JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 7)
Let $$f:R \to R$$ be defined as
$$f(x) = \left\{ {\matrix{ { - 55x,} & {if\,x < - 5} \cr {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr } } \right.$$
Let A = {x $$ \in $$ R : f is increasing}. Then A is equal to :
$$f(x) = \left\{ {\matrix{ { - 55x,} & {if\,x < - 5} \cr {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr } } \right.$$
Let A = {x $$ \in $$ R : f is increasing}. Then A is equal to :
$$( - 5,\infty )$$
$$( - \infty , - 5) \cup (4,\infty )$$
$$( - 5, - 4) \cup (4,\infty )$$
$$( - \infty , - 5) \cup ( - 4,\infty )$$
Explanation
$$f(x) = \left\{ {\matrix{
{ - 55x,} & {if\,x < - 5} \cr
{2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr
{2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr
} } \right.$$
Now, $$f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6({x^2} - x - 20)} & ; & { - 5 < x < 4} \cr {6({x^2} - x - 6)} & ; & {x > 4} \cr } } \right.$$
$$f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6(x - 5)(x + 4)} & ; & { - 5 < x < 4} \cr {6(x - 3)(x + 2)} & ; & {x > 4} \cr } } \right.$$
Hence, f(x) is monotonically increasing in interval $$( - 5, - 4) \cup (4,\infty )$$
Now, $$f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6({x^2} - x - 20)} & ; & { - 5 < x < 4} \cr {6({x^2} - x - 6)} & ; & {x > 4} \cr } } \right.$$
$$f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6(x - 5)(x + 4)} & ; & { - 5 < x < 4} \cr {6(x - 3)(x + 2)} & ; & {x > 4} \cr } } \right.$$
Hence, f(x) is monotonically increasing in interval $$( - 5, - 4) \cup (4,\infty )$$
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