CUET UG 2025 — Mathematics Algebra

The function $f(x) = 4x^2 + 12x + 7$f(x)=4x2+12x+7 is a quadratic function where the coefficient of $x^2$x2 is positive.

Since the coefficient of $x^2$x2 is $+4$+4 (positive), the parabola opens upward, meaning the function has a minimum value.

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To find the minimum value, complete the square:

$$f(x) = 4x^2 + 12x + 7$$f(x)=4x2+12x+7

$$f(x) = 4(x^2 + 3x) + 7$$f(x)=4(x2+3x)+7

$$f(x) = 4(x^2 + 3x + \frac{9}{4} - \frac{9}{4}) + 7$$f(x)=4(x2+3x+49​−49​)+7

$$f(x) = 4(x^2 + 3x + \frac{9}{4}) - 4(\frac{9}{4}) + 7$$f(x)=4(x2+3x+49​)−4(49​)+7

$$f(x) = 4(x + \frac{3}{2})^2 - 9 + 7$$f(x)=4(x+23​)2−9+7

$$f(x) = 4(x + \frac{3}{2})^2 - 2$$f(x)=4(x+23​)2−2

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From the completed square form:

The term $(x + \frac{3}{2})^2 \geq 0$(x+23​)2≥0 for all real $x$x.

Therefore, $4(x + \frac{3}{2})^2 \geq 0$4(x+23​)2≥0 for all real $x$x.

This gives $f(x) = 4(x + \frac{3}{2})^2 - 2 \geq -2$f(x)=4(x+23​)2−2≥−2.

The minimum value is $-2$−2, occurring when $x = -\frac{3}{2}$x=−23​.

Since the parabola opens upward, the function takes all values from $-2$−2 to infinity.

Therefore, the range of $f(x)$f(x) is $[-2, \infty)$[−2,∞).