The range of the expression y=cot2⁡θ+5cot2⁡θ+10cot2⁡θ+1

Given y=cot2⁡θ+5cot2⁡θ+10cot2⁡θ+1 Let cot2⁡θ=x Then y=(x+5)(x+10)(x+1)⇒ y=x2+15x+50x+1⇒ x2+15x+50−xy−y=0⇒ x2+(15−y)x+(50−y)=0∴ D≥0⇒ (15−y)2−4(50−y)≥0⇒ 225−30y+y2−200+4y≥0⇒ y2−26y+25≥0⇒ (y−1)(y−25)≥0⇒ y≤1 and y≥25Thus y≥25.

The range of the expression y=cot2⁡θ+5cot2⁡θ+10cot2⁡θ+1

Given y=cot2⁡θ+5cot2⁡θ+10cot2⁡θ+1 Let cot2⁡θ=x Then y=(x+5)(x+10)(x+1)⇒ y=x2+15x+50x+1⇒ x2+15x+50−xy−y=0⇒ x2+(15−y)x+(50−y)=0∴ D≥0⇒ (15−y)2−4(50−y)≥0⇒ 225−30y+y2−200+4y≥0⇒ y2−26y+25≥0⇒ (y−1)(y−25)≥0⇒ y≤1 and y≥25Thus y≥25.

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The range of the expression $y = \frac{(\cot^{2} θ + 5) (\cot^{2} θ + 10)}{(\cot^{2} θ + 1)}$

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zero

$y \geq 10$

$y \geq 25$

$y \geq 25 / 2$

answer is C.

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Detailed Solution

Given $y = \frac{(\cot^{2} θ + 5) (\cot^{2} θ + 10)}{(\cot^{2} θ + 1)}$ Let $\cot^{2} θ = x$

Then $y = \frac{(x + 5) (x + 10)}{(x + 1)}$

$\begin{array}{ll} \Rightarrow y = \frac{x^{2} + 15 x + 50}{x + 1} \\ \Rightarrow x^{2} + 15 x + 50 - x y - y = 0 \\ \Rightarrow x^{2} + (15 - y) x + (50 - y) = 0 \\ ∴ D \geq 0 \\ \Rightarrow (15 - y)^{2} - 4 (50 - y) \geq 0 \\ \Rightarrow 225 - 30 y + y^{2} - 200 + 4 y \geq 0 \\ \Rightarrow y^{2} - 26 y + 25 \geq 0 \\ \Rightarrow (y - 1) (y - 25) \geq 0 \end{array}$