I- 4 x 2-49II- 9 y 2-66 y -121-0A- If x - yB- If xy

The correct option is B If x lt; yEquation I 4x^2=49 ⇒ x^2 = 49/4 ⇒ x= ±7/2 Equation II 9y^2 66y+121=0 ⇒ 9y^2 33y 33y+121=0 ⇒ 3y(3y 11) 11(3y 11)=0 ⇒ (3y 1 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. 4 x 2=49II...

Question

I. $4 x^{2} = 49$
II. $9 y^{2} - 66 y + 121 = 0$

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
$4 x^{2} = 49$
$\Rightarrow x^{2} = \frac{49}{4}$
$\Rightarrow x = \pm \frac{7}{2}$

Equation II
$9 y^{2} - 66 y + 121 = 0$
$\Rightarrow 9 y^{2} - 33 y - 33 y + 121 = 0$
$\Rightarrow 3 y (3 y - 11) - 11 (3 y - 11) = 0$
$\Rightarrow (3 y - 11) (3 y - 11) = 0$
$\Rightarrow y = \frac{11}{3}$
Hence $x < y$

Suggest Corrections

I. 4x2=49 II. 9y2−66y+121=0

The correct option is A If x<yEquation I 4x2=49 ⇒x2=494 ⇒x=±72 Equation II 9y2−66y+121=0 ⇒9y2−33y−33y+121=0 ⇒3y(3y−11)−11(3y−11)=0 ⇒(3y−11)(3y−11)=0 ⇒y=113 Hence x<y

I. $4 x^{2} = 49$
II. $9 y^{2} - 66 y + 121 = 0$