JEE 2024 — Mathematics PYQ
JEE | Mathematics | 2024The function f(x)=x2−6x−16x, x∈R∖{−2,8}
Choose the correct answer:
- A.
decreases in (−2,8) and increases in (−∞,−2)∪(8,∞)
- B.
decreases in (−∞,−2)∪(−2,8)∪(8,∞)
(Correct Answer) - C.
decreases in (−∞,−2) and increases in (8,∞)
decreases in (−∞,−2)∪(−2,8)∪(8,∞)
Explanation
f(x)=x2−6x−16xx∈R{−2,8}
f′(x)=(x2−6x−16)2(x2−6x−16)(1)−x(2x−6)
=(x2−6x−16)2x2−6x−16−2x2+6x
=(x2−6x−16)2−x2−16
=\frac{-\left(x^{2}-16\right)}{\left(x^{2}-6 x-16\right)^{2}}<0
\therefore f^{\prime}(x)<0 for x∈ domain
∴ It is a decreasing function f4 in its domain
x∈(−∞,−2)∪(2,8)∪(8,∞)
Explanation
f(x)=x2−6x−16xx∈R{−2,8}
f′(x)=(x2−6x−16)2(x2−6x−16)(1)−x(2x−6)
=(x2−6x−16)2x2−6x−16−2x2+6x
=(x2−6x−16)2−x2−16
=\frac{-\left(x^{2}-16\right)}{\left(x^{2}-6 x-16\right)^{2}}<0
\therefore f^{\prime}(x)<0 for x∈ domain
∴ It is a decreasing function f4 in its domain
x∈(−∞,−2)∪(2,8)∪(8,∞)