If $y = 2 x - 7 - 5 x^{2}$ has maximum value at $x = a$ , then the value of $a$ is

- \frac{1}{5}

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\frac{1}{5}

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\frac{34}{5}

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- \frac{34}{5}

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Solution

The correct option is B $\frac{1}{5}$
$y = 2 x - 7 - 5 x^{2}$
$y = - (x - \frac{1}{5})^{2} - 4$ , which is maximum at $x = \frac{1}{5}$
Hence, option 'B' is correct .

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Similar questions

2 x - 7 - 5 x^{2}

x = a

a = . . . . . . . . . .

\frac{2 + \frac{1}{3 \frac{4}{5}}}{2 + \frac{1}{3 + \frac{1}{1 + \frac{1}{4}}}} = x

(5 x^{2} + \frac{3}{4} y^{2}) (5 x^{2} - \frac{3}{4} y^{2})

(\frac{4 x}{5} - \frac{5 y}{3}) (\frac{4 x}{5} + \frac{5 y}{3})

(x + \frac{1}{x}) (x - \frac{1}{x})

(\frac{1}{x} + \frac{1}{y}) (\frac{1}{x} - \frac{1}{y})

A = (5 \sqrt{2} + 6 \sqrt{3}) (4 \sqrt{2} - 5 \sqrt{3})

B = (7 \sqrt{2} + 5 \sqrt{3}) (8 \sqrt{2} - 5 \sqrt{3})

A + B

x = {(\frac{5}{7})}^{- 5} \times {(\frac{7}{5})}^{- 7}

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