IPMAT Indore 2020 (MCQ) - Consider the following statements: (i) When 0 < x < 1, then 1/1+x < 1 - x + x^2 (ii) When 0 < x < 1, then 1/1+x > 1 - x + x^2 (iii) When -1 < x < 0, then 1/1+x < 1 - x + x^2 (iv) When -1 < x < 0, then 1/1+x > 1 - x + x^2 Then the correct statements are: | PYQs + Solutions

IPMAT Indore 2020

Algebra

Inequalities

Medium

Consider the following statements:
(i) When $0 < x < 1$ , then $\frac{1}{1+x} < 1 - x + x^2$
(ii) When $0 < x < 1$ , then $\frac{1}{1+x} > 1 - x + x^2$
(iii) When $-1 < x < 0$ , then $\frac{1}{1+x} < 1 - x + x^2$
(iv) When $-1 < x < 0$ , then $\frac{1}{1+x} > 1 - x + x^2$
Then the correct statements are:

(i) and (ii)

(ii) and (iv)

(i) and (iv)

(ii) and (iii)

Correct Option: 3

The easiest way to solve this question is by value-putting. Let's plug in some values in each of the given options and check which of the following statements are correct.

Statement (i):

\newline

0<x<1

, let

x=\dfrac{1}{2}

\Rightarrow \dfrac{1}{1 + \frac{1}{2}} < 1 - \dfrac{1}{2} + \dfrac{1}{4}

\Rightarrow \dfrac {2}{3} < \dfrac{3}{4}

which is true.

Therefore, statement (i) is correct.

Statement (ii):

\newline

Using the result obtained in statement (i), it is clear that statement (ii) is false.

Statement (iii):

-1<x<0

, let

x=-\dfrac{1}{2}

\Rightarrow \dfrac{1}{1 - \frac{1}{2}} < 1 - (-\dfrac{1}{2}) + \dfrac{1}{4}

\Rightarrow 2 < \dfrac{7}{4}

which is false.

Statement (iv):

\newline

Using the result obtained in statement (iii), it is clear that statement (iv) is true (as

2 > \dfrac{7}{4}

Hence, the correct option is

(3)

IPMAT Indore 2020 (MCQ) PYQs

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