Let f : R® R and g : R ® R be two functions defined by f(x) = loge (x2 + 1) – e-x + 1 and <math> <mrow> <mi>g</mi> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mo>−</mo> <mn>2</mn> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mn>2</mn> <mi>x</mi> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mi>x</mi> </mrow> </msup> </mrow> </mfrac> <mo>.</mo> </mrow> </math> Then, for which of the following range of a, the inequality <math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>g</mi> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>α</mi> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mo>)</mo> </mrow> <mo>></mo> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>g</mi> <mrow> <mo>(</mo> <mrow> <mi>α</mi> <mo>−</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mo>)</mo> </mrow> </mrow> </math> holds?

f ( x ) = l n ( x 2 + 1 ) − e − x + 1 , g ( x ) = e − x − 2 e x for f ( g ( a ) ) > f ( g ( b ) ) g ' ( x ) = e x ¯ − 2 e x < 0 ∀ x ∈ R f ' ( x ) = 2 x x 2 + 1 + e − x { < 1 f o r n ≥ 0 , g ( x ) ↓ ≥ 1 f o r n < 0 g(a) > g(b) ⇒ a < b ( a s g ↓ ) ⇒ α 2 − 5 α + 6 < 0 ( α − 3 ) ( α − 2 ) < 0

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Maths NCERT Exemplar Solutions Class 11th Chapter Two Maths Class 12th Maths Continuity and Differentiability

Let f : R® R and g : R ® R be two functions defined by f(x) = log_e (x² + 1) – e^-x + 1 and $g (x) = \frac{1 - 2 e^{2 x}}{e^{x}} .$ Then, for which of the following range of a, the inequality $f (g (\frac{{(α - 1)}^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

Option 1 -

(2, 3)

Option 2 -

(-2, -1)

Option 3 -

(1, 2)

Option 4 -

(-1, 1)

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 4

Detailed Solution:

$f (x) = l n (x^{2} + 1) - e^{- x} + 1, g (x) = e^{- x} - 2 e^{x}$

for $f (g (a)) > f (g (b)) g' (x) = \bar{e^{x}} - 2 e^{x} < 0 \forall x \in R$

$f' (x) = \frac{2 x}{x^{2} + 1} + e^{- x}$

${\begin{cases} < 1 f o r n \geq 0, g (x) ↓ \\ \geq 1 f o r n < 0 \end{cases}$

g(a) > g(b)

$\Rightarrow a < b (a s g ↓)$

$\begin{array}{l} \Rightarrow α^{2} - 5 α + 6 < 0 \\ (α - 3) (α - 2) < 0 \end{array}$

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Let f : R® R and g : R ® R be two functions defined by f(x) = log_e (x² + 1) – e^-x + 1 and $g (x) = \frac{1 - 2 e^{2 x}}{e^{x}} .$ Then, for which of the following range of a, the inequality $f (g (\frac{{(α - 1)}^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

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Let f : R® R and g : R ® R be two functions defined by f(x) = loge (x2 + 1) – e-x + 1 and g ( x ) = 1 − 2 e 2 x e x . Then, for which of the following range of a, the inequality f ( g ( ( α − 1 ) 2 3 ) ) > f ( g ( α − 5 3 ) ) holds?

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Let f : R® R and g : R ® R be two functions defined by f(x) = log_e (x² + 1) – e^-x + 1 and $g (x) = \frac{1 - 2 e^{2 x}}{e^{x}} .$ Then, for which of the following range of a, the inequality $f (g (\frac{{(α - 1)}^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?