I- 9 x2-18 x-5-0II- 2 y 2-9 y -10-0A- If xy

The correct option is A If x lt; yEquation I 9x^2 18x+5=0 ⇒ 9x^2 15x 3x+5=0 ⇒ 3x(3x 5) (3x 5)=0 ⇒ (3x 5)(3x 1)=0 ⇒ x= 5/3 or x= 1/3 Equation II 2y^2 9y+10 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. 9 x2-18 x+...

Question

I. $9 x^{2} - 18 x + 5 = 0$
II. $2 y^{2} - 9 y + 10 = 0$

x < y

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x \leq y

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x = y

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x > y

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x \geq y

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Solution

The correct option is A If $x < y$
Equation I
$9 x^{2} - 18 x + 5 = 0$
$\Rightarrow 9 x^{2} - 15 x - 3 x + 5 = 0$
$\Rightarrow 3 x (3 x - 5) - (3 x - 5) = 0$
$\Rightarrow (3 x - 5) (3 x - 1) = 0$
$\Rightarrow x = \frac{5}{3} o r x = \frac{1}{3}$

Equation II
$2 y^{2} - 9 y + 10 = 0$
$\Rightarrow 2 y^{2} - 5 y - 4 y + 10 = 0$
$\Rightarrow y (2 y - 5) - 2 (2 y - 5) = 0$
$\Rightarrow (2 y - 5) (y - 2) = 0$
$\Rightarrow y = \frac{5}{2} o r y = 2$
Hence, $x < y$

Suggest Corrections

I. 9x2−18x+5=0 II. 2y2−9y+10=0

The correct option is A If x<yEquation I 9x2−18x+5=0 ⇒9x2−15x−3x+5=0 ⇒3x(3x−5)−(3x−5)=0 ⇒(3x−5)(3x−1)=0 ⇒x=53 or x=13 Equation II 2y2−9y+10=0 ⇒2y2−5y−4y+10=0 ⇒y(2y−5)−2(2y−5)=0 ⇒(2y−5)(y−2)=0 ⇒y=52 or y=2 Hence, x<y

I. $9 x^{2} - 18 x + 5 = 0$
II. $2 y^{2} - 9 y + 10 = 0$