Test Index

MHTCET 23 Apr 2025 Shift 1 Paper

Question : 108 of 150

Marks: +1, -0

\lim _{x \rightarrow 2} \frac{x+3 x^{2}+5 x^{3}+7 x^{4}-166}{x-2}=

Solution:

Concept:

The limit is of the form

\frac{0}{0}

, so factor the numerator and cancel the common factor

(x-2)

Explanation:

x = 2

, the numerator becomes

2 + 3(4) + 5(8) + 7(16) - 166 = 0

(x-2)

is a factor of the numerator.

Factor the numerator:

x + 3x^2 + 5x^3 + 7x^4 - 166 = (x-2)(7x^3 + 19x^2 + 41x + 83)

Therefore,

\lim_{x \to 2} \frac{(x-2)(7x^3 + 19x^2 + 41x + 83)}{x-2} = \lim_{x \to 2} (7x^3 + 19x^2 + 41x + 83)

Substitute

x = 2

7(8) + 19(4) + 41(2) + 83 = 56 + 76 + 82 + 83 = 297

Answer:

297

, i.e. option D.

Go to Question:

Prev Question

Next Question