CUET UG 2025 — Mathematics Calculus

The range of $\sin(3x)$sin(3x) is $[-1, 1]$[−1,1], so:

$$-1 \leq \sin(3x) \leq 1$$−1≤sin(3x)≤1

Maximum of $f(x) = \sin(3x) + 6$f(x)=sin(3x)+6 occurs when $\sin(3x) = 1$sin(3x)=1:

$$f(x)_{\max} = 1 + 6 = 7 \longrightarrow \textbf{(III)}$$f(x)max​=1+6=7⟶(III)

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Since $|x + 2| \geq 0$∣x+2∣≥0, we have $-|x + 2| \leq 0$−∣x+2∣≤0.

Maximum of $f(x) = -|x + 2| + 4$f(x)=−∣x+2∣+4 occurs when $|x + 2| = 0$∣x+2∣=0, i.e. $x = -2$x=−2:

$$f(x)_{\max} = 0 + 4 = 4 \longrightarrow \textbf{(IV)}$$f(x)max​=0+4=4⟶(IV)

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Since $(3x + 1)^2 \geq 0$(3x+1)2≥0, the minimum of $f(x) = (3x + 1)^2 + 5$f(x)=(3x+1)2+5 occurs when $(3x + 1)^2 = 0$(3x+1)2=0, i.e. $x = -\dfrac{1}{3}$x=−31​:

$$f(x)_{\min} = 0 + 5 = 5 \longrightarrow \textbf{(II)}$$f(x)min​=0+5=5⟶(II)

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The range of $\cos x$cosx is $[-1, 1]$[−1,1], so:

$$-2 \leq 2\cos x \leq 2$$−2≤2cosx≤2

Minimum of $f(x) = 2\cos x + 4$f(x)=2cosx+4 occurs when $\cos x = -1$cosx=−1:

$$f(x)_{\min} = 2(-1) + 4$$f(x)min​=2(−1)+4

$$= -2 + 4$$=−2+4

$$= 2 \longrightarrow \textbf{(I)}$$=2⟶(I)

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$(A) \rightarrow (III)$(A)→(III), $(B) \rightarrow (IV)$(B)→(IV), $(C) \rightarrow (II)$(C)→(II), $(D) \rightarrow (I)$(D)→(I)