Let f:(0,1)\to\mathbbR be a function defined f(x) = 1 \over 1 - e^ - x , and g(x) = ( f( - x) - f(x) ) . Consider two statements (I) g is an increasing function in (0, 1) (II) g is one-one in (0, 1) Then,

Application of Derivatives PYQ — JEE Main 2023 · 25 Jan · Shift 1 · Q38 with Solution

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Correct answer: A

Given functions

We have $f(x)=\frac{1}{1-e^{-x}}$ and $g(x)=f(-x)-f(x).$

We need to examine whether:

(I) $g$ is increasing on $(0,1)$
(II) $g$ is one-one on $(0,1)$

Compute $f(-x)$

Substitute $-x$ into $f$ : $f(-x)=\frac{1}{1-e^{-(-x)}}=\frac{1}{1-e^x}.$

So, $g(x)=\frac{1}{1-e^x}-\frac{1}{1-e^{-x}}.$

Simplify $g(x)$

Let $t=e^x$ . Then $e^{-x}=\frac1t$ . Hence $\frac{1}{1-e^{-x}}=\frac{1}{1-\frac1t}=\frac{1}{\frac{t-1}{t}}=\frac{t}{t-1}.$ Also, $\frac{1}{1-e^x}=\frac{1}{1-t}=-\frac{1}{t-1}.$

Therefore,

Replacing $t=e^x$ ,

Check monotonicity of $g$

Differentiate: $g(x)=-\frac{1+e^x}{e^x-1}.$

Let $u=1+e^x,\qquad v=e^x-1.$ Then $u'=e^x,\qquad v'=e^x.$

By quotient rule,

=-\frac{e^x(e^x-1)-(1+e^x)e^x}{(e^x-1)^2}.$$ Simplify the numerator: $$e^x(e^x-1)-(1+e^x)e^x=e^{2x}-e^x-e^x-e^{2x}=-2e^x.$$ Thus, $$g'(x)=-\frac{-2e^x}{(e^x-1)^2}=\frac{2e^x}{(e^x-1)^2}.$$ Now for $x\in(0,1)$, - $e^x>0$ - $(e^x-1)^2>0$ Hence $$g'(x)>0 \quad \forall x\in(0,1).$$ So $g$ is **strictly increasing** on $(0,1)$. Therefore, **Statement (I) is true**. --- 5. **Check whether $g$ is one-one** A strictly increasing function on an interval is automatically one-one. Since $g$ is strictly increasing on $(0,1)$, it is **injective (one-one)** on $(0,1)$. Therefore, **Statement (II) is true**. --- 6. **Conclusion** Both statements (I) and (II) are true. So the correct option is $$\boxed{\text{A}}.$$ --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.

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View written solutionFree

Correct answer: A

Given functions

We have $f(x)=\frac{1}{1-e^{-x}}$ and $g(x)=f(-x)-f(x).$

We need to examine whether:

(I) $g$ is increasing on $(0,1)$
(II) $g$ is one-one on $(0,1)$

Compute $f(-x)$

Substitute $-x$ into $f$ : $f(-x)=\frac{1}{1-e^{-(-x)}}=\frac{1}{1-e^x}.$

So, $g(x)=\frac{1}{1-e^x}-\frac{1}{1-e^{-x}}.$

Simplify $g(x)$

Let $t=e^x$ . Then $e^{-x}=\frac1t$ . Hence $\frac{1}{1-e^{-x}}=\frac{1}{1-\frac1t}=\frac{1}{\frac{t-1}{t}}=\frac{t}{t-1}.$ Also, $\frac{1}{1-e^x}=\frac{1}{1-t}=-\frac{1}{t-1}.$

Therefore,

Replacing $t=e^x$ ,

Check monotonicity of $g$

Differentiate: $g(x)=-\frac{1+e^x}{e^x-1}.$

Let $u=1+e^x,\qquad v=e^x-1.$ Then $u'=e^x,\qquad v'=e^x.$

By quotient rule,

=-\frac{e^x(e^x-1)-(1+e^x)e^x}{(e^x-1)^2}.$$ Simplify the numerator: $$e^x(e^x-1)-(1+e^x)e^x=e^{2x}-e^x-e^x-e^{2x}=-2e^x.$$ Thus, $$g'(x)=-\frac{-2e^x}{(e^x-1)^2}=\frac{2e^x}{(e^x-1)^2}.$$ Now for $x\in(0,1)$, - $e^x>0$ - $(e^x-1)^2>0$ Hence $$g'(x)>0 \quad \forall x\in(0,1).$$ So $g$ is **strictly increasing** on $(0,1)$. Therefore, **Statement (I) is true**. --- 5. **Check whether $g$ is one-one** A strictly increasing function on an interval is automatically one-one. Since $g$ is strictly increasing on $(0,1)$, it is **injective (one-one)** on $(0,1)$. Therefore, **Statement (II) is true**. --- 6. **Conclusion** Both statements (I) and (II) are true. So the correct option is $$\boxed{\text{A}}.$$ --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.

Application of Derivatives question

Application of Derivatives question

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