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Solution for Find a Point on the Curve Y = X2 + X, Where the Tangent is Parallel to the Chord Joining (0, 0) and (1, 2) ? - CBSE (Commerce) Class 12 - Mathematics

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Question

Find a point on the curve y = x2 + x, where the tangent is parallel to the chord joining (0, 0) and (1, 2) ?

Solution

​Let: \[f\left( x \right) = x^2 + x\] 

The tangent to the curve is parallel to the chord joining the points \[\left( 0, 0 \right)\] and \[\left( 1, 2 \right)\] .

Assume that the chord joins the points

\[\left( a, f\left( a \right) \right)\] and \[\left( b, f\left( b \right) \right)\] .
\[\therefore\] \[a = 0, b = 1\]
The polynomial function is everywhere continuous and differentiable.
So, 
\[f\left( x \right) = x^2 + x\] is continuous on \[\left[ 0, 1 \right]\] and differentiable on \[\left( 0, 1 \right)\] .
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists  \[c \in \left( 0, 1 \right)\] such that 
\[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\].
Now, 
\[f\left( x \right) = x^2 + x\]
\[\Rightarrow\] \[f'\left( x \right) = 2x + 1\],
\[f\left( 1 \right) = 2, f\left( 0 \right) = 0\]
\[\therefore\] \[f'\left( x \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\]
\[\Rightarrow\]  \[2x + 1 = \frac{2 - 0}{1 - 0} \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}\]
Thus, 
\[c = \frac{1}{2} \in \left( 0, 1 \right)\] such that ​\[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\] .
Clearly,
\[f\left( c \right) = \left( \frac{1}{2} \right)^2 + \frac{1}{2} = \frac{3}{4}\].
Thus, 
\[\left( c, f\left( c \right) \right)\] i.e.​  
\[\left( \frac{1}{2}, \frac{3}{4} \right)\] is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).
  Is there an error in this question or solution?
Solution Find a Point on the Curve Y = X2 + X, Where the Tangent is Parallel to the Chord Joining (0, 0) and (1, 2) ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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