CUET UG 2025 — Mathematics Geometry

Given: $f(x) = 3\sin x - 4\cos x$f(x)=3sinx−4cosx where $x \in [-4\pi, 4\pi]$x∈[−4π,4π]

For any function of the form $a\sin x + b\cos x$asinx+bcosx, the maximum and minimum values can be found using:

Maximum value $= \sqrt{a^2 + b^2}$=a2+b2​

Minimum value $= -\sqrt{a^2 + b^2}$=−a2+b2​

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Identifying the coefficients:

$a = 3$a=3 (coefficient of $\sin x$sinx)

$b = -4$b=−4 (coefficient of $\cos x$cosx)

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Calculate $\sqrt{a^2 + b^2}$a2+b2​:

$$\sqrt{a^2 + b^2} = \sqrt{3^2 + (-4)^2}$$a2+b2​=32+(−4)2​

$$= \sqrt{9 + 16}$$=9+16​

$$= \sqrt{25}$$=25​

$$= 5$$=5

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The function $f(x) = 3\sin x - 4\cos x$f(x)=3sinx−4cosx can be rewritten as $R\sin(x + \phi)$Rsin(x+ϕ) where $R = \sqrt{a^2 + b^2} = 5$R=a2+b2​=5.

Since $\sin(x + \phi)$sin(x+ϕ) ranges from $-1$−1 to $+1$+1, the function $f(x)$f(x) ranges from $-5$−5 to $+5$+5.

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The interval $[-4\pi, 4\pi]$[−4π,4π] spans $8\pi$8π units, which covers 4 complete cycles of the trigonometric functions (one cycle $= 2\pi$=2π). This is sufficient for the function to achieve both its maximum and minimum values.

Maximum value of $f(x) = 5$f(x)=5

Minimum value of $f(x) = -5$f(x)=−5

Therefore, the minimum value of the function is $-5$−5.