If Pe^x=Qe^-x for all real x, what are P and Q in GATE CS 2025 Set-2 Q3?

P=Q=0. Multiplying both sides by e^x gives Pe^(2x)=Q. Since e^(2x) varies with x and Q is constant, the only way this holds for ALL x is if P=0, which forces Q=0.

Why can't P and Q both be 1 in Pe^x=Qe^-x?

If P=Q=1, then e^x=e^-x, which means e^(2x)=1, which only holds when x=0, not for all real x. The equation must hold for every value of x, so P=Q=1 is not a valid solution.

What is the general approach for 'equation holds for all x' problems?

When an equation must hold for all values of x, collect and group terms by their x-dependent parts. Each coefficient of a distinct x-dependent function must independently equal zero. Since e^(2x) is a non-constant function of x, its coefficient P must be zero.

Pe^x = Qe^-x for All Real x | GATE CS 2025 Set-2 Q3 Algebra

Computer Sciences > GATE 2025 SET-2 > Algebra

If Pe^x=Qe^-x for all real values of x, which one of the following statements is true?

P=Q=0

P=Q=1

P=1; Q=-1

P/Q=0

Correct : a

We are given the equation P·e^x = Q·e^-x for all real x.

Multiplying both sides by e^x gives P·e^2x = Q.

Since this must hold true for all real values of x, and e^2x changes with x, the only way the left-hand side remains a constant equal to Q is if P = 0. Substituting P = 0 into the equation gives Q = 0.

✅ Final Answer: Option (a) — P = Q = 0.

Reason: The equation can only hold for all x if both P and Q are zero, since e^2x is not constant.

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