JEE MainMathematicsFunctionsNumerical+4 / −1
If the range of is , then the sum of the infinite G.P., whose first term is 64 and the common ratio is , is equal to .
Numerical answer
View written solutionFree
Correct answer: 96
- Rewrite the function in one variable
Let Since , we have Also,
So
=\frac{(1-x)^2+3x}{(1-x)^2+x}.$$ Simplify: $$f(x)=\frac{1-2x+x^2+3x}{1-2x+x^2+x} =\frac{x^2+x+1}{x^2-x+1}, \quad 0\le x\le 1.$$ 2. **Find the range of** $f(x)$ We need the maximum and minimum of $$f(x)=\frac{x^2+x+1}{x^2-x+1}.$$ Differentiate: $$f'(x)=\frac{(2x+1)(x^2-x+1)-(x^2+x+1)(2x-1)}{(x^2-x+1)^2}.$$ Now simplify the numerator:(2x+1)(x^2-x+1)=2x^3-x^2+x+1,
(x^2+x+1)(2x-1)=2x^3+x^2+x-1.
(2x+1)(x^2-x+1)-(x^2+x+1)(2x-1) =(2x^3-x^2+x+1)-(2x^3+x^2+x-1) =2-2x^2=2(1-x^2).
Hence $$f'(x)=\frac{2(1-x^2)}{(x^2-x+1)^2}.$$ For $0\le x\le 1$, the denominator is always positive, and $$1-x^2\ge 0,$$ with equality only at $x=1$. So $f(x)$ is increasing on $[0,1]$. Thus, - minimum at $x=0$, - maximum at $x=1$. Compute: $$f(0)=\frac{1}{1}=1,$$ $$f(1)=\frac{3}{1}=3.$$ So the range is $$[\alpha,\beta]=[1,3].$$ Thus, $$\alpha=1,\quad \beta=3.$$ 3. **Find the sum of the infinite G.P.** First term: $$a=64$$ Common ratio: $$r=\frac{\alpha}{\beta}=\frac{1}{3}.$$ Since $|r|<1$, the sum to infinity is $$S_\infty=\frac{a}{1-r}=rac{64}{1-\frac13} =\frac{64}{\frac23}=64\cdot\frac32=96.$$ 4. **Final answer** $$\boxed{96}$$ This matches the stored correct answer.More from Functions
- Let where and …2024 · MCQ
- If the domain of the function is , then is equal to :2024 · MCQ
- If a function satisfies for all and , then the largest natural number such that …2024 · Numerical
- Let the range of the function be . If and ar respectively the A.M. and the G.M. of and , then is equal to2024 · MCQ
- Let and . Then the number of one-one functions from to is equal to .2024 · Numerical
- The function ; defined by the highest prime factor of , is :2024 · MCQ
- Let and be defined as and . Then, the domain of…2024 · MCQ
- If , then range…2024 · MCQ