JEE Challenger
More from Application of Derivatives
MathMedAdv 2007 • Paper 2

Condition on k for a Single Root of ke^x - x = 0

Comprehension Passage

If a continuous function ff defined on the real line R\mathbf{R}, assumes positive and negative values in R\mathbf{R} then the equation f(x)=0f(x) = 0 has a root in R\mathbf{R}. For example, if it is known that a continuous function ff on R\mathbf{R} is positive at some point and its minimum value is negative then the equation f(x)=0f(x) = 0 has a root in R\mathbf{R}.

Consider f(x)=kex−xf(x) = ke^x - x for all real xx where kk is a real constant.

The positive value of kk for which kex−x=0ke^x - x = 0 has only one root is

Options

A

1e\frac{1}{e}

Correct
B

1

C

ee

D

log⁡e2\log_e 2

Topics & Concepts

Application of DerivativesMaxima and MinimaRoots of EquationsExponential Function

Step-by-Step Solution

Consider f(x)=kex−xf(x) = ke^x - x with k>0k > 0.

  1. Critical points:
    f′(x)=kex−1=0  ⟹  ex=1k  ⟹  x=−ln⁡kf'(x) = ke^x - 1 = 0 \implies e^x = \frac{1}{k} \implies x = -\ln k
    Since f′′(x)=kex>0f''(x) = ke^x > 0, the function attains a global minimum at x=−ln⁡kx = -\ln k.

  2. The minimum value is:
    fmin=ke−ln⁡k−(−ln⁡k)=k(1k)+ln⁡k=1+ln⁡kf_{\text{min}} = k e^{-\ln k} - (-\ln k) = k\left(\frac{1}{k}\right) + \ln k = 1 + \ln k

  3. For kex−x=0ke^x - x = 0 to have exactly one real root, the graph must touch the xx-axis at its minimum (fmin=0f_{\text{min}} = 0):
    1+ln⁡k=0  ⟹  ln⁡k=−1  ⟹  k=e−1=1e1 + \ln k = 0 \implies \ln k = -1 \implies k = e^{-1} = \frac{1}{e}

Correct Answer: A

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