If fx is continuous at x=3 where f( x )={ ax+1, for x≤3 bx+3, for x>3 , then

Correct answer: (C) a-b=2/3. We have, fx is continuous at x=3 Where fx= ax+1 for x≤3 bx+3 for x>3 ∴limx→3^(-)fx=limx→3^(+)fx=f3 …i Now, limx→3^(-)fx=limh→0f3-h =limh→0f3-h+1 =3a+1 …ii limx→3^(+)fx=limh→0f3-h =limh→0b3+h+3 =3b+3 … (iii) and f3=3a+1 … (iv) Since, fx is continuous at x=3 ∴ From Eqs. (i), (ii), (iii) and (iv) we get 3a+1=3b+3 ⇒3a-3b=2 ⇒a-b=2/3

Q76 / 150 · MHT CET 2019 (Shift 1) · Mathematics

If $f (x)$ is continuous at $x = 3$ where $f (x) = {\begin{matrix} a x + 1, f o r x \leq 3 \\ b x + 3, f o r x > 3 \end{matrix}$ , then

A
$a + b = \frac{- 2}{3}$
B
$a - b = \frac{- 2}{3}$
Correct answer
C
$a - b = \frac{2}{3}$
D
$a + b = \frac{2}{3}$

Solution

We have,

f (x)

is continuous at

x = 3

Where

f (x) = \{\begin{matrix} a x + 1 f o r & x \leq 3 \\ b x + 3 f o r & x > 3 \end{matrix}

∴ \underset{x \to 3^{-}}{l i m} f (x) = \underset{x \to 3^{+}}{l i m} f (x) = f (3) \dots (i)

Now,

\underset{x \to 3^{-}}{l i m} f (x) = \underset{h \to 0}{l i m} f (3 - h)

= \underset{h \to 0}{l i m} f (3 - h) + 1

= 3 a + 1 \dots (i i)

\underset{x \to 3^{+}}{l i m} f (x) = \underset{h \to 0}{l i m} f (3 - h)

= \underset{h \to 0}{l i m} b (3 + h) + 3

= 3 b + 3 \dots

(iii)
and

f (3) = 3 a + 1 \dots

(iv)
Since,

f (x)

is continuous at

x = 3

∴

From Eqs. (i), (ii), (iii) and (iv)
we get

3 a + 1 = 3 b + 3

\Rightarrow 3 a - 3 b = 2

\Rightarrow a - b = \frac{2}{3}

Hence, the correct answer is option: 'C'.

If fx is continuous at x=3 where f( x )={ ax+1, for x≤3 bx+3, for x>3 , then

Solution

Relevant Questions

If $f (x)$ is continuous at $x = 3$ where $f (x) = {\begin{matrix} a x + 1, f o r x \leq 3 \\ b x + 3, f o r x > 3 \end{matrix}$ , then