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# Solution for Find the Points on the Curve Y = X3 − 3x, Where the Tangent to the Curve is Parallel to the Chord Joining (1, −2) and (2, 2) ? - CBSE (Science) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

Find the points on the curve y = x3 − 3x, where the tangent to the curve is parallel to the chord joining (1, −2) and (2, 2) ?

#### Solution

​Let: $f\left( x \right) = x^3 - 3x$

The tangent to the curve is parallel to the chord joining the points  $\left( 1, - 2 \right)$ and $\left( 2, 2 \right)$.

Assume that the chord joins the points

$\left( a, f\left( a \right) \right)$ and $\left( b, f\left( b \right) \right)$ .

$\therefore$ $a = 1, b = 2$

The polynomial function is everywhere continuous and differentiable.
So,

$f\left( x \right) = x^3 - 3x$ is continuous on $\left[ 1, 2 \right]$ and differentiable on $\left( 1, 2 \right)$ .

Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists

$c \in \left( 1, 2 \right)$ such that

$f'\left( c \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1}$ .

Now,

$f\left( x \right) = x^3 - 3x$

$\Rightarrow$ $f'\left( x \right) = 3 x^2 - 3$,

$f\left( 1 \right) = - 2, f\left( 2 \right) = 2$

$\therefore$ $f'\left( x \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1}$

$\Rightarrow$ $3 x^2 - 3 = \frac{2 + 2}{2 - 1} \Rightarrow 3 x^2 = 7 \Rightarrow x = \pm \sqrt{\frac{7}{3}}$

Thus,

$c = \pm \sqrt{\frac{7}{3}}$ such that ​

$f'\left( c \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1}$.

Clearly,

$f\left( \sqrt{\frac{7}{3}} \right) = \left[ \left( \frac{7}{3} \right)^\frac{3}{2} - 3\sqrt{\frac{7}{3}} \right] = \sqrt{\frac{7}{3}}\left[ \frac{7}{3} - 3 \right] = \sqrt{\frac{7}{3}}\left[ \frac{- 2}{3} \right] = \frac{- 2}{3}\sqrt{\frac{7}{3}}$ and $f\left( - \sqrt{\frac{7}{3}} \right) = \frac{2}{3}\sqrt{\frac{7}{3}}$

∴ $f\left( c \right) = \mp \frac{2}{3}\sqrt{\frac{7}{3}}$

Thus,

$\left( c, f\left( c \right) \right)$ , i.e.​

$\left( \pm \sqrt{\frac{7}{3}}, \mp \frac{2}{3}\sqrt{\frac{7}{3}} \right)$ , are points on the given curve where the tangent is parallel to the chord joining the points $\left( 1, - 2 \right)$ and $\left( 2, 2 \right)$ .

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#### APPEARS IN

Solution for question: Find the Points on the Curve Y = X3 − 3x, Where the Tangent to the Curve is Parallel to the Chord Joining (1, −2) and (2, 2) ? concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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