JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 17)
If for all real triplets (a, b, c), ƒ(x) = a + bx + cx2;
then $$\int\limits_0^1 {f(x)dx} $$ is equal to :
$${1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$$
$$2\left\{ 3{f(1) + 2f\left( {{1 \over 2}} \right)} \right\}$$
$${1 \over 3}\left\{ {f(0) + f\left( {{1 \over 2}} \right)} \right\}$$
$${1 \over 2}\left\{ {f(1) + 3f\left( {{1 \over 2}} \right)} \right\}$$
Explanation
ƒ(x) = a + bx + cx2
$$\int\limits_0^1 {f\left( x \right)dx} $$ = $$\left[ {ax + {{b{x^2}} \over 2} + {{c{x^3}} \over 3}} \right]_0^1$$
= $${a + {b \over 2} + {c \over 3}}$$
= $${1 \over 6}\left[ {6a + 3b + c} \right]$$
f(1) = a + b + c
f(0) = a
$$f\left( {{1 \over 2}} \right) = a + {b \over 2} + {c \over 4}$$
By checking each option you have to find the solution.
$${1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$$
= $${1 \over 6}\left[ {a + a + b + c + 4\left( {a + {b \over 2} + {c \over 4}} \right)} \right]$$
= $${1 \over 6}\left[ {6a + 3b + c} \right]$$
$$ \therefore $$ Option (A) is correct option.
$$\int\limits_0^1 {f\left( x \right)dx} $$ = $$\left[ {ax + {{b{x^2}} \over 2} + {{c{x^3}} \over 3}} \right]_0^1$$
= $${a + {b \over 2} + {c \over 3}}$$
= $${1 \over 6}\left[ {6a + 3b + c} \right]$$
f(1) = a + b + c
f(0) = a
$$f\left( {{1 \over 2}} \right) = a + {b \over 2} + {c \over 4}$$
By checking each option you have to find the solution.
$${1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$$
= $${1 \over 6}\left[ {a + a + b + c + 4\left( {a + {b \over 2} + {c \over 4}} \right)} \right]$$
= $${1 \over 6}\left[ {6a + 3b + c} \right]$$
$$ \therefore $$ Option (A) is correct option.
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