Let a- b - R-a-0- If the function f defined as f x - begin cases dfrac 2 x - 2 a - 0 -e x -12 b - 2-4 b x - 3- sqrt2le x - infty lend cases is continuous in the interval -0- - then an ordered pair a - b is-A- -2- 1-3B- -2- 1-3- -2- 1-3D- -2-1-3

The correct option is B (√(2),1 √(3))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 s ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Continuity at a Boundary

Let a, b ∈ R,...

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})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$(a, b) = (\sqrt{2}, 1 - \sqrt{3})$

Suggest Corrections

Similar questions

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:

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

(a, b)

is:

Q. If the function

f

defined as

f (x) = \frac{1}{x} - \frac{k - 1}{e^{2 x} - 1}, x \neq 0,

is continuous at

x = 0

,then the ordered pair

(k, f (0))

is equal to:

Q. Let

f (x) = \frac{x}{1 + x^{2}}

. Then range of

f

Q. Let

f : R \to R

be a function defined as

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{sin (a + 1) x + sin 2 x}{2 x}, & if x < 0 b, & if x = 0 \frac{\sqrt{x + b x^{3}} - \sqrt{x}}{b x^{5 / 2}}, & if x > 0 \end{matrix}

f

is continuous at

x = 0,

then the value of

a + b

is equal

Join BYJU'S Learning Program

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
$(a, b) = (\sqrt{2}, 1 - \sqrt{3})$