Verified Solution • 2025–26 Session moderate

QJEEM16Ma7aa6: JEE Main 2016 Mathematics - Class 12 JEEM16Ma7aa6 Solution: Let a, b R , (a 0). If the function f defined as f ( x ) = { 2…

Let a, b ∈\in R, (a ≠\ne 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, ∞\infty ), then an ordered pair ( a, b) is :

Multiple Choice Options

  • A (2,1−3)\left( {\sqrt 2 ,1 - \sqrt 3 } \right)
  • B (−2,1+3)\left( { - \sqrt 2 ,1 + \sqrt 3 } \right)
  • C (2,−1+3)\left( {\sqrt 2 , - 1 + \sqrt 3 } \right)
  • D (−2,1−3)\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)

Step-by-Step Verified Solution

f(x) is continuous at x = 1

∴\therefore     2(1)2a=a{{2{{\left( 1 \right)}^2}} \over a} = a

⇒\Rightarrow    a2 = 2

⇒\Rightarrow    a = ±\pm 2\sqrt 2

Also f(x) is continuous at x = 2\sqrt 2

∴\therefore    a = 2b2−4b22{{2{b^2} - 4b} \over {2\sqrt 2 }}

Now, when a = 2,\sqrt 2 , then

4 = 2b2 −- 4b

⇒\Rightarrow    b2 −- 2b = 2

⇒\Rightarrow    b2 −- 2b −- 2 = 0

⇒\Rightarrow    b = 2∓4+4.22{{2 \mp \sqrt {4 + 4.2} } \over 2}

= 1 ±\pm 3\sqrt 3

∴\therefore    (a, b) = (2,1±3)\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)

When a = −- 2\sqrt 2 , then

−- 2\sqrt 2 = 2b2−4b22{{2{b^2} - 4b} \over {2\sqrt 2 }}

⇒\Rightarrow     −- 4 = 2b2 −- 4b

⇒\Rightarrow    b2 −- 2b + 2 = 0

∴\therefore     b = 2±4−82{{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) = (2,1+3)\left( {\sqrt 2 ,1 + \sqrt 3 } \right)   or   (2,1−3)\left( {\sqrt 2 ,1 - \sqrt 3 } \right)

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