- If the function f defined asfx-2x2a -0- x-1a -1- x-22b2-4bx3 -2- x-is continuous in the interval -0-then an ordered pair a-b is-

This function to be continuous in the interval (0,∞) nbsp; nbsp;lim_x → 12x^2/a=a=lim_x →√(2)2b^2 4b/x^3 nbsp; By solving the above equation (a,b)=(√(2), ...

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Standard XIII

Mathematics

Continuity at a Boundary

Let a,b∈ R,a0...

Question

Let $a, b \in R, (a \neq 0) .$ If the function $f$ defined as
$f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{2 x^{2}}{a}, 0 \leq x < 1 a, 1 \leq x < \sqrt{2} \frac{2 b^{2} - 4 b}{x^{3}}, \sqrt{2} \leq x < \infty \end{matrix}$
is continuous in the interval $[0, \infty),$ ,then an ordered pair $(a, b)$ is:

(\sqrt{2}, 1 - \sqrt{3})

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(- \sqrt{2}, 1 + \sqrt{3})

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(- \sqrt{2}, 1 - \sqrt{3})

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(\sqrt{2}, - 1 + \sqrt{3})

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Solution

The correct option is A $(\sqrt{2}, 1 - \sqrt{3})$
This function to be continuous in the interval $(0, \infty)$
$lim x \to 1 \frac{2 x^{2}}{a} = a = lim x \to \sqrt{2} \frac{2 b^{2} - 4 b}{x^{3}}$
By solving the above equation
$(a, b) = (\sqrt{2}, 1 - \sqrt{3})$

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Similar questions

Q. Let

a, b ϵ R, (a \neq 0)

. If the function f defined as

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{2 x^{2}}{a}, & 0 \leq x < 1 a, & 1 \leq x < \sqrt{2} \frac{2 b^{2} - 4 b}{x^{3}}, & \sqrt{2} \leq x < \infty \end{matrix}

is continuous in the interval

[0, \infty)

, then an ordered pair

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is:

Q. The function

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x^{2} / a, & 0 \leq x < 1 a, & 1 \leq x < \sqrt{2} \frac{2 b^{2} - 4 b}{x^{2}}, & \sqrt{2} \leq x < \infty \end{matrix}

is continuous for

0 \leq x < \infty

, then the most suitable values of

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and

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are

Q. The function

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{x^{2}}{a} & 0 \leq x < 1 a & 1 \leq x < \sqrt{2} (2 b^{2} - 4 b) / x^{2} & \sqrt{2} \leq x < \infty \end{matrix}

is continuous for

0 \leq x < \infty

, then the most suitable values of

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and

b

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Q. Let

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{x^{2}}{a}; 0 \leq x < 1 a; 1 \leq x < \sqrt{2} \frac{2 b^{2} - 4 b}{x^{2}}; \sqrt{2} \leq x < \infty \end{matrix}

f (x)

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Q. Let

a, b \in R, (a \neq 0) .

If the function

f

defined as

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{2 x^{2}}{a}, 0 \leq x < 1 a, 1 \leq x < \sqrt{2} \frac{2 b^{2} - 4 b}{x^{3}}, \sqrt{2} \leq x < \infty \end{matrix}

is continuous in the interval

[0, \infty),

,then an ordered pair

(a, b)

is:

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Let a,b∈R,(a≠0). If the function f defined as f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩2x2a ,0≤x<1a ,1≤x<√22b2−4bx3 ,√2≤x<∞ is continuous in the interval [0,∞),,then an ordered pair (a,b) is:

The correct option is A (√2,1−√3)This function to be continuous in the interval (0,∞) limx→12x2a=a=limx→√22b2−4bx3 By solving the above equation (a,b)=(√2,1−√3)

The correct option is A $(\sqrt{2}, 1 - \sqrt{3})$
This function to be continuous in the interval $(0, \infty)$
$lim x \to 1 \frac{2 x^{2}}{a} = a = lim x \to \sqrt{2} \frac{2 b^{2} - 4 b}{x^{3}}$
By solving the above equation
$(a, b) = (\sqrt{2}, 1 - \sqrt{3})$