The explanation of the correct option :Let g (x) = log (1 + x ) x theng'(x) = 1/1+x 1= x/1+x g (x)is decreasing on (0, 1). ∴x > 0We have g (x) ⇒log_e(1 + x) ...

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Differentiability

For all x ∈ 0...

Question

For all $x \in (0, 1)$

$e^{x} < 1 + x$

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$\log_{e} (1 + x) < x$

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$\sin x > x$

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$\log_{e} x > x$

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Solution

The correct option is B
$\log_{e} (1 + x) < x$

The explanation of the correct option :
Let $g (x) = \log (1 + x) - x$ then
$g' (x) = \frac{1}{1 + x} - 1 = \frac{- x}{1 + x} < 0, \forall x \in (0, 1)$
$g (x)$ is decreasing on $(0, 1)$ .
$∴ x > 0$
We have $g (x) < g (0)$
$\Rightarrow \log_{e} (1 + x) - x < 0 \Rightarrow \log_{e} (1 + x) < x$
Hence, the correct option is (B).

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Which of the following is necessarily true ?

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For all x∈(0,1)

The correct option is B loge(1+x)<xThe explanation of the correct option :Let g(x)=log(1+x)−x theng'(x)=11+x-1=-x1+x<0,∀x∈(0,1)g(x) is decreasing on (0,1).∴x>0We have g(x)<g(0)⇒loge(1+x)−x<0⇒loge(1+x)<xHence, the correct option is (B).

For all $x \in (0, 1)$