Which of the following describes a true relationship between the functions f x - x-3 2 -2 and g x - 1 2 x-1 graphed below in the standard x-y coordinate plane-

You visited us 2 times! Enjoying our articles? Unlock Full Access!

f(x) Transforms to |f(x)|

Which of the ...

Question

Which of the following describes a true relationship between the functions $f (x) = {(x - 3)}^{2} + 2$ and $g (x) = \frac{1}{2} x + 1$ graphed below in the standard $(x, y)$ coordinate plane?

f (x) = g (x)

for exactly

2

values of

x

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

f (x) = g (x)

for exactly

1

values of

x

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

f (x) < g (x)

for all

x

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

f (x) > g (x)

for all

x

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

f (x)

is the inverse of

g (x)

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

Open in App

Solution

The correct option is B $f (x) = g (x)$ for exactly $2$ values of $x$
Given $f (x) = {(x - 3)}^{2} + 2, g (x) = \frac{x}{2} + 1$
$f (x) - g (x) = {(x - 3)}^{2} + 2 - \frac{x}{2} - 1$
$= x^{2} - 6 x + 9 + 1 - \frac{x}{2}$
$= x^{2} - \frac{13 x}{2} + 10$
$= {(x - \frac{13}{4})}^{2} - \frac{9}{16}$
$∴ f (x) - g (x) = 0$ for exactly two values of x
$∴ f (x) = g (x)$ for exactly two values of x.

Suggest Corrections

Similar questions

Q. Which of the following functions are not identical?

f (x) = x / x, g (x) = 1

f (x) = 1, g (x) = {sin}^{2} x + {cos}^{2} x

f (x) = x, g (x) = {(\sqrt{x})}^{2}

f (x) = log (x - 2) + log (x - 3), g (x) = log (x - 2) (x - 3)

Q. Find fog and gof if
(i)

f (x) = e^{x}, g (x) = \log_{e} x

(ii)

f (x) = x^{2}, g (x) = \cos x

(iii)

f (x) = | x |, g (x) = \sin x

(iv)

f (x) = x + 1, g (x) = e^{x}

(v)

f (x) = \sin^{- 1} x, g (x) = x^{2}

(vi)

f (x) = x + 1, g (x) = \sin x

(vii)

f (x) = x + 1, g (x) = 2 x + 3

(viii)

f (x) = c, c \in R, g (x) = \sin x^{2}

(ix)

f (x) = x^{2} + 2, g (x) = 1 - \frac{1}{1 - x}

Q. Let

f, g : R \to R

be defined by

f (x) = x^{2} - \frac{cos x}{2}

and

g (x) = \frac{x sin x}{2}

. Then

Q. Assertion :Let the function

f (x) = x^{2} - x + 1 \forall x \geq \frac{1}{2}

and

g (x) = \frac{1}{2} + \sqrt{x - \frac{3}{4}}

, then

f (x) = g (x)

has two solutions. Reason:

f (x)

and

g (x)

are inverse of each other.

How many of the following pairs of function are identcal

(a) $f (x) = \frac{x^{2}}{x}, g (x) = x$

(b) $f (x) = \frac{x^{2} - 1}{x - 1}, g (x) = x + 1$

(d) $f (x) = \frac{x}{x^{2}}, g (x) = \frac{1}{x}$

Join BYJU'S Learning Program

Which of the following describes a true relationship between the functions f(x)=(x−3)2+2 and g(x)=12x+1 graphed below in the standard (x,y) coordinate plane?

The correct option is B f(x)=g(x) for exactly 2 values of xGiven f(x)=(x−3)2+2,g(x)=x2+1f(x)−g(x)=(x−3)2+2−x2−1=x2−6x+9+1−x2=x2−13x2+10=(x−134)2−916∴f(x)−g(x)=0 for exactly two values of x∴f(x)=g(x) for exactly two values of x.

Which of the following describes a true relationship between the functions $f (x) = {(x - 3)}^{2} + 2$ and $g (x) = \frac{1}{2} x + 1$ graphed below in the standard $(x, y)$ coordinate plane?