CUET UG 2025 — Mathematics Calculus

A function is increasing when $f'(x) \geq 0$f′(x)≥0.

Given: $f(x) = x^{15} + 5x^9 + 10$f(x)=x15+5x9+10

Using the power rule:

$$f'(x) = 15x^{14} + 45x^8$$f′(x)=15x14+45x8

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Factoring out common terms:

$$f'(x) = 15x^8(x^6 + 3)$$f′(x)=15x8(x6+3)

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Analyzing the sign of $f'(x) = 15x^8(x^6 + 3)$f′(x)=15x8(x6+3):

The factor $15$15 is always positive.

The factor $x^8$x8 is always non-negative for all real $x$x (even power).

The factor $(x^6 + 3)$(x6+3) is always positive since $x^6 \geq 0$x6≥0, which means $x^6 + 3 \geq 3 > 0$x6+3≥3>0.

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Therefore:

$$f'(x) = (\text{positive}) \times (\text{non-negative}) \times (\text{positive}) \geq 0$$f′(x)=(positive)×(non-negative)×(positive)≥0

This holds for all real values of $x$x.

Note that $f'(x) = 0$f′(x)=0 only when $x = 0$x=0, and $f'(x) > 0$f′(x)>0 for all $x \neq 0$x=0.

Since $f'(x) \geq 0$f′(x)≥0 for all real $x$x, the function is increasing for all real values of $x$x.