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If for all real triplets (a, b, c), ƒ(x) = a + bx + cx2; then $$\int\limits_0^1 {f(x)dx} $$ is equal to :

JEE · Math · previous-year question

  1. A.$${1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$$correct
  2. B.$$2\left\{ 3{f(1) + 2f\left( {{1 \over 2}} \right)} \right\}$$
  3. C.$${1 \over 3}\left\{ {f(0) + f\left( {{1 \over 2}} \right)} \right\}$$
  4. D.$${1 \over 2}\left\{ {f(1) + 3f\left( {{1 \over 2}} \right)} \right\}$$

Answer

A. $${1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$$

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.

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