Verified Solution • 2025–26 Session moderate

QJEEM20Mac82b: JEE Main 2020 Mathematics - Class 11 JEEM20Mac82b Solution: If g(x) = x 2 + x - 1 and (goƒ) (x) = 4x 2 - 10x + 5, then ƒ is …

If g(x) = x2 + x - 1 and
(goƒ) (x) = 4x2 - 10x + 5, then ƒ (54)\left( {{5 \over 4}} \right) is equal to:

Multiple Choice Options

  • A 12{1 \over 2}
  • B 32{3 \over 2}
  • C - 12{1 \over 2}
  • D - 32{3 \over 2}

Step-by-Step Verified Solution

Given, (goƒ) (x) = 4x2 - 10x + 5

⇒\Rightarrow g(f(x)) = 4x2 - 10x + 5

∴\therefore g(f( 54{5 \over 4} )) = 4×2516−504+54 \times {{25} \over {16}} - {{50} \over 4} + 5 = −54- {5 \over 4} ...(1)

Also given, g(x) = x2 + x - 1

∴\therefore g(f(x)) = f2(x) + f(x) –1

⇒\Rightarrow g(f( 54{5 \over 4} )) = f2( 54{5 \over 4} ) + f( 54{5 \over 4} ) –1 ....(2)

from (1) & (2)

f2( 54{5 \over 4} ) + f( 54{5 \over 4} ) –1 = −54- {5 \over 4}

⇒\Rightarrow f2( 54{5 \over 4} ) + f( 54{5 \over 4} ) + 14{1 \over 4} = 0

⇒\Rightarrow (f( 54{5 \over 4} ) + 12{1 \over 2} )2 = 0

⇒\Rightarrow f( 54{5 \over 4} ) = - 12{1 \over 2}

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