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Questions  >>  CBSE XII  >>  Math  >>  Integrals
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Q)

Find the point on the curve $y = x^3 - 11 x + 5$ at which the target is $y = x -11.$


$(A)\; 2, 9$
$(B)\; 2,-9$
$(C)\; -2, 9 $
$(D)\; -2,-9$

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A)
Toolbox:
  • If $y = f(x)$, then $(\frac{dy}{dx})P =$ slope of the tangent to $y = f(x)$ at that point $P$.
  • Equation of the line with given two points is $\frac{x-x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1}$
STEP 1:
Given : $y = x^3-11x+5$
Differentiate w.r.t $x$ we get,
$\frac{dy}{dx} = 3x^2-11 ------(1)$
The equation of the given tangent is $y = x-11$
STEP 2:
Hence the slope of the tangent is 1
Now Equating this slope to equ (1)
$\begin{align*} 3x^2 -11 & = 1 \\ 3x^2 & = 12 \\ x^2 & = 4 \\ x & = \pm 2 \end{align*}$
STEP 3:
When $x = 2, y = -9$
When $x = -2, y = -13$
Hence the point is $(2, -9)$
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