I- - X - 2-20 x-91-0 1II- - y - 2-32 y -247-0 A- relationship between x and y cannot be establishedB- x - yC- x-y D- x - yE- x - y

The correct option is D x ≤ yEquation I x^2 20x+91=0 ⇒ x^2 13x 7x+91=0 ⇒ x(x 13) 7(x 13)=0 ⇒ (x 7)(x 13)=0 ∴ x=13, 7 Equation II y^2 32y+247=0 ⇒ y^2 19y 13y+24 ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

I. ∼ X ∧ 2-20...

Question

$I . x^{2} - 20 x + 91 = 0 I I . y^{2} - 32 y + 247 = 0$

x > y

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relationship between x and y cannot be established

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x \geq y

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x \leq y

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x < y

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Solution

The correct option is D $x \leq y$
Equation I
$x^{2} - 20 x + 91 = 0$
$\Rightarrow x^{2} - 13 x - 7 x + 91 = 0$
$\Rightarrow x (x - 13) - 7 (x - 13) = 0$
$\Rightarrow (x - 7) (x - 13) = 0$
$∴ x = 13, 7$
Equation II
$y^{2} - 32 y + 247 = 0$
$\Rightarrow y^{2} - 19 y - 13 y + 247 = 0$
$\Rightarrow y (y - 19) - 13 (y - 19) = 0$
$\Rightarrow (y - 13) (1 - 19) = 0$
$∴ y = 13, 19$
Hence, $x \leq y$

Suggest Corrections

I. x2−20x+91=0II. y2−32y+247=0

The correct option is D x≤yEquation I x2−20x+91=0 ⇒x2−13x−7x+91=0 ⇒x(x−13)−7(x−13)=0 ⇒(x−7)(x−13)=0 ∴x=13,7 Equation II y2−32y+247=0 ⇒y2−19y−13y+247=0 ⇒y(y−19)−13(y−19)=0 ⇒(y−13)(1−19)=0 ∴y=13,19 Hence, x≤y

$I . x^{2} - 20 x + 91 = 0 I I . y^{2} - 32 y + 247 = 0$