Find the value of $x$ such that $f (x) = 2 x + 1$ and $f (x + 1) + f (x) = x$ .

- \frac{4}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- \frac{4}{3}$
$f (x) = 2 x + 1$ and $f (x + 1) = 2 (x + 1) + 1 = 2 x + 3$
$f (x) + f (x + 1) = 2 x + 1 + 2 x + 3 = 4 x + 4$
$b u t f (x) + f (x + 1) = x$ [given]
$\Rightarrow 4 x + 4 = x$
$3 x = - 4$ which gives $x = - 4 / 3$

Suggest Corrections

Similar questions

Q. If a function,

f (x) = 3 - 2 x

, then find the value of

x

such that

f (x^{2}) = [f (x)]^{2}

Q. If a function 'f' defined by f(x) = 2x + 3, then find the value of 'x' such that f(x) = f(1-x).

Let f(X)={1-sinπx÷1+cos2πx. X<1÷2. { P . X=1÷2. { √2x-1÷√4+√2x-1-2. X>1÷2. Determine the value of p , so that the function is continuous at X=1÷2 . { Is over the whole f(X) and √ is on complete √4+√2x-1

Q. Find the value of

k

f (x)

is continuous at

x = 2.

f (x) = ⎧ ⎨ ⎩ \begin{matrix} 2 x + 1; i f x < 2 k; x = 2 3 x - 1; x > 2 \end{matrix}

Q. If

f (x) = \frac{1}{2 x + 1}, x \neq - \frac{1}{2}, f (f (x)) = \frac{2 x + 1}{2 x + 3}

provided that

x \neq - \frac{3}{2}

Join BYJU'S Learning Program

Find the value of x such that f(x)=2x+1 and f(x+1)+f(x)=x.

The correct option is A −43f(x)=2x+1 and f(x+1)=2(x+1)+1=2x+3f(x)+f(x+1)=2x+1+2x+3=4x+4butf(x)+f(x+1)=x [given]⇒4x+4=x3x=−4 which gives x=−4/3

Find the value of $x$ such that $f (x) = 2 x + 1$ and $f (x + 1) + f (x) = x$ .

The correct option is A $- \frac{4}{3}$
$f (x) = 2 x + 1$ and $f (x + 1) = 2 (x + 1) + 1 = 2 x + 3$
$f (x) + f (x + 1) = 2 x + 1 + 2 x + 3 = 4 x + 4$
$b u t f (x) + f (x + 1) = x$ [given]
$\Rightarrow 4 x + 4 = x$
$3 x = - 4$ which gives $x = - 4 / 3$