If \( y^3=e^{2x+1} \) such that \( a\displaystyle\frac{d^2y}{dx^2}=by \) (where \( a \) & \( b \) are co-prime), then which one of the following is TRUE?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( a-b=5 \)
Note: The given answer is as per the official NTA Answer Key. However, the question does not say that a and b are positive, and with that reading Option (3) could also be a possible answer, as explained below.
Taking the cube root,
\( \displaystyle y=e^{\frac{2x+1}{3}} \)
\( \displaystyle \frac{dy}{dx}=\frac{2}{3}e^{\frac{2x+1}{3}}=\frac{2}{3}y \)
\( \displaystyle \frac{d^2y}{dx^2}=\frac{2}{3}\cdot\frac{dy}{dx}=\frac{4}{9}y \)
\( \displaystyle 9\frac{d^2y}{dx^2}=4y \)
Comparing with \( a\frac{d^2y}{dx^2}=by \), the co-prime pair is a = 9, b = 4 when a and b are natural numbers (HCF of 9 and 4 is 1). If negative integers are allowed, multiplying both sides by −1 gives
\( \displaystyle -9\frac{d^2y}{dx^2}=-4y \)
so a = −9, b = −4 also fit, and their HCF is still 1.
Option (1) \( a-b=5 \) — Correct. Taking a and b as co-prime natural numbers, a = 9 and b = 4, so a − b = 9 − 4 = 5.
Option (2) \( b-a=2 \) — Incorrect. With a = 9, b = 4, b − a = −5; with a = −9, b = −4, b − a = 5. Neither is 2.
Option (3) \( a-b=-5 \) — Should also be correct, though not the answer given by NTA. If a and b may be negative, a = −9 and b = −4 also satisfy the equation and are co-prime, and then a − b = −9 − (−4) = −5.
Option (4) \( b-a=-1 \) — Incorrect. b − a is −5 (for a = 9, b = 4) or 5 (for a = −9, b = −4), never −1.