Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Definite Integration question

2021 · 18 Mar · Shift 1 · Q41
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Definite Integration
  5. /2021 · 18 Mar · Shift 1 · Q41

Definite Integration question

2021 · 18 Mar · Shift 1 · Q41

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
Let f(x) and g(x) be two functions satisfying f(x2) + g(4 −-− x) = 4x3 and g(4 −-− x) + g(x) = 0, then the value of ∫−44f(x)2dx\int\limits_{ - 4}^4 {f{{(x)}^2}dx}−4∫4​f(x)2dx is
Numerical answer
View written solutionFree

Correct answer: 512

  1. We are given f(x2)+g(4−x)=4x3f(x^2)+g(4-x)=4x^3f(x2)+g(4−x)=4x3 and g(4−x)+g(x)=0.g(4-x)+g(x)=0.g(4−x)+g(x)=0.

We need to find I=∫−44f(x2) dx.I=\int_{-4}^{4} f(x^2)\,dx.I=∫−44​f(x2)dx.

  1. From the second relation, g(4−x)=−g(x).g(4-x)=-g(x).g(4−x)=−g(x). Substitute this into the first relation: f(x2)−g(x)=4x3.f(x^2)-g(x)=4x^3.f(x2)−g(x)=4x3. So, f(x2)=4x3+g(x).f(x^2)=4x^3+g(x).f(x2)=4x3+g(x).

  2. Therefore,

=\int_{-4}^{4}4x^3\,dx+\int_{-4}^{4}g(x)\,dx.$$ 4. Now, $$\int_{-4}^{4}4x^3\,dx=0$$ because $4x^3$ is an odd function over the symmetric interval $[-4,4]$. So, $$I=\int_{-4}^{4}g(x)\,dx.$$ 5. Use the relation $$g(4-x)=-g(x).$$ Integrate both sides from $x=-4$ to $x=4$: $$\int_{-4}^{4} g(4-x)\,dx=-\int_{-4}^{4}g(x)\,dx.$$ Let $$u=4-x \Rightarrow dx=-du.$$ When $x=-4$, $u=8$; when $x=4$, $u=0$. Thus, $$\int_{-4}^{4} g(4-x)\,dx=\int_{0}^{8} g(u)\,du.$$ Hence, $$\int_{0}^{8} g(u)\,du=-\int_{-4}^{4}g(x)\,dx.$$ This alone does not directly determine $\int_{-4}^{4}g(x)\,dx$. 6. Instead, use the first equation more cleverly. Replace $x$ by $-x$: $$f(x^2)+g(4+x)=-4x^3.$$ Now add this to the original equation $$f(x^2)+g(4-x)=4x^3.$$ We get $$2f(x^2)+g(4-x)+g(4+x)=0.$$ From the second relation, for input $-x$: $$g(4-(-x))+g(-x)=0 \Rightarrow g(4+x)=-g(-x).$$ Also, $$g(4-x)=-g(x).$$ So the above becomes $$2f(x^2)-g(x)-g(-x)=0,$$ or $$f(x^2)=\frac{g(x)+g(-x)}{2}.$$ 7. Now integrate from $-4$ to $4$: $$I=\int_{-4}^{4} f(x^2)\,dx =\frac12\int_{-4}^{4}\big(g(x)+g(-x)\big)\,dx.$$ But $$\int_{-4}^{4}g(-x)\,dx=\int_{-4}^{4}g(x)\,dx$$ (by substituting $t=-x$ over symmetric limits). Therefore, $$I=\frac12\left(\int_{-4}^{4}g(x)\,dx+\int_{-4}^{4}g(x)\,dx\right) =\int_{-4}^{4}g(x)\,dx.$$ This is consistent but still not enough. 8. A better direct method: from $$f(x^2)+g(4-x)=4x^3,$$ put $x=\sqrt{t}$ and $x=-\sqrt{t}$ for $t\ge 0$: $$f(t)+g(4-\sqrt{t})=4t\sqrt{t},$$ $$f(t)+g(4+\sqrt{t})=-4t\sqrt{t}.$$ Add them: $$2f(t)+g(4-\sqrt{t})+g(4+\sqrt{t})=0.$$ Using $$g(4-y)=-g(y),$$ with $y=\sqrt{t}$ and $y=-\sqrt{t}$, $$g(4-\sqrt{t})=-g(\sqrt{t}), \qquad g(4+\sqrt{t})=-g(-\sqrt{t}).$$ Thus, $$2f(t)=g(\sqrt{t})+g(-\sqrt{t}).$$ Again this does not uniquely determine $f$ numerically. 9. Let us derive a direct functional equation for $f$. From $$f(x^2)+g(4-x)=4x^3,$$ and using $g(4-x)=-g(x)$, $$g(x)=f(x^2)-4x^3.$$ Now apply the condition $g(4-x)+g(x)=0$: $$\big(f((4-x)^2)-4(4-x)^3\big)+\big(f(x^2)-4x^3\big)=0.$$ Hence, $$f((4-x)^2)+f(x^2)=4\big((4-x)^3+x^3\big).$$ 10. Integrate this from $x=0$ to $x=4$: $$\int_0^4 f((4-x)^2)\,dx+\int_0^4 f(x^2)\,dx =4\int_0^4\big((4-x)^3+x^3\big) \,dx.$$ In the first integral, let $u=4-x$. Then $$\int_0^4 f((4-x)^2)\,dx=\int_0^4 f(u^2)\,du.$$ So the left side is $$2\int_0^4 f(x^2)\,dx.$$ Therefore, $$2\int_0^4 f(x^2)\,dx=4\left(\int_0^4 (4-x)^3\,dx+\int_0^4 x^3\,dx\right).$$ But $$\int_0^4 (4-x)^3\,dx=\int_0^4 x^3\,dx=\left[\frac{x^4}{4}\right]_0^4=64.$$ Thus, $$2\int_0^4 f(x^2)\,dx=4(64+64)=512,$$ so $$\int_0^4 f(x^2)\,dx=256.$$ 11. Since $f(x^2)$ is an even function of $x$, $$\int_{-4}^{4} f(x^2)\,dx=2\int_0^4 f(x^2)\,dx=2\cdot 256=512.$$ Therefore, the required integer is $$\boxed{512}.$$
PreviousNext

