JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 20)
Let f : R $$ \to $$ R be a function such that f(x) = x3 + x2f'(1) + xf''(2) + f'''(3), x $$ \in $$ R. Then f(2) equals -
30
$$-$$ 2
$$-$$ 4
8
Explanation
f(x) = x3 + x2f '(1) + xf ''(2) + f '''(3)
$$ \Rightarrow $$ f '(x) = 3x2 + 2xf '(1) + f ''(x) . . . . . (1)
$$ \Rightarrow $$ f ''(x) = 6x + 2f '(1) . . . . . . (2)
$$ \Rightarrow $$ f '''(x) = 6 . . . . . .(3)
put x = 1 in equation (1) :
f '(1) = 3 + 2f '(1) + f ''(2) . . . . .(4)
put x = 2 in equation (2) :
f ''(2) = 12 + 2f '(1) . . . . .(5)
from equation (4) & (5) :
$$-$$3 $$-$$ f '(1) = 12 + 2f'(1)
$$ \Rightarrow $$ 3f '(1) = $$-$$ 15
$$ \Rightarrow $$ f '(1) = $$-$$ 5 $$ \Rightarrow $$ f ''(2) = 2 . . . . .(2)
put x = 3 in equation (3) :
f ''' (3) = 6
$$ \therefore $$ f(x) = x3 $$-$$ 5x2 + 2x + 6
f(2) = 8 $$-$$ 20 + 4 + 6 = $$-$$ 2
$$ \Rightarrow $$ f '(x) = 3x2 + 2xf '(1) + f ''(x) . . . . . (1)
$$ \Rightarrow $$ f ''(x) = 6x + 2f '(1) . . . . . . (2)
$$ \Rightarrow $$ f '''(x) = 6 . . . . . .(3)
put x = 1 in equation (1) :
f '(1) = 3 + 2f '(1) + f ''(2) . . . . .(4)
put x = 2 in equation (2) :
f ''(2) = 12 + 2f '(1) . . . . .(5)
from equation (4) & (5) :
$$-$$3 $$-$$ f '(1) = 12 + 2f'(1)
$$ \Rightarrow $$ 3f '(1) = $$-$$ 15
$$ \Rightarrow $$ f '(1) = $$-$$ 5 $$ \Rightarrow $$ f ''(2) = 2 . . . . .(2)
put x = 3 in equation (3) :
f ''' (3) = 6
$$ \therefore $$ f(x) = x3 $$-$$ 5x2 + 2x + 6
f(2) = 8 $$-$$ 20 + 4 + 6 = $$-$$ 2
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