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

Set of Values of k for Two Real Roots

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.

For k>0k > 0, the set of all values of kk for which kex−x=0ke^x - x = 0 has two distinct roots is

Options

A

(0,1e)\left(0, \frac{1}{e}\right)

Correct
B

(1e,1)\left(\frac{1}{e}, 1\right)

C

(1e,∞)\left(\frac{1}{e}, \infty\right)

D

(0,1)(0, 1)

Topics & Concepts

Application of DerivativesMaxima and MinimaRoots of EquationsIntervals

Step-by-Step Solution

From the previous question, the global minimum of f(x)=kex−xf(x) = ke^x - x for k>0k > 0 is:
fmin=1+ln⁡kf_{\text{min}} = 1 + \ln k

Since lim⁡x→±∞f(x)=+∞\lim_{x \to \pm\infty} f(x) = +\infty, the equation f(x)=0f(x) = 0 has two distinct real roots if and only if the minimum value is strictly negative:
fmin<0  ⟹  1+ln⁡k<0  ⟹  ln⁡k<−1  ⟹  k<1ef_{\text{min}} < 0 \implies 1 + \ln k < 0 \implies \ln k < -1 \implies k < \frac{1}{e}

Since k>0k > 0, the admissible range is:
k∈(0,1e)k \in \left(0, \frac{1}{e}\right)

Correct Answer: A

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