The function f ( x ) = x x 2 - 6 x - 16 , x ? R - { - 2 , 8 } [2024] Differential Calculus jee mains questions

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Q.
The function $f (x) = \frac{x}{x^{2} - 6 x - 16}, x \in R - {- 2, 8}$ [2024]

1 decreases in $(- 2, 8)$ and increases in $(- \infty, - 2) \cup (8, \infty)$

2 decreases in $(- \infty, - 2)$ and increases in $(8, \infty)$

3 increases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$

4 decreases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$

Ans.
(4)

$f (x) = \frac{x}{x^{2} - 6 x - 16}$

$\Rightarrow f' (x) = \frac{(x^{2} - 6 x - 16) (1) - x (2 x - 6)}{{(x^{2} - 6 x - 16)}^{2}}$

$= \frac{- x^{2} - 16}{{(x^{2} - 6 x - 16)}^{2}} < 0 \forall x$

$∴ f (x) is a decreasing function in its domain.$

$\Rightarrow f (x) decreases in (- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$

Solution

Q.
The function $f (x) = \frac{x}{x^{2} - 6 x - 16}, x \in R - {- 2, 8}$ [2024]

1 decreases in $(- 2, 8)$ and increases in $(- \infty, - 2) \cup (8, \infty)$

2 decreases in $(- \infty, - 2)$ and increases in $(8, \infty)$

3 increases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$

4 decreases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$

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Solution

Q. The function f(x)=xx2-6x-16, x∈R-{-2,8} [2024]

1 decreases in (-2,8) and increases in (-∞,-2)∪(8,∞)

2 decreases in (-∞,-2) and increases in (8,∞)

3 increases in (-∞,-2)∪(-2,8)∪(8,∞)

4 decreases in (-∞,-2)∪(-2,8)∪(8,∞)

Ans. (4) f(x)=xx2-6x-16 ⇒f'(x)=(x2-6x-16)(1)-x(2x-6)(x2-6x-16)2 =-x2-16(x2-6x-16)2<0 ∀ x ∴ f(x) is a decreasing function in its domain. ⇒ f(x) decreases in (-∞,-2)∪(-2,8)∪(8,∞)

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Q.
The function $f (x) = \frac{x}{x^{2} - 6 x - 16}, x \in R - {- 2, 8}$ [2024]

1 decreases in $(- 2, 8)$ and increases in $(- \infty, - 2) \cup (8, \infty)$

2 decreases in $(- \infty, - 2)$ and increases in $(8, \infty)$

3 increases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$

4 decreases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$