I- 4-x-3-x-xII- 3-2-7 1-2-y-0

The correct option is E x = y or relationship cannot be established7/√(x) = √(x) x = 7 (y^35/2 7^5/2)/√(y) = 0 y^5/2 = 7^5/2 y = 7 x = y ...

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Standard XII

Mathematics

Implicit Differentiation

I. 4/√x+3/√x=...

Question

I. $\frac{4}{\sqrt{x}} + \frac{3}{\sqrt{x}} = \sqrt{x}$
II. $\frac{(y^{\frac{35}{2}} - 7^{\frac{5}{2}})}{\sqrt{y}} = 0$

x > y

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x < y

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x \geq y

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x \leq y

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x = y

or relationship cannot be established

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Solution

The correct option is E $x = y$ or relationship cannot be established
$\frac{7}{\sqrt{x}} = \sqrt{x} x = 7 \frac{(y^{\frac{35}{2}} - 7^{\frac{5}{2}})}{\sqrt{y}} = 0 y^{\frac{5}{2}} = 7^{\frac{5}{2}} y = 7 x = y$

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(xii)

2 x^{2} - (3 + 7 i) x + (9 i - 3) = 0

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I. 4√x+3√x=√x II. (y352−752)√y=0

The correct option is E x=y or relationship cannot be established7√x=√xx=7(y352−752)√y=0y52=752y=7x=y

I. $\frac{4}{\sqrt{x}} + \frac{3}{\sqrt{x}} = \sqrt{x}$
II. $\frac{(y^{\frac{35}{2}} - 7^{\frac{5}{2}})}{\sqrt{y}} = 0$

The correct option is E $x = y$ or relationship cannot be established
$\frac{7}{\sqrt{x}} = \sqrt{x} x = 7 \frac{(y^{\frac{35}{2}} - 7^{\frac{5}{2}})}{\sqrt{y}} = 0 y^{\frac{5}{2}} = 7^{\frac{5}{2}} y = 7 x = y$