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

2019 · 10 Jan · Shift 2 · Q23
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. /2019 · 10 Jan · Shift 2 · Q23

Definite Integration question

2019 · 10 Jan · Shift 2 · Q23

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If ∫0x \int\limits_0^x \,0∫x​ f(t) dt = x2 + ∫x1 \int\limits_x^1 \,x∫1​ t2f(t) dt then f '(12)\left( {{1 \over 2}} \right)(21​) is -
  1. A
    1825{{18} \over {25}}2518​
  2. B
    625{{6} \over {25}}256​
  3. C
    2425{{24} \over {25}}2524​
  4. D
    45{{4} \over {5}}54​
View written solutionFree

Correct answer: C

  1. Interpret the equation correctly

The given relation is

∫0xf(t) dt=x2+∫x1t2f(t) dt.\int_0^x f(t)\,dt = x^2 + \int_x^1 t^2 f(t)\,dt.∫0x​f(t)dt=x2+∫x1​t2f(t)dt.

We need to find

f′(12).f'\left(\frac12\right).f′(21​).
  1. Differentiate both sides with respect to xxx

Using the Fundamental Theorem of Calculus:

ddx(∫0xf(t) dt)=f(x).\frac{d}{dx}\left(\int_0^x f(t)\,dt\right)=f(x).dxd​(∫0x​f(t)dt)=f(x).

Also,

ddx(x2)=2x.\frac{d}{dx}(x^2)=2x.dxd​(x2)=2x.

For the last term,

ddx(∫x1t2f(t) dt)=−x2f(x)\frac{d}{dx}\left(\int_x^1 t^2 f(t)\,dt\right) = -x^2 f(x)dxd​(∫x1​t2f(t)dt)=−x2f(x)

because the lower limit is variable.

So differentiating the whole equation gives:

f(x)=2x−x2f(x).f(x)=2x-x^2f(x).f(x)=2x−x2f(x).

Hence,

f(x)+x2f(x)=2xf(x)+x^2f(x)=2xf(x)+x2f(x)=2x f(x)(1+x2)=2xf(x)(1+x^2)=2xf(x)(1+x2)=2x f(x)=2x1+x2.f(x)=\frac{2x}{1+x^2}.f(x)=1+x22x​.
  1. Differentiate f(x)f(x)f(x)

Now,

f(x)=2x1+x2.f(x)=\frac{2x}{1+x^2}.f(x)=1+x22x​.

Using the quotient rule,

f′(x)=2(1+x2)−2x(2x)(1+x2)2.f'(x)=\frac{2(1+x^2)-2x(2x)}{(1+x^2)^2}.f′(x)=(1+x2)22(1+x2)−2x(2x)​.

Simplify:

f′(x)=2+2x2−4x2(1+x2)2f'(x)=\frac{2+2x^2-4x^2}{(1+x^2)^2}f′(x)=(1+x2)22+2x2−4x2​ f′(x)=2−2x2(1+x2)2=2(1−x2)(1+x2)2.f'(x)=\frac{2-2x^2}{(1+x^2)^2}=\frac{2(1-x^2)}{(1+x^2)^2}.f′(x)=(1+x2)22−2x2​=(1+x2)22(1−x2)​.
  1. Evaluate at x=12x=\frac12x=21​
f′(12)=2(1−14)(1+14)2.f'\left(\frac12\right)=\frac{2\left(1-\frac14\right)}{\left(1+\frac14\right)^2}.f′(21​)=(1+41​)22(1−41​)​.

Now,

1−14=34,1+14=54.1-\frac14=\frac34, \qquad 1+\frac14=\frac54.1−41​=43​,1+41​=45​.

So,

f′(12)=2⋅34(54)2=322516.f'\left(\frac12\right)=\frac{2\cdot \frac34}{\left(\frac54\right)^2} =\frac{\frac32}{\frac{25}{16}}.f′(21​)=(45​)22⋅43​​=1625​23​​.

Thus,

f′(12)=32⋅1625=2425.f'\left(\frac12\right)=\frac32\cdot \frac{16}{25}=\frac{24}{25}.f′(21​)=23​⋅2516​=2524​.
  1. Match with the options
2425\boxed{\frac{24}{25}}2524​​

which is Option C.

PreviousNext

More from Definite Integration

  • The value of −π/2∫π/2​[x]+[sinx]+4dx​, where [t] denotes the greatest integer less than or equal to t, is2019 · MCQ
  • The value of the integral −2∫2​[πx​]+21​sin2x​dx (where [x] denotes the greatest integer less than or equal to x) is2019 · MCQ
  • The integral π/6∫π/4​sin2x(tan5x+cot5x)dx​ equals :2019 · MCQ
  • If 0∫π/2​cotx+cscxcotx​dx=m(π+n), then m⋅n is equal to2019 · MCQ
  • Let f : R → R be a continuously differentiable function such that f(2) = 6 and f'(2) = 481​. If 6∫f(x)​4t3dt= (x - 2)g(x), then x→2lim​g(x) is…2019 · MCQ
  • A value of α such that α∫α+1​(x+α)(x+α+1)dx​=loge​(89​) is :2019 · MCQ
  • Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = 4, then 0∫a​ f(x) g(x) dx is equal to :2019 · MCQ
  • The integral 1∫e​{(ex​)2x−(xe​)x} loge x dx is equal to :2019 · MCQ