More from Definite Integration

  • Let g(x) = ∫0x​f(t)dt, where f is continuous function in [ 0, 3 ] such that 31​≤ f(t) ≤ 1 for all t ∈[0, 1] and 0 ≤ f(t) ≤21​ for all t ∈ (1, 3]. The largest possible interval in which…2021 · MCQ
  • Let P(x) be a real polynomial of degree 3 which vanishes at x = − 3. Let P(x) have local minima at x = 1, local maxima at x = − 1 and −1∫1​P(x)dx = 18, then the sum of all the coefficients of the polynomial P(x) is…2021 · Numerical
  • Let a be a positive real number such that ∫0a​ex−[x]dx=10e−9 where [ x ] is the greatest integer less than or equal to x. Then a is equal to:2021 · MCQ
  • The value of the integral −1∫1​loge​(1−x​+1+x​)dx is equal to:2021 · MCQ
  • If [x] denotes the greatest integer less than or equal to x, then the value of the integral ∫−π/2π/2​[[x]−sinx]dx is equal to :2021 · MCQ
  • If the real part of the complex number (1−cosθ+2isinθ)−1 is 51​ for θ∈(0,π), then the value of the integral ∫0θ​sinxdx is equal to:2021 · MCQ
  • Let g(t)=∫−π/2π/2​cos(4π​t+f(x))dx, where f(x)=loge​(x+x2+1​),x∈R. Then which one of the following is correct?2021 · MCQ
  • If 0∫100π​e(πx​−[πx​])sin2x​dx=1+4π2απ3​,α∈R where [x] is the greatest integer less than or…2021 · MCQ