JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Analyzing +4 | -1

The domain of definition of

f\left( x \right) = {\log _{1.7}}{\left( {\frac{{2 - \phi '\left( x \right)}}{{x + 1}}} \right)^{1/2}},

where

\phi \left( x \right) = \frac{{{x^3}}}{3} - \frac{3}{2}{x^2} - 2x + \frac{3}{2},

View answer and explanation

Correct option: C

\left( { - \infty , - 1} \right) \cup \left( { - 1,\;4} \right)

Option C is correct. The domain requires

\frac{{2 - \phi'(x)}}{{x + 1}} > 0

. After finding

\phi'(x)

and simplifying, we get

\frac{{(x - 4)(x + 1)}}{{x + 1}} < 0

, which leads to

x < 4

and

x \neq -1

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