If fx-1-x2sin2x-x2-R and gx-1x-2-t-1-t-1 -ln t- ft dt- x- 1- - which of the following is correct-

Step 1: Given informationg(x)=∫_1^x|(2[t 1])/(t+1) lnt| f(t) dt x∈ (1,∞ )Step 2: Applying the Newton Leibnitz theoremAccording to the Newton Leibnitz theorem,⇒I ...

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Byju's Answer

Standard XII

Mathematics

Theorems for Continuity

If fx=1-x2sin...

Question

If $f (x) = {(1 - x)}^{2} \sin^{2} x + x^{2}$ , $\forall \in R$ and $g (x) = \int_{1}^{x} |\frac{(2 [t - 1])}{(t + 1)} - l n t| f (t) dt \forall x \in (1, \infty) .$ . which of the following is correct?

g is increasing on (1,∞)

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g is decreasing on (1,∞)

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g is increasing on (1,2) and decreasing on (2,∞)

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g is decreasing on (1,2) and increasing on (2,∞)

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Solution

The correct option is B
g is decreasing on (1,∞)

Step 1: Given information
$g (x) = \int_{1}^{x} |\frac{(2 [t - 1])}{(t + 1)} - \ln t| f (t) dt x \in (1, \infty)$
Step 2: Applying the Newton-Leibnitz theorem
According to the Newton-Leibnitz theorem,
$\Rightarrow I (x) = \int_{\emptyset (x)}^{\emptyset (x)} f (t) . dt \Rightarrow I' (x) = f ((x)) . \emptyset' (x) - f (\emptyset (x)) . \emptyset' (x)$
$g' (x) = (\frac{2 (x - 1)}{(x + 1)} - \ln x) f (x)$
For $x \in (1, \infty), f (x) > 0$
Let $h (x) = (\frac{2 (x - 1)}{(x + 1)} - \ln x)$
$\begin{array}{rcl} ∴ h' (x) & = & [\frac{4}{{(x + 1)}^{2}} - \frac{1}{x}] \\ = & \frac{- {(x - 1)}^{2}}{{(x + 1)}^{2}} < 0 \end{array}$
Also $h (1) = 0 .$
Therefore, $h (x) < 0 \forall x > 1$
$\Rightarrow g (x)$ is decreasing on $(1, \infty)$
Hence, the correct option is option(B) g(x) is decreasing on $(1, \infty) .$

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If f(x)=(1-x)2sin2x+x2,∀∈R and g(x)=∫1x2t-1(t+1)-lntf(t)dt∀x∈(1,∞).. which of the following is correct?

If $f (x) = {(1 - x)}^{2} \sin^{2} x + x^{2}$ , $\forall \in R$ and $g (x) = \int_{1}^{x} |\frac{(2 [t - 1])}{(t + 1)} - l n t| f (t) dt \forall x \in (1, \infty) .$ . which of the following is correct?