JEE MAIN - Mathematics (2017 - 9th April Morning Slot - No. 17)
The function f defined by
f(x) = x3 $$-$$ 3x2 + 5x + 7 , is :
f(x) = x3 $$-$$ 3x2 + 5x + 7 , is :
increasing in R.
decreasing in R.
decreasing in (0, $$\infty $$) and increasing in ($$-$$ $$\infty $$, 0)
increasing in (0, $$\infty $$) and decreasing in ($$-$$ $$\infty $$, 0)
Explanation
The given function is
$$f(x) = {x^2} - 3{x^2} + 5x + 7$$
$$f'(x) = 3{x^2} - 6x + 5$$
The discriminant of the above quadratic equation is
$$\Delta = 36 - 4(3)(5) = 36 - 60 < 0$$
Therefore, $$f'(x) > 0\,\forall x \in {R^ + }$$
Also, $$f'(x) > 0\,\forall x \in {R^ - }$$
Therefore, the given function f is increasing in R.
Comments (0)
