Verified Solution • 2025–26 Session moderate

QJEEM19Ma0e2a: JEE Main 2019 Mathematics - Class 12 JEEM19Ma0e2a Solution: If   f(t) dt = x 2 + t 2 f(t) dt then f ' is -

If   ∫0x \int\limits_0^x \, f(t) dt = x2 + ∫x1 \int\limits_x^1 \, t2f(t) dt then f ' (12)\left( {{1 \over 2}} \right) is -

Multiple Choice Options

  • A 1825{{18} \over {25}}
  • B 625{{6} \over {25}}
  • C 2425{{24} \over {25}}
  • D 45{{4} \over {5}}

Step-by-Step Verified Solution

∫0x \int\limits_0^x \, f(t) dt = x2 + ∫x1 \int\limits_x^1 \, t2f(t) dt           f ' (12)\left( {{1 \over 2}} \right) = ?

Differentiate w.r.t. 'x'

f(x) = 2x + 0 −- x2 f(x)

f(x) = 2x1+x2{{2x} \over {1 + {x^2}}} ⇒\Rightarrow f '(x) = (1+x2)2−2x(2x)(1+x2)2{{\left( {1 + {x^2}} \right)2 - 2x\left( {2x} \right)} \over {{{\left( {1 + {x^2}} \right)}^2}}}

f '(x) = 2x2−4x2+2(1+x2)2{{2{x^2} - 4{x^2} + 2} \over {{{\left( {1 + {x^2}} \right)}^2}}}

f ' (12)=2−2(14)(1+14)2=(32)2516=4850=2425\left( {{1 \over 2}} \right) = {{2 - 2\left( {{1 \over 4}} \right)} \over {{{\left( {1 + {1 \over 4}} \right)}^2}}} = {{\left( {{3 \over 2}} \right)} \over {{{25} \over {16}}}} = {{48} \over {50}} = {{24} \over {25}}

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