I- 10 x 2-7 x -1-0II- 35 y 2-12 y -1-0A- x - y or no relationship can be established between x and yB- x yD- x-y E- x - y

The correct option is E x ≥ yEquation I 10x^2 7x+1=0 ⇒ 10x^2 5x 2x+1=0 ⇒ 5x(2x 1) 1(2x 1)=0 ⇒ (5x 1)(2x 1)=0 ∴ x=1/5, 1/2 Equation II 35y^2 12y+1=0 ⇒ 35y^2 7y ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. 10 x 2-7 x...

Question

$I . 10 x^{2} - 7 x + 1 = 0$
$I I . 35 y^{2} - 12 y + 1 = 0$

x < y

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x > y

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x = y

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x \geq y

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x \leq y

or no relationship can be established between x and y

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Solution

The correct option is D $x \geq y$
Equation I
$10 x^{2} - 7 x + 1 = 0$
$\Rightarrow 10 x^{2} - 5 x - 2 x + 1 = 0$
$\Rightarrow 5 x (2 x - 1) - 1 (2 x - 1) = 0$
$\Rightarrow (5 x - 1) (2 x - 1) = 0$
$∴ x = \frac{1}{5}, \frac{1}{2}$
Equation II
$35 y^{2} - 12 y + 1 = 0$
$\Rightarrow 35 y^{2} - 7 y - 5 y + 1 = 0$
$\Rightarrow 7 y (5 y - 1) - 1 (5 y - 1) = 0$
$\Rightarrow (5 y - 1) (7 y - 1) = 0$
$∴ y = \frac{1}{5}, \frac{1}{7}$
Hence, $x \geq y$

Suggest Corrections

I. 10x2−7x+1=0 II. 35y2−12y+1=0

The correct option is D x≥yEquation I 10x2−7x+1=0 ⇒10x2−5x−2x+1=0 ⇒5x(2x−1)−1(2x−1)=0 ⇒(5x−1)(2x−1)=0 ∴x=15,12 Equation II 35y2−12y+1=0 ⇒35y2−7y−5y+1=0 ⇒7y(5y−1)−1(5y−1)=0 ⇒(5y−1)(7y−1)=0 ∴y=15,17 Hence, x≥y

$I . 10 x^{2} - 7 x + 1 = 0$
$I I . 35 y^{2} - 12 y + 1 = 0$