Let a,b∈R,a≠0. If the function f, defined as fx= (2x^2)/a, 0≤x<1 a, 1≤x<√2 2b^2-4bx^3, √2≤x<8 ,is continuous in the interval 0, ∞, then an ordered pair a,b can be

Correct answer: (C) √2,1-√3. Here, fx is continuous at x=1 ⇒limx→1^(-)fx=limx→1^(+)fx ⇒2/a=a ⇒a= ± √2 Since, f(x) is continuous at x=√2 ⇒limx→2^(-)fx=limx→2^(+)fx ⇒a=(2b^2-4b)/(2√2) Put a= √2 ⇒2=b^2-2b ⇒b^2-2b-2=0 ⇒b=(2± √(4+4.2))/2=1± √3 Hence, a,b≡√2,1±√3

Q33 / 90 · JEE Main 2016 (10 Apr Online) · Mathematics

Let $a, b \in R, (a \neq 0) .$ If the function $f$ , defined as

$f (x) = \{\begin{matrix} \frac{2 x^{2}}{a}, 0 \leq x < 1 \\ a, 1 \leq x < \sqrt{2} \\ \frac{2 b^{2} - 4 b}{x^{3}}, \sqrt{2} \leq x < 8 \end{matrix}$ ,is continuous in the interval $[0, \infty)$ , then an ordered pair $(a, b)$ can be

A
$(- \sqrt{2}, 1 - \sqrt{3})$
B
$(\sqrt{2}, - 1 + \sqrt{3})$
Correct answer
C
$(\sqrt{2}, 1 - \sqrt{3})$
D
$(- \sqrt{2}, 1 + \sqrt{3})$

Solution

Here, $f (x)$ is continuous at $x = 1$ $\Rightarrow \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x)$
$\Rightarrow \frac{2}{a} = a \Rightarrow a = \pm \sqrt{2}$

Since, $f (x)$ is continuous at $x = \sqrt{2}$

$\Rightarrow \lim_{x \to {\sqrt{2}}^{-}} f (x) = \lim_{x \to {\sqrt{2}}^{+}} f (x)$

$\Rightarrow a = \frac{2 b^{2} - 4 b}{2 \sqrt{2}}$

Put $a = \sqrt{2}$

$\Rightarrow 2 = b^{2} - 2 b \Rightarrow b^{2} - 2 b - 2 = 0$

$\Rightarrow b = \frac{2\pm \sqrt{4 + 4.2}}{2} = 1 \pm \sqrt{3}$

Hence, $(a, b) \equiv (\sqrt{2}, 1 \pm \sqrt{3})$

Hence, the correct answer is option: 'C'.

Let a,b∈R,a≠0. If the function f, defined asfx=2x2a, 0≤x<1a, 1≤x<22b2-4bx3, 2≤x<8,is continuous in the interval 0, ∞, then an ordered pair a,b can be

Solution

Relevant Questions

Let $a, b \in R, (a \neq 0) .$ If the function $f$ , defined as

$f (x) = \{\begin{matrix} \frac{2 x^{2}}{a}, 0 \leq x < 1 \\ a, 1 \leq x < \sqrt{2} \\ \frac{2 b^{2} - 4 b}{x^{3}}, \sqrt{2} \leq x < 8 \end{matrix}$ ,is continuous in the interval $[0, \infty)$ , then an ordered pair $(a, b)$ can be