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Question

If the line 3x4y=0 is tangent in the first quadrant to the curve y=x3+k, then value of k is

A
12
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B
14
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C
18
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D
12
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Solution

The correct option is B 14
y=x3+k
dydx=3x2=34 x=±12
When x=12 ,y=38, and (12,38) lies on the curve y=x3+k
So, 38=18+k k=14
When x=12,y=38, and (12,38) lies on the curve
So, 38=18+kk=14
But we need the point lying in the first quadrant. Hence, k=14

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