CUET UG 2025 — Mathematics Calculus

To find where the function $f(x) = 4x^3 - 6x^2 - 72x + 30$f(x)=4x3−6x2−72x+30 is strictly decreasing, the derivative must be negative.

$$f'(x) = \frac{d}{dx}(4x^3 - 6x^2 - 72x + 30)$$f′(x)=dxd​(4x3−6x2−72x+30)

$$f'(x) = 12x^2 - 12x - 72$$f′(x)=12x2−12x−72

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For strictly decreasing, $f'(x) < 0$f′(x)<0:

$$12x^2 - 12x - 72 < 0$$12x2−12x−72<0

Dividing by 12:

$$x^2 - x - 6 < 0$$x2−x−6<0

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Factoring the quadratic:

$$x^2 - x - 6 = (x - 3)(x + 2)$$x2−x−6=(x−3)(x+2)

The inequality becomes:

$$(x - 3)(x + 2) < 0$$(x−3)(x+2)<0

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The critical points are $x = -2$x=−2 and $x = 3$x=3.

A product is negative when one factor is positive and the other is negative.

For $x < -2$x<−2: both factors are negative, product is positive

For $-2 < x < 3$−2<x<3: $(x + 2) > 0$(x+2)>0 and $(x - 3) < 0$(x−3)<0, product is negative

For $x > 3$x>3: both factors are positive, product is positive

The inequality $(x - 3)(x + 2) < 0$(x−3)(x+2)<0 is satisfied when $-2 < x < 3$−2<x<3

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The strictly decreasing interval is $(a, b) = (-2, 3)$(a,b)=(−2,3)

Therefore:

$$a = -2$$a=−2

$$b = 3$$b=3

$$a + b = -2 + 3 = 1$$a+b=−2+3=1