Question

Find the equation of curve whose roots gives the point which lies in the curve f(x) = xSin(x) in the interval [0, ^π⁄2] where slope of a tangent to a curve is equals to the slope of a line joining (0, ^π⁄2)

(a) c = -Sec(c) – Tan(c)

(b) c = -Sec(c) – Tan(c)

(c) c = Sec(c) +Tan(c)

(d) c = Sec(c) – Tan(c)

The question was asked during an interview for a job.

I would like to ask this question from Lagrange’s Mean Value Theorem topic in division Differential Calculus of Engineering Mathematics

Answer 1

Correct option is (d) c = Sec(c) – Tan(c)

For explanation: f(x) = xSin(x)

Since f1(x) = x and f2(x)=Sin(x) both are continuous in interval [0, ^π⁄2], the curve f(x)=f1(x)f2(x) is also continuous.

f’(x) = xCos(x) + Sin(x)

f’(x) always have finite value in interval [0, ^π⁄2] hence it is differentiable in interval (0, ^π⁄2).

f(0) = 0

f(^π⁄2) = ^π⁄2

By mean value theorem,

f’(c) = cCos(c) + Sin(c) = (^π⁄2 – 0)/(^π⁄2 – 0)=1

Hence, c = Sec(c) – Tan(c) is the required curve.