If $l^{2} + m^{2} = 1$ , then find the maximum value of $l + m$ .

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Solution

$l^{2} + m^{2} = 1$

$\Rightarrow m^{2} = 1 - l^{2}$

$\Rightarrow m = \pm \sqrt{1 - l^{2}}$

Let $N = l + m \Rightarrow N = l \pm \sqrt{1 - l^{2}}$

$\Rightarrow \frac{d N}{d l} = 1 \pm \frac{1}{2 \sqrt{1 - l^{2}}} \times - 2 l$

$= 1 \pm \frac{l}{\sqrt{1 - l^{2}}}$

For maximum value we have $\frac{d N}{d l} = 0$

$\Rightarrow 1 \pm \frac{l}{\sqrt{1 - l^{2}}} = 0$

$\Rightarrow 1 - l^{2} = l^{2}$

$\Rightarrow 2 l^{2} = 1$

$∴ l = \pm \frac{1}{\sqrt{2}}$

$\Rightarrow \frac{d^{2} N}{d l^{2}} = \pm \frac{\sqrt{1 - l^{2}} - l (\frac{- 2 l}{2 \sqrt{1 - l^{2}}})}{\sqrt{1 - l^{2}}}$

$= \pm \frac{1 - l^{2} + l^{2}}{(1 - l^{2}) \sqrt{1 - l^{2}}}$

$= \pm \frac{1}{{(1 - l^{2})}^{\frac{3}{2}}}$ from above $l = \pm \frac{1}{\sqrt{2}}$

$= \pm \frac{1}{{⎛ ⎝ 1 - {(\frac{1}{\sqrt{2}})}^{2} ⎞ ⎠}^{\frac{3}{2}}} = 2 \sqrt{2} > 0$

$N = l + \sqrt{1 - l^{2}}$

$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \sqrt{2}$

Hence the maximum value of $l + m = \sqrt{2}$

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List-I	List-II
(A) Maximum value of $x y$ subject to $x + y = 7$ is	1) $72$
(B) If $l^{2} + m^{2} = 1$ , then the maximum value of $l + m$ is	2) $1$
(C) If $x + y = 12$ , then the minimum Value of $x^{2} + y^{2}$ is	3) $\sqrt{2}$
(D) Minimum value $x^{2} - 8 x + 17$ is	4) $\frac{49}{4}$
	5) $0$

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x t a n^{- 1} x + s e c^{- 1} \frac{1}{x}, x ϵ (- 1, 1) - {0} \frac{π}{2}, x = 0 \end{matrix}

f^{'} (0^{+}) = l

f^{'} (0^{-}) = m

(l^{2} + m^{2}) .

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