If f is an increasing function - x - R and u - -c - 1 - -c- v - -c - -c - 1 - c-1 then

The correct option is A f(u) u = √(c+1) √(c)⇒u = 1 √(c+1)+√(c) v = √(c) √(c 1)⇒v = 1 √(c)+√(c 1) Clearly v gt; u Hence f(u) lt;f(v) ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

If f is an ...

Question

If $f$ is an increasing function $\forall x ϵ R$ and $u = \sqrt{c + 1} - \sqrt{c}; v = \sqrt{c} - \sqrt{c - 1}; (c > 1)$ then

f (u) > f (v)

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f (u) < f (v)

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f (u) = f (v)

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nothing can be said

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Solution

The correct option is A $f (u) < f (v)$
$u = \sqrt{c + 1} - \sqrt{c} \Rightarrow u = \frac{1}{\sqrt{c + 1} + \sqrt{c}}$
$v = \sqrt{c} - \sqrt{c - 1} \Rightarrow v = \frac{1}{\sqrt{c} + \sqrt{c - 1}}$
Clearly $v > u$
Hence $f (u) < f (v)$

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If f is an increasing function ∀xϵR and u=√c+1−√c;v=√c−√c−1;(c>1) then

The correct option is A f(u)<f(v)u=√c+1−√c⇒u=1√c+1+√cv=√c−√c−1⇒v=1√c+√c−1Clearly v>uHence f(u)<f(v)

If $f$ is an increasing function $\forall x ϵ R$ and $u = \sqrt{c + 1} - \sqrt{c}; v = \sqrt{c} - \sqrt{c - 1}; (c > 1)$ then

The correct option is A $f (u) < f (v)$
$u = \sqrt{c + 1} - \sqrt{c} \Rightarrow u = \frac{1}{\sqrt{c + 1} + \sqrt{c}}$
$v = \sqrt{c} - \sqrt{c - 1} \Rightarrow v = \frac{1}{\sqrt{c} + \sqrt{c - 1}}$
Clearly $v > u$
Hence $f (u) < f (v)$