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Standard XII

Mathematics

Double Ordinate of Ellipse

If t1 and t...

Question

If $t_{1} a n d t_{2}$ are abscissae of two points on the curve $f (x) = x - x^{2}$ in the interval $(0, 1)$ , then find the maximum value of the expression $(t_{1} + t_{2}) - (t_{1}^{2} + t_{2}^{2})$ .

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Solution

$f (x) = x - x^{2}$
$f (t_{1}) = t_{1} - t_{1}^{2}$
$f (t_{2}) = t_{2} = t_{2}^{2}$
$\Rightarrow f (t_{1}) + f (t_{2}) = (t_{1} + t_{2}) - (t_{1}^{2} + t_{2}^{2})$
$f (x) = x - x^{2}$
$f^{'} (x) = 1 - 2 x$
$f^{'} (x) = 0$
$1 - 2 x = 0 \Rightarrow x = \frac{1}{2}$
$f^{''} (x) = - 2 < 0$
Hence, $x = \frac{1}{2}$ is pt. of maxima.
$f (\frac{1}{2}) = \frac{1}{2} - {[\frac{1}{2}]}^{2} = \frac{1}{4}$
$f (t_{1}) + f (t_{2}) ∣ ∣ ∣_{max} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

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If t1andt2 are abscissae of two points on the curve f(x)=x−x2 in the interval (0,1), then find the maximum value of the expression (t1+t2)−(t21+t22).

f(x)=x−x2f(t1)=t1−t21f(t2)=t2=t22⇒f(t1)+f(t2)=(t1+t2)−(t21+t22)f(x)=x−x2f′(x)=1−2xf′(x)=01−2x=0⇒x=12f′′(x)=−2<0Hence, x=12 is pt. of maxima.f(12)=12−[12]2=14f(t1)+f(t2)∣∣∣max=14+14=12

If $t_{1} a n d t_{2}$ are abscissae of two points on the curve $f (x) = x - x^{2}$ in the interval $(0, 1)$ , then find the maximum value of the expression $(t_{1} + t_{2}) - (t_{1}^{2} + t_{2}^{2})$ .