Match the following inequalities with intervals given such that inequalities are valid Column IA) 1 + x x

Question

Match the following inequalities with intervals given such that inequalities are valid

Column IA) $\frac{x}{1 + x} < ln (1 + x)$ (B) $x - \frac{x ^{2}}{2} < ln (1 + x)$ (C) $ln (1 + x) < x$ (D) $\frac{1}{l n ( 1 + x )} - \frac{1}{x} < 1$ Column IIP) $(0, \infty)$ (Q) $(- 1, 0)$ (R) $(1, \infty)$ (S) $(- 1, 0) \cup (0, 1)$

TestHub · Accepted Answer

(A) Let g(x) = ln(1+x) - x/(1+x); g'(x) = x/(1+x)^2, so g decreases on (-1,0) and increases on (0,inf) with g(0) = 0. Hence g(x) > 0 on both (-1,0) and (0,inf), so (A) is valid on P, Q, R and S. (B) Let h(x) = ln(1+x) - x + x^2/2; h'(x) = x^2/(1+x) >= 0 for x > -1 and h(0) = 0, so h(x) > 0 exactly when x > 0: valid on P and R only. (C) Let k(x) = x - ln(1+x); k'(x) = x/(1+x), with minimum k(0) = 0, so k(x) > 0 for every x > -1, x != 0: valid on P, Q, R and S. (D) 1/ln(1+x) - 1/x < 1 rearranges to x - ln(1+x) < x ln(1+x), i.e. ln(1+x) > x/(1+x) (the sign of x ln(1+x) is positive for all x > -1, x != 0), which is exactly inequality (A); so (D) is valid on P, Q, R and S. Answer: A -> PQRS, B -> PR, C -> PQRS, D -> PQRS (option A).

Mathematics - Application of Derivative Question with Solution | TestHub

Options:

Doubts & Discussion