Search any question & find its solution
Question:
Answered & Verified by Expert
Let $R$ be the set of all real numbers and let $f$ be a fucntion $R$ to $R$ such that $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$, for all $x \in R .$ Then $2 f(0)+3 f(1)$ is equal to.
Options:
Solution:
2837 Upvotes
Verified Answer
The correct answer is:
$-2$
Given $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1 \quad ......\text {(1)}$
but $x=0$
$\begin{array}{l}
f(0)+\frac{1}{2} f(1)=1 \\
\Rightarrow 2 f(0)+f(1)=2 \quad ......\text {(2)}\end{array}$
$\begin{array}{l}
\text { put } x=1 \text { in }(1) \\
\Rightarrow f(1)+\frac{3}{2} f(0)=1 \\
\Rightarrow 2 f(1)+3 f(0)=2 \quad ......\text {(3)}
\end{array}$
Solving $(2) \&(3)$ we have
$\begin{array}{l}
F(0)=2 \& f(1)=-2 \\
\therefore 2 f(0)+f(1)=4-6=-2
\end{array}$
but $x=0$
$\begin{array}{l}
f(0)+\frac{1}{2} f(1)=1 \\
\Rightarrow 2 f(0)+f(1)=2 \quad ......\text {(2)}\end{array}$
$\begin{array}{l}
\text { put } x=1 \text { in }(1) \\
\Rightarrow f(1)+\frac{3}{2} f(0)=1 \\
\Rightarrow 2 f(1)+3 f(0)=2 \quad ......\text {(3)}
\end{array}$
Solving $(2) \&(3)$ we have
$\begin{array}{l}
F(0)=2 \& f(1)=-2 \\
\therefore 2 f(0)+f(1)=4-6=-2
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.