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The slope of the tangent to the curve y = 3x2 + 2kx - 5 at x = 1 is 9. The value of k is: |
$-\frac{3}{2}$ $\frac{1}{2}$ $\frac{5}{2}$ $\frac{3}{2}$ |
$\frac{3}{2}$ |
$y = 3x^2+2kx-5⇒\frac{dy}{dx}=6x+2k$ ∴ Slope of tangent at x = 1 is 9 $6(1)+2k=9$ $6+2k=9⇒2k=9-6⇒2k=3$ $k=\frac{3}{2}$ |
