JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 2)
Let f : R $$\to$$ R be defined as
$$f(x) = \left\{ {\matrix{ { - {4 \over 3}{x^3} + 2{x^2} + 3x,} & {x > 0} \cr {3x{e^x},} & {x \le 0} \cr } } \right.$$. Then f is increasing function in the interval
$$f(x) = \left\{ {\matrix{ { - {4 \over 3}{x^3} + 2{x^2} + 3x,} & {x > 0} \cr {3x{e^x},} & {x \le 0} \cr } } \right.$$. Then f is increasing function in the interval
$$\left( { - {1 \over 2},2} \right)$$
(0,2)
$$\left( { - 1,{3 \over 2}} \right)$$
($$-$$3, $$-$$1)
Explanation
$$f'(x)\left\{ {\matrix{
{ - 4{x^2} + 4x + 3} & {x > 0} \cr
{3{e^x}(1 + x)} & {x \le 0} \cr
} } \right.$$
_22th_July_Evening_Shift_en_2_1.png)
For x > 0, $$f'(x) = - 4{x^2} + 4x + 3$$
f(x) is increasing in $$\left( { - {1 \over 2},{3 \over 2}} \right)$$
For x $$\le$$ 0, f'(x) = 3ex(1 + x)
f'(x) > 0 $$\forall$$ x $$\in$$($$-$$1, 0)
$$\Rightarrow$$ f(x) is increasing in ($$-$$1, 0)
So, in complete domain, f(x) is increasing in $$\left( { - 1,{3 \over 2}} \right)$$
For x > 0, $$f'(x) = - 4{x^2} + 4x + 3$$
f(x) is increasing in $$\left( { - {1 \over 2},{3 \over 2}} \right)$$
For x $$\le$$ 0, f'(x) = 3ex(1 + x)
f'(x) > 0 $$\forall$$ x $$\in$$($$-$$1, 0)
$$\Rightarrow$$ f(x) is increasing in ($$-$$1, 0)
So, in complete domain, f(x) is increasing in $$\left( { - 1,{3 \over 2}} \right)$$
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