JEE MAIN - Mathematics (2022 - 26th July Morning Shift - No. 8)

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 :

($$-$$6, $$-$$2)
(2, 6)
$$[ - 6, - 2) \cup (2,6]$$
$$\left[ {-\sqrt 6 , - 2} \right) \cup \left( {2,\sqrt 6 } \right]$$

Explanation

$$f(x) = \left\{ {\matrix{ {{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr } } \right.$$

If $$f(x)$$ has maximum value at $$x = 1$$ then $$f(1 + ) \le f(1)$$

$$ - 2 + {\log _2}({b^2} - 4) \le 1 - 1 + 10 - 7$$

$${\log _2}({b^2} - 4) \le 5$$

$$0 < {b^2} - 4 \le 32$$

(i) $${b^2} - 4 > 0 \Rightarrow b \in ( - \infty , - 2) \cup (2,\infty )$$

(ii) $${b^2} - 36 \le 0 \Rightarrow b \in [ - 6,6]$$

Intersection of above two sets

$$b \in [ - 6, - 2) \cup (2,6]$$

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