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प्रश्न
Find the point on the curve y = x2 – 5x + 4 at which the tangent is parallel to the line 3x + y = 7
उत्तर
y = x2 – 5x + 4
Differentiating w.r.t. ‘x’
Slope of the tangent `("d")/("d"x)` = 2x – 5
Given line 3x + y = 7
Slope of the line = `- 3/1` = – 3
Since the tangent is parallel to the line, their slopes are equal.
∴ ("d"y)/("d"x)` = – 3
⇒ 2x – 5 = – 3
2x = 2
x = 1
When x = 1, y = (1)2 – 5(1) + 4 = 0
∴ Point on the curve is (1, 0).
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