JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 15)
If g(x) = x2 + x - 1 and
(goƒ) (x) = 4x2 - 10x + 5, then ƒ$$\left( {{5 \over 4}} \right)$$ is equal to:
(goƒ) (x) = 4x2 - 10x + 5, then ƒ$$\left( {{5 \over 4}} \right)$$ is equal to:
$${1 \over 2}$$
$${3 \over 2}$$
-$${1 \over 2}$$
-$${3 \over 2}$$
Explanation
Given, (goƒ) (x) = 4x2 - 10x + 5
$$ \Rightarrow $$ g(f(x)) = 4x2 - 10x + 5
$$ \therefore $$ g(f($${5 \over 4}$$)) = $$4 \times {{25} \over {16}} - {{50} \over 4} + 5$$ = $$ - {5 \over 4}$$ ...(1)
Also given, g(x) = x2 + x - 1
$$ \therefore $$ g(f(x)) = f2(x) + f(x) –1
$$ \Rightarrow $$ g(f($${5 \over 4}$$)) = f2($${5 \over 4}$$) + f($${5 \over 4}$$) –1 ....(2)
from (1) & (2)
f2($${5 \over 4}$$) + f($${5 \over 4}$$) –1 = $$ - {5 \over 4}$$
$$ \Rightarrow $$ f2($${5 \over 4}$$) + f($${5 \over 4}$$) + $${1 \over 4}$$ = 0
$$ \Rightarrow $$ (f($${5 \over 4}$$) + $${1 \over 2}$$)2 = 0
$$ \Rightarrow $$ f($${5 \over 4}$$) = -$${1 \over 2}$$
$$ \Rightarrow $$ g(f(x)) = 4x2 - 10x + 5
$$ \therefore $$ g(f($${5 \over 4}$$)) = $$4 \times {{25} \over {16}} - {{50} \over 4} + 5$$ = $$ - {5 \over 4}$$ ...(1)
Also given, g(x) = x2 + x - 1
$$ \therefore $$ g(f(x)) = f2(x) + f(x) –1
$$ \Rightarrow $$ g(f($${5 \over 4}$$)) = f2($${5 \over 4}$$) + f($${5 \over 4}$$) –1 ....(2)
from (1) & (2)
f2($${5 \over 4}$$) + f($${5 \over 4}$$) –1 = $$ - {5 \over 4}$$
$$ \Rightarrow $$ f2($${5 \over 4}$$) + f($${5 \over 4}$$) + $${1 \over 4}$$ = 0
$$ \Rightarrow $$ (f($${5 \over 4}$$) + $${1 \over 2}$$)2 = 0
$$ \Rightarrow $$ f($${5 \over 4}$$) = -$${1 \over 2}$$
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