If $f(y)=2 y^2+b y+c$ and $f(0)=3$ and $f(2)=1$, then the value of $f(1)$ is

JEE MATHEMATICS MCQ: RELATIONS

1 Easy

1 of 1

Application +4 | -1

f(y)=2 y^2+b y+c

and

f(0)=3

and

f(2)=1

, then the value of

f(1)

View answer and explanation

Correct option: A

Option A (0) is correct because substituting the given values into the function allows us to solve for b and c, and then find f(1) = 0.

Practice All Chapter MCQs →

\mathrm{Z}

R

\mathrm{N}

\mathrm{R}

\mathrm{N} \times \mathrm{N}

(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})

a d(b+c)=b c(a+d)

\mathrm{R}

W

R

R=\{(x, y) \in W \times W \mid

x

y

\}

S

R = \{ \left( {a,\;b} \right):1 + ab > 0\}

S

X

R

X

A

B

R

R

N

R = \left\{ {\left( {x,\;y} \right):4x + 3y = 20} \right\}

R