Q.24. Let f: RR be a function such that f(x) + 3f(-x) = sinx, x R. Let maximum value of f on R be .

Q: What is the correct answer for JEE Main Mathematics Q24?

The correct answer is 16. Sol. f(x) + 3f($\frac{\pi}{2}$-x) = sinx ...(1) Replace x with $\frac{\pi}{2}$-x: f($\frac{\pi}{2}$-x) + 3f(x) = sin($\frac{\pi}{2}$-x) = cosx ...(2) From (1) and (2): Multiply (2) by 3: 3f($\frac{\pi}{2}$-x) + 9f(x) = 3cosx Subtract (1) from this: (9f(x) + 3f($\frac{\pi}{2}$-x)) - (f(x) + 3f($\frac{\pi}{2}$-x)) = 3cosx - sinx 8f(x) = 3cosx - sinx f(x) = $\frac{1}{8}$(3cosx - sinx) To find the max

Q: Q.24. Let f: RR be a function such that f(x) + 3f(-x) = sinx, x R. Let maximum value of f on R be . If the area of the region bounded by the curves g(x) = x² and h(x) = x³, > 0, is ², then 30³ is…

The correct answer is 16. Sol. f(x) + 3f($\frac{\pi}{2}$-x) = sinx ...(1) Replace x with $\frac{\pi}{2}$-x: f($\frac{\pi}{2}$-x) + 3f(x) = sin($\frac{\pi}{2}$-x) = cosx ...(2) From (1) and (2): Multiply (2) by 3: 3f($\frac{\pi}{2}$-x) + 9f(x) = 3cosx Subtract (1) from this: (9f(x) + 3f($\frac{\pi}{2}$-x)) - (f(x) + 3f($\frac{\pi}{2}$-x)) = 3cosx - sinx 8f(x) = 3cosx - sinx f(x) = $\frac{1}{8}$(3cosx - sinx) To find the max

Question

Q.24. Let f: RR be a function such that f(x) + 3f(-x) = sinx, x  R. Let maximum value of f on R be . If the area of the region bounded by the curves g(x) = x² and h(x) = x³,  > 0, is ², then 30³ is equal to ...... [Figure: A diagram showing the region bounded by g(x)=x^2 and h(x)=beta*x^3 is present]

FirstInTest · Accepted Answer

Sol. f(x) + 3f($\frac{\pi}{2}$-x) = sinx ...(1)
Replace x with $\frac{\pi}{2}$-x:
f($\frac{\pi}{2}$-x) + 3f(x) = sin($\frac{\pi}{2}$-x) = cosx ...(2)
From (1) and (2):
Multiply (2) by 3: 3f($\frac{\pi}{2}$-x) + 9f(x) = 3cosx
Subtract (1) from this: (9f(x) + 3f($\frac{\pi}{2}$-x)) - (f(x) + 3f($\frac{\pi}{2}$-x)) = 3cosx - sinx
8f(x) = 3cosx - sinx
f(x) = $\frac{1}{8}$(3cosx - sinx)
To find the maximum value of f(x), we use the amplitude formula for Acosx + Bsinx:
Maximum value = $\sqrt{A^2 + B^2}$
So, $\alpha = \frac{1}{8}\sqrt{3^2 + (-1)^2} = \frac{1}{8}\sqrt{9+1} = \frac{\sqrt{10}}{8}$
Area bounded by g(x) = x² and h(x) = $\beta$x³:
Find intersection points: x² = $\beta$x³ $\Rightarrow$ x²(1 - $\beta$x) = 0 $\Rightarrow$ x = 0 or x = $\frac{1}{\beta}$
Area = $\int_0^{1/\beta} (\beta x^3 - x^2) dx$ (assuming $\beta x^3$ is above $x^2$ for $x \in (0, 1/\beta)$)
If $\beta x^3 > x^2$, then $\beta x > 1$. This means $x > 1/\beta$. So the order should be $x^2 - \beta x^3$.
Area = $\int_0^{1/\beta} (x^2 - \beta x^3) dx = [\frac{x^3}{3} - \frac{\beta x^4}{4}]_0^{1/\beta} = \frac{1}{3\beta^3} - \frac{\beta}{4\beta^4} = \frac{1}{3\beta^3} - \frac{1}{4\beta^3} = \frac{4-3}{12\beta^3} = \frac{1}{12\beta^3}$
Given Area = $\alpha^2$
$\frac{1}{12\beta^3} = (\frac{\sqrt{10}}{8})^2 = \frac{10}{64} = \frac{5}{32}$
$12\beta^3 = \frac{32}{5}$
$\beta^3 = \frac{32}{60} = \frac{8}{15}$
We need to find 30$\beta$³:
30$\beta$³ = 30 * $\frac{8}{15} = 2 * 8 = 16$.

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions