Let $$f(x) = \left\{ {\matrix{ {{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr } } \right.$$.
Then the set of all values of b, for which f(x) has maximum value at x = 1, is :
Solution
<p>$$f(x) = \left\{ {\matrix{
{{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr
{ - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr
} } \right.$$</p>
<p>If $f(x)$ has maximum value at $x = 1$ then $f(1 + ) \le f(1)$</p>
<p>$- 2 + {\log _2}({b^2} - 4) \le 1 - 1 + 10 - 7$</p>
<p>${\log _2}({b^2} - 4) \le 5$</p>
<p>$0 < {b^2} - 4 \le 32$</p>
<p>(i) ${b^2} - 4 > 0 \Rightarrow b \in ( - \infty , - 2) \cup (2,\infty )$</p>
<p>(ii) ${b^2} - 36 \le 0 \Rightarrow b \in [ - 6,6]$</p>
<p>Intersection of above two sets</p>
<p>$b \in [ - 6, - 2) \cup (2,6]$</p>
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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