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प्रश्न
Find the point on the curve y = x3 − 11x + 5 at which the tangent is y = x − 11.
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उत्तर
The equation of the given curve is y = x3 − 11x + 5.
The equation of the tangent to the given curve is given as y = x − 11 (which is of the form y = mx + c).
∴Slope of the tangent = 1
When x = 2, y = (2)3 − 11 (2) + 5 = 8 − 22 + 5 = −9.
When x = −2, y = (−2)3 − 11 (−2) + 5 = −8 + 22 + 5 = 19.
Hence, the required points are (2, −9) and (−2, 19).
But, both these points should satisfy the equation of the tangent as there would be point of contact between tangent and the curve.
∴ (2, −9) is the required point as (−2, 19) is not satisfying the given equation of tangent.
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