JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 21)
Let a, b $$ \in $$ R, (a $$ \ne $$ 0). If the function f defined as
$$f\left( x \right) = \left\{ {\matrix{ {{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \cr {a\,\,\,,} & {1 \le x < \sqrt 2 } \cr {{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \cr } } \right.$$
is continuous in the interval [0, $$\infty $$), then an ordered pair ( a, b) is :
$$f\left( x \right) = \left\{ {\matrix{ {{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \cr {a\,\,\,,} & {1 \le x < \sqrt 2 } \cr {{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \cr } } \right.$$
is continuous in the interval [0, $$\infty $$), then an ordered pair ( a, b) is :
$$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$
$$\left( { - \sqrt 2 ,1 + \sqrt 3 } \right)$$
$$\left( {\sqrt 2 , - 1 + \sqrt 3 } \right)$$
$$\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)$$
Explanation
f(x) is continuous at x = 1
$$ \therefore $$ $${{2{{\left( 1 \right)}^2}} \over a} = a$$
$$ \Rightarrow $$ a2 = 2
$$ \Rightarrow $$ a = $$ \pm $$ $$\sqrt 2 $$
Also f(x) is continuous at x = $$\sqrt 2 $$
$$ \therefore $$ a = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$
Now, when a = $$\sqrt 2 ,$$ then
4 = 2b2 $$-$$ 4b
$$ \Rightarrow $$ b2 $$-$$ 2b = 2
$$ \Rightarrow $$ b2 $$-$$ 2b $$-$$ 2 = 0
$$ \Rightarrow $$ b = $${{2 \mp \sqrt {4 + 4.2} } \over 2}$$
= 1 $$ \pm $$ $$\sqrt 3 $$
$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$$
When a = $$-$$ $$\sqrt 2 $$, then
$$-$$ $$\sqrt 2 $$ = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$
$$ \Rightarrow $$ $$-$$ 4 = 2b2 $$-$$ 4b
$$ \Rightarrow $$ b2 $$-$$ 2b + 2 = 0
$$ \therefore $$ b = $${{2 \pm \sqrt {4 - 8} } \over 2}$$ = 1 $$ \pm $$ i
As b $$ \in $$ Real number so,
b = 1 $$ \pm $$ i is not accepted.
$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 + \sqrt 3 } \right)$$ or $$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$
$$ \therefore $$ $${{2{{\left( 1 \right)}^2}} \over a} = a$$
$$ \Rightarrow $$ a2 = 2
$$ \Rightarrow $$ a = $$ \pm $$ $$\sqrt 2 $$
Also f(x) is continuous at x = $$\sqrt 2 $$
$$ \therefore $$ a = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$
Now, when a = $$\sqrt 2 ,$$ then
4 = 2b2 $$-$$ 4b
$$ \Rightarrow $$ b2 $$-$$ 2b = 2
$$ \Rightarrow $$ b2 $$-$$ 2b $$-$$ 2 = 0
$$ \Rightarrow $$ b = $${{2 \mp \sqrt {4 + 4.2} } \over 2}$$
= 1 $$ \pm $$ $$\sqrt 3 $$
$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$$
When a = $$-$$ $$\sqrt 2 $$, then
$$-$$ $$\sqrt 2 $$ = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$
$$ \Rightarrow $$ $$-$$ 4 = 2b2 $$-$$ 4b
$$ \Rightarrow $$ b2 $$-$$ 2b + 2 = 0
$$ \therefore $$ b = $${{2 \pm \sqrt {4 - 8} } \over 2}$$ = 1 $$ \pm $$ i
As b $$ \in $$ Real number so,
b = 1 $$ \pm $$ i is not accepted.
$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 + \sqrt 3 } \right)$$ or $$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$
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