If the tangent to the curve y=x log x at (c,f(x)) is parallel to the line-segment joining A(1,0) and B(e,e), then c=.........,Select Proper option from the given options.
(a) e-1/e
(b) log e-1/e
(c) 1/e¹⁻ᵉ
(d) 1/eᵉ⁻¹
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In the attachment I have answered this problem. Concept: If two lines are parallel their slopes are equal. See the attachment for detailed solution.
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slope of tangent of curve, y = xlogx = m
m = dy/dx at (c , f(x))
now, dy/dx = x × 1/x + logx = 1 + logx
so, m = 1 + logc ......(1)
slope of line joining the points A (1,0) and B(e,e) = M
M = (e - 0)/(e - 1) = e/(e - 1)
now, tangent of curve is parallel to line joining the points A and B.
so, slope of tangent of curve = slope of line AB
m = M
1 + logc = e/(e - 1)
logc = e/(e - 1) - 1 = (e - e + 1)/(e - 1) = 1/(e - 1)
c = e^{1/(e - 1)}
m = dy/dx at (c , f(x))
now, dy/dx = x × 1/x + logx = 1 + logx
so, m = 1 + logc ......(1)
slope of line joining the points A (1,0) and B(e,e) = M
M = (e - 0)/(e - 1) = e/(e - 1)
now, tangent of curve is parallel to line joining the points A and B.
so, slope of tangent of curve = slope of line AB
m = M
1 + logc = e/(e - 1)
logc = e/(e - 1) - 1 = (e - e + 1)/(e - 1) = 1/(e - 1)
c = e^{1/(e - 1)}
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