If f(x) = (\tan 1^\circ )x + \log _e(123) \over x\log _e(1234) - (\tan 1^\circ ),x \gt 0 , then the least value of f(f(x)) + f( f( 4 \over x ) ) is :

Functions PYQ — JEE Main 2023 · 10 Apr · Shift 1 · Q28 with Solution

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Correct answer: B

Let $a=\tan 1^\circ>0,\quad b=\ln 123>0,\quad c=\ln 1234>0.$ Then $f(x)=\frac{ax+b}{cx-a},\qquad x>0.$ We need the least value of

left(f\left(\frac4x\right)\right).$$ ## 1. Compute $f(f(x))$ Given a linear fractional function $$f(x)=\frac{ax+b}{cx-a},$$ we substitute $f(x)$ into itself: $$f(f(x))=\frac{a\,f(x)+b}{c\,f(x)-a}.$$ Now $$a\,f(x)+b=a\cdot\frac{ax+b}{cx-a}+b =\frac{a(ax+b)+b(cx-a)}{cx-a} =\frac{a^2x+ab+bcx-ab}{cx-a} =\frac{(a^2+bc)x}{cx-a}.$$ Also, $$c\,f(x)-a=c\cdot\frac{ax+b}{cx-a}-a =\frac{c(ax+b)-a(cx-a)}{cx-a} =\frac{acx+bc-acx+a^2}{cx-a} =\frac{a^2+bc}{cx-a}.$$ Hence $$f(f(x))=\frac{(a^2+bc)x}{a^2+bc}=x.$$ So $f$ is self-inverse: $$f(f(x))=x.$$ ## 2. Compute $f\left(f\left(\frac4x\right)\right)$ Using the same property, $$f\left(f\left(\frac4x\right)\right)=\frac4x.$$ Therefore the required expression becomes $$x+\frac4x,\qquad x>0.$$ ## 3. Find its least value For $x>0$, by AM-GM, $$x+\frac4x\ge 2\sqrt{x\cdot \frac4x}=2\sqrt4=4.$$ Equality holds when $$x=\frac4x \implies x^2=4 \implies x=2$$ (since $x>0$). Thus the least value is $$\boxed{4}.$$ ## 4. Check options - A: $2$ — incorrect - B: $4$ — correct - C: $0$ — incorrect - D: $8$ — incorrect

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Correct answer: B

Let $a=\tan 1^\circ>0,\quad b=\ln 123>0,\quad c=\ln 1234>0.$ Then $f(x)=\frac{ax+b}{cx-a},\qquad x>0.$ We need the least value of

left(f\left(\frac4x\right)\right).$$ ## 1. Compute $f(f(x))$ Given a linear fractional function $$f(x)=\frac{ax+b}{cx-a},$$ we substitute $f(x)$ into itself: $$f(f(x))=\frac{a\,f(x)+b}{c\,f(x)-a}.$$ Now $$a\,f(x)+b=a\cdot\frac{ax+b}{cx-a}+b =\frac{a(ax+b)+b(cx-a)}{cx-a} =\frac{a^2x+ab+bcx-ab}{cx-a} =\frac{(a^2+bc)x}{cx-a}.$$ Also, $$c\,f(x)-a=c\cdot\frac{ax+b}{cx-a}-a =\frac{c(ax+b)-a(cx-a)}{cx-a} =\frac{acx+bc-acx+a^2}{cx-a} =\frac{a^2+bc}{cx-a}.$$ Hence $$f(f(x))=\frac{(a^2+bc)x}{a^2+bc}=x.$$ So $f$ is self-inverse: $$f(f(x))=x.$$ ## 2. Compute $f\left(f\left(\frac4x\right)\right)$ Using the same property, $$f\left(f\left(\frac4x\right)\right)=\frac4x.$$ Therefore the required expression becomes $$x+\frac4x,\qquad x>0.$$ ## 3. Find its least value For $x>0$, by AM-GM, $$x+\frac4x\ge 2\sqrt{x\cdot \frac4x}=2\sqrt4=4.$$ Equality holds when $$x=\frac4x \implies x^2=4 \implies x=2$$ (since $x>0$). Thus the least value is $$\boxed{4}.$$ ## 4. Check options - A: $2$ — incorrect - B: $4$ — correct - C: $0$ — incorrect - D: $8$ — incorrect