Let f, g:(1, \infty) arrow \mathbbR be defined as f(x)=\frac2 x+35 x+2 and g(x)=\frac2-3 x1-x . If the range of the function fog: [2,4] arrow \mathbbR is [\alpha, \beta] , then \frac1\beta-\alpha is equal to

Functions PYQ — JEE Main 2025 · 4 Apr · Shift 1 · Q41 with Solution

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Correct answer: A

We need the range of the composite function $(f\circ g)(x)=f(g(x))$ for $x\in[2,4]$ , where $f(x)=\frac{2x+3}{5x+2},\qquad g(x)=\frac{2-3x}{1-x}.$
First simplify $g(x)$ : $g(x)=\frac{2-3x}{1-x}=\frac{3x-2}{x-1}.$ For $x\in[2,4]$ , we have $x>1$ , so this is well-defined.
Now compute $f(g(x))$ : $f(g(x))=\frac{2g(x)+3}{5g(x)+2}.$ Substitute $g(x)=\frac{3x-2}{x-1}$ : $f(g(x))=\frac{2\cdot \frac{3x-2}{x-1}+3}{5\cdot \frac{3x-2}{x-1}+2}.$
Simplify numerator and denominator separately:

=\frac{6x-4+3x-3}{x-1} =\frac{9x-7}{x-1}. $$ Similarly, $$ 5\cdot \frac{3x-2}{x-1}+2= rac{15x-10}{x-1}+2 =\frac{15x-10+2x-2}{x-1} =\frac{17x-12}{x-1}. $$ Hence, $$ (f\circ g)(x)=\frac{\frac{9x-7}{x-1}}{\frac{17x-12}{x-1}}=\frac{9x-7}{17x-12}. $$ 5. So we need the range of $$ h(x)=\frac{9x-7}{17x-12},\qquad x\in[2,4]. $$ 6. Check monotonicity: For a function of the form $$ \frac{ax+b}{cx+d}, $$ its derivative is $$ \frac{ad-bc}{(cx+d)^2}. $$ Here, $$ a=9,\ b=-7,\ c=17,\ d=-12. $$ So $$ h'(x)=\frac{9(-12)-(-7)(17)}{(17x-12)^2} =\frac{-108+119}{(17x-12)^2} =\frac{11}{(17x-12)^2}>0. $$ Thus $h(x)$ is strictly increasing on $[2,4]$. 7. Therefore the minimum value is at $x=2$ and the maximum value is at $x=4$. Compute: $$ h(2)=\frac{18-7}{34-12}=\frac{11}{22}=\frac12, $$ $$ h(4)=\frac{36-7}{68-12}=\frac{29}{56}. $$ Since $h$ is increasing, $$ \alpha=\frac12,\qquad \beta=\frac{29}{56}. $$ 8. Now find $$ \beta-\alpha=\frac{29}{56}-\frac12=\frac{29}{56}-\frac{28}{56}=\frac{1}{56}. $$ Hence, $$ \frac{1}{\beta-\alpha}=56. $$ 9. Therefore the correct option is: $$ \boxed{56} $$ which is option A.

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Correct answer: A

We need the range of the composite function $(f\circ g)(x)=f(g(x))$ for $x\in[2,4]$ , where $f(x)=\frac{2x+3}{5x+2},\qquad g(x)=\frac{2-3x}{1-x}.$
First simplify $g(x)$ : $g(x)=\frac{2-3x}{1-x}=\frac{3x-2}{x-1}.$ For $x\in[2,4]$ , we have $x>1$ , so this is well-defined.
Now compute $f(g(x))$ : $f(g(x))=\frac{2g(x)+3}{5g(x)+2}.$ Substitute $g(x)=\frac{3x-2}{x-1}$ : $f(g(x))=\frac{2\cdot \frac{3x-2}{x-1}+3}{5\cdot \frac{3x-2}{x-1}+2}.$
Simplify numerator and denominator separately:

=\frac{6x-4+3x-3}{x-1} =\frac{9x-7}{x-1}. $$ Similarly, $$ 5\cdot \frac{3x-2}{x-1}+2= rac{15x-10}{x-1}+2 =\frac{15x-10+2x-2}{x-1} =\frac{17x-12}{x-1}. $$ Hence, $$ (f\circ g)(x)=\frac{\frac{9x-7}{x-1}}{\frac{17x-12}{x-1}}=\frac{9x-7}{17x-12}. $$ 5. So we need the range of $$ h(x)=\frac{9x-7}{17x-12},\qquad x\in[2,4]. $$ 6. Check monotonicity: For a function of the form $$ \frac{ax+b}{cx+d}, $$ its derivative is $$ \frac{ad-bc}{(cx+d)^2}. $$ Here, $$ a=9,\ b=-7,\ c=17,\ d=-12. $$ So $$ h'(x)=\frac{9(-12)-(-7)(17)}{(17x-12)^2} =\frac{-108+119}{(17x-12)^2} =\frac{11}{(17x-12)^2}>0. $$ Thus $h(x)$ is strictly increasing on $[2,4]$. 7. Therefore the minimum value is at $x=2$ and the maximum value is at $x=4$. Compute: $$ h(2)=\frac{18-7}{34-12}=\frac{11}{22}=\frac12, $$ $$ h(4)=\frac{36-7}{68-12}=\frac{29}{56}. $$ Since $h$ is increasing, $$ \alpha=\frac12,\qquad \beta=\frac{29}{56}. $$ 8. Now find $$ \beta-\alpha=\frac{29}{56}-\frac12=\frac{29}{56}-\frac{28}{56}=\frac{1}{56}. $$ Hence, $$ \frac{1}{\beta-\alpha}=56. $$ 9. Therefore the correct option is: $$ \boxed{56} $$ which is option A.