Let f: \mathbbR arrow \mathbbR be a function defined by f(x)=\fracx(1+x^4)^1 / 4 , and g(x)=f(f(f(f(x)))) . Then, 18 \int_0^\sqrt2 \sqrt5 x^2 g(x) d x is equal to

Definite Integration PYQ — JEE Main 2024 · 30 Jan · Shift 2 · Q31 with Solution

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Correct answer: C

Given functions

We have $f(x)=\frac{x}{(1+x^4)^{1/4}},$ and $g(x)=f(f(f(f(x)))).$

We need to compute $18\int_0^{\sqrt{2\sqrt{5}}} x^2 g(x)\,dx.$

Find a useful pattern for repeated application of $f$

Let $y=f(x)=\frac{x}{(1+x^4)^{1/4}}.$ Then $y^4=\frac{x^4}{1+x^4}.$ So, $1-y^4=1-\frac{x^4}{1+x^4}=\frac{1}{1+x^4}.$ Hence, $(1-y^4)^{1/4}=\frac{1}{(1+x^4)^{1/4}}.$ Therefore, $y=x(1-y^4)^{1/4}.$

But an even better way is to track the fourth power. If $T(u)=\frac{u}{(1+u^4)^{1/4}},$ then $T(u)^4=\frac{u^4}{1+u^4}.$

Define $u_0=x,\quad u_{n+1}=f(u_n).$ Let $v_n=u_n^4.$ Then $v_{n+1}=\frac{v_n}{1+v_n}.$ Now invert: $\frac{1}{v_{n+1}}=\frac{1+v_n}{v_n}=\frac{1}{v_n}+1.$ So after 4 iterations, $\frac{1}{v_4}=\frac{1}{v_0}+4=\frac{1}{x^4}+4.$ Thus, $v_4=\frac{x^4}{1+4x^4}.$ Hence, $g(x)=u_4=\frac{x}{(1+4x^4)^{1/4}}.$

Set up the integral

So,

=\int_0^{\sqrt{2\sqrt{5}}} \frac{x^3}{(1+4x^4)^{1/4}}\,dx.$$ We need $$18I.$$ --- 4. **Evaluate the integral** Use substitution: $$t=1+4x^4 \implies dt=16x^3dx,$$ so $$x^3dx=\frac{dt}{16}.$$ Therefore, $$I=\frac{1}{16}\int t^{-1/4}dt.$$ Now change limits: - When $x=0$, $t=1$. - When $x=\sqrt{2\sqrt5}$, $$x^2=2\sqrt5,\quad x^4=20,$$ so $$t=1+4(20)=81.$$ Thus, $$I=\frac{1}{16}\int_1^{81} t^{-1/4}dt.$$ Integrate: $$\int t^{-1/4}dt=\frac{t^{3/4}}{3/4}=\frac{4}{3}t^{3/4}.$$ Hence, $$I=\frac{1}{16}\cdot\frac{4}{3}\left[t^{3/4}\right]_1^{81} =\frac{1}{12}\left(81^{3/4}-1\right).$$ Now, $$81^{1/4}=3 \quad \Rightarrow \quad 81^{3/4}=27.$$ So, $$I=\frac{1}{12}(27-1)=\frac{26}{12}=\frac{13}{6}.$$ Therefore, $$18I=18\cdot\frac{13}{6}=3\cdot 13=39.$$ --- 5. **Check with options** The value is $$\boxed{39}.$$ So the correct option is **C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.

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View written solutionFree

Correct answer: C

Given functions

We have $f(x)=\frac{x}{(1+x^4)^{1/4}},$ and $g(x)=f(f(f(f(x)))).$

We need to compute $18\int_0^{\sqrt{2\sqrt{5}}} x^2 g(x)\,dx.$

Find a useful pattern for repeated application of $f$

Let $y=f(x)=\frac{x}{(1+x^4)^{1/4}}.$ Then $y^4=\frac{x^4}{1+x^4}.$ So, $1-y^4=1-\frac{x^4}{1+x^4}=\frac{1}{1+x^4}.$ Hence, $(1-y^4)^{1/4}=\frac{1}{(1+x^4)^{1/4}}.$ Therefore, $y=x(1-y^4)^{1/4}.$

But an even better way is to track the fourth power. If $T(u)=\frac{u}{(1+u^4)^{1/4}},$ then $T(u)^4=\frac{u^4}{1+u^4}.$

Set up the integral

So,

=\int_0^{\sqrt{2\sqrt{5}}} \frac{x^3}{(1+4x^4)^{1/4}}\,dx.$$ We need $$18I.$$ --- 4. **Evaluate the integral** Use substitution: $$t=1+4x^4 \implies dt=16x^3dx,$$ so $$x^3dx=\frac{dt}{16}.$$ Therefore, $$I=\frac{1}{16}\int t^{-1/4}dt.$$ Now change limits: - When $x=0$, $t=1$. - When $x=\sqrt{2\sqrt5}$, $$x^2=2\sqrt5,\quad x^4=20,$$ so $$t=1+4(20)=81.$$ Thus, $$I=\frac{1}{16}\int_1^{81} t^{-1/4}dt.$$ Integrate: $$\int t^{-1/4}dt=\frac{t^{3/4}}{3/4}=\frac{4}{3}t^{3/4}.$$ Hence, $$I=\frac{1}{16}\cdot\frac{4}{3}\left[t^{3/4}\right]_1^{81} =\frac{1}{12}\left(81^{3/4}-1\right).$$ Now, $$81^{1/4}=3 \quad \Rightarrow \quad 81^{3/4}=27.$$ So, $$I=\frac{1}{12}(27-1)=\frac{26}{12}=\frac{13}{6}.$$ Therefore, $$18I=18\cdot\frac{13}{6}=3\cdot 13=39.$$ --- 5. **Check with options** The value is $$\boxed{39}.$$ So the correct option is **C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.

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