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MHTCET 5 May 2025 Shift 2 Paper

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Question : 108 of 150

Marks: +1, -0

Define

f(x)=\left\{\begin{array}{cll}b-a x & , \text { if } & x < 2 \\ 3 & , \text { if } & x=2 \\ a+2 b x & , \text { if } & x > 2\end{array}\right.

\lim _{x \rightarrow 2} f(x)

\frac{a}{b}=

1
-1
$\frac{2}{3}$
$\frac{3}{2}$

Solution:

Concept:

For

\lim _{x \rightarrow 2} f(x)

to exist, the left-hand limit and right-hand limit must be equal.

Explanation:

Left-hand limit:

\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}(b-a x)=b-2a

Right-hand limit:

\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(a+2b x)=a+4b

Since the limit exists at

x=2

:

\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)

b-2a=a+4b

-2a-a=4b-b

-3a=3b

\frac{a}{b}=-1

Answer:

\frac{a}{b}=-1

Correct option: B.

-1

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