CUET UG 2025 — Mathematics Calculus

Given function: $f(x) = \frac{-3}{4}x^4 - 8x^3 - \frac{45}{2}x^2 + 163$f(x)=4−3​x4−8x3−245​x2+163

The first derivative is:

$$f'(x) = \frac{-3}{4} \times 4x^3 - 8 \times 3x^2 - \frac{45}{2} \times 2x$$f′(x)=4−3​×4x3−8×3x2−245​×2x

$$f'(x) = -3x^3 - 24x^2 - 45x$$f′(x)=−3x3−24x2−45x

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To find critical points, set $f'(x) = 0$f′(x)=0:

$$-3x^3 - 24x^2 - 45x = 0$$−3x3−24x2−45x=0

$$-3x(x^2 + 8x + 15) = 0$$−3x(x2+8x+15)=0

This gives either $x = 0$x=0 or $x^2 + 8x + 15 = 0$x2+8x+15=0

Factoring the quadratic:

$$(x + 3)(x + 5) = 0$$(x+3)(x+5)=0

$x = -3$x=−3 or $x = -5$x=−5

Critical points: $x = 0, -3, -5$x=0,−3,−5

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The second derivative is:

$$f''(x) = -9x^2 - 48x - 45$$f′′(x)=−9x2−48x−45

Testing each critical point:

At $x = 0$x=0:

$$f''(0) = -9(0)^2 - 48(0) - 45$$f′′(0)=−9(0)2−48(0)−45

$$f''(0) = -45 < 0$$f′′(0)=−45<0

Therefore $x = 0$x=0 is a local maximum.

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At $x = -3$x=−3:

$$f''(-3) = -9(9) - 48(-3) - 45$$f′′(−3)=−9(9)−48(−3)−45

$$f''(-3) = -81 + 144 - 45$$f′′(−3)=−81+144−45

$$f''(-3) = 18 > 0$$f′′(−3)=18>0

Therefore $x = -3$x=−3 is a local minimum.

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At $x = -5$x=−5:

$$f''(-5) = -9(25) - 48(-5) - 45$$f′′(−5)=−9(25)−48(−5)−45

$$f''(-5) = -225 + 240 - 45$$f′′(−5)=−225+240−45

$$f''(-5) = -30 < 0$$f′′(−5)=−30<0

Therefore $x = -5$x=−5 is a local maximum.

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The function has local maxima at $x = 0$x=0 and $x = -5$x=−5.