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Limits Continuity and Differentiability question

2018 · Shift 1 · Q23
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  4. /Limits Continuity and Differentiability
  5. /2018 · Shift 1 · Q23

Limits Continuity and Differentiability question

2018 · Shift 1 · Q23

JEE AdvancedMathematicsLimits Continuity and DifferentiabilityMultiple correct+4 / −1
Let f : R →\to→ R and g : R →\to→ R be two non-constant differentiable functions. If f'(x) = (e(f(x) −-− g(x))) g'(x) for all x ∈\in∈ R and f(1) = g(2) = 1, then which of the following statement(s) is (are) TRUE?
  1. A
    f(2) < 1 −-− loge 2
  2. B
    f(2) > 1 −-− loge 2
  3. C
    g(1) > 1 −-− loge 2
  4. D
    g(1) < 1 −-− loge 2
View written solutionFree

Correct answer: B, C

  1. Given relation

We are given

\qquad \text{for all }x\in\mathbb R.$$ Also, $$f(1)=1,\qquad g(2)=1.$$ We must compare $f(2)$ and $g(1)$ with $1-\ln 2$. --- 2. **Try to find a conserved quantity** Let $$h(x)=e^{-f(x)}+e^{-g(x)}.$$ Differentiate: $$h'(x)=-f'(x)e^{-f(x)}-g'(x)e^{-g(x)}.$$ Using $$f'(x)=e^{f(x)-g(x)}g'(x),$$ we get $$f'(x)e^{-f(x)}=e^{f(x)-g(x)}g'(x)e^{-f(x)}=e^{-g(x)}g'(x).$$ Hence $$h'(x)=-e^{-g(x)}g'(x)-e^{-g(x)}g'(x)=-2e^{-g(x)}g'(x).$$ This does not simplify to zero, so let us instead try another combination. Consider $$F(x)=e^{-f(x)}-e^{-g(x)}.$$ Then $$F'(x)=-f'(x)e^{-f(x)}+g'(x)e^{-g(x)}.$$ Again using the given relation, $$f'(x)e^{-f(x)}=e^{-g(x)}g'(x).$$ So $$F'(x)=-e^{-g(x)}g'(x)+e^{-g(x)}g'(x)=0.$$ Therefore, $$e^{-f(x)}-e^{-g(x)}=\text{constant}.$$ --- 3. **Determine the constant** Use the given values at two different points: - At $x=1$, $f(1)=1$, so $$e^{-f(1)}-e^{-g(1)}=e^{-1}-e^{-g(1)}.$$ - At $x=2$, $g(2)=1$, so $$e^{-f(2)}-e^{-g(2)}=e^{-f(2)}-e^{-1}.$$ Since the quantity is constant, $$e^{-1}-e^{-g(1)}=e^{-f(2)}-e^{-1}.$$ Thus, $$e^{-f(2)}+e^{-g(1)}=2e^{-1}. $$ Multiplying by $e$, $$e^{1-f(2)}+e^{1-g(1)}=2. $$ Equivalently, $$e^{-f(2)}+e^{-g(1)}=\frac{2}{e}. $$ --- 4. **Use non-constancy** Suppose $f(2)=1-\ln 2$. Then $$e^{-f(2)}=e^{-(1-\ln 2)}=e^{-1}e^{\ln 2}=\frac{2}{e}.$$ From $$e^{-f(2)}+e^{-g(1)}=\frac{2}{e},$$ we get $$e^{-g(1)}=0,$$ which is impossible. So $f(2)\neq 1-\ln 2$. Similarly, $g(1)\neq 1-\ln 2$. Now note that both terms $e^{-f(2)}$ and $e^{-g(1)}$ are positive and their sum is $\frac{2}{e}$. Hence each one is strictly less than $\frac{2}{e}$: $$e^{-f(2)}<\frac{2}{e},\qquad e^{-g(1)}<\frac{2}{e}. $$ Taking natural logarithms carefully: From $$e^{-f(2)}<\frac{2}{e}=e^{\ln 2-1},$$ we get $$-f(2)<\ln 2-1,$$ so $$f(2)>1-\ln 2.$$ Similarly, $$e^{-g(1)}<\frac{2}{e}=e^{\ln 2-1}$$ gives $$-g(1)<\ln 2-1,$$ so $$g(1)>1-\ln 2.$$ --- 5. **Check options** - **A:** $f(2)<1-\ln 2$ is **false**. - **B:** $f(2)>1-\ln 2$ is **true**. - **C:** $g(1)>1-\ln 2$ is **true**. - **D:** $g(1)<1-\ln 2$ is **false**. --- 6. **Final answer** The true statements are $$\boxed{\text{B and C}}.$$ This matches the stored correct answer.
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