Q 11 :    

Let I(x)=dx(x11)1113(x+15)1513. If I(37)I(24)=14(1b1131c113), b, cN, then 3(b + c) is equal to          [2025]

  • 39

     

  • 22

     

  • 40

     

  • 26

     

(1)

Given, I(x)=dx(x11)11/13(x+15)15/13

Let x11x+15=t  26dx(x+15)2=dt

  I(x)=126dt(t)11/13

                  =126×t2/13213+C

                   =14t2/13+C

  I(x)=14(x11x+15)2/13+C

  I(37)I(24)=14[(12)2/13(13)2/13]

14(1b1/131c1/13)=14(1(4)1/13191/13)

   b = 4 and c = 9

   3(b + c) = 3(4 + 9) = 39.



Q 12 :    

Let x3 sin xdx=g(x)+C, where C is the constant of integration. If 8(g(π2)+g'(π2))=απ3+βπ2+γ, α, β, γZ, then α+βγ equals:          [2025]

  • 47

     

  • 62

     

  • 48

     

  • 55

     

(4)

x3 sin xdx=x3 cos x+3x2 cos xdx

                             =x3 cos x +3x2 sin x6x sin xdx

                             =x3 cos x +3x2 sin x+6x cos x6 sin x+C

So, g(x)=x3 cos x +3x2 sin x+6x cos x6 sin x

and g'(x)=3x2 cos x+x3 sin x+3x2 cos x+6x sin x6x sin x+6 cos x6 cos x

=x3 sin x

 g(π2)=3π246 and g'(π2)=π38

  8(g(π2)+g'(π2))=π3+6π248

 α=1, β=6, γ=48

So, α+βγ=55.



Q 13 :    

Let f:(0,)R be a function which is differentiable at all points of its domain and satisfies the condition x2f'(x)=2xf(x)+3, with f(1) = 4. Then 2f(2) is equal to :          [2025]

  • 23

     

  • 19

     

  • 29

     

  • 39

     

(4)

Given, x2f'(x)=2xf(x)+3 and f(1) = 4

 x2f'(x)2xf(x)=3

 x2f'(x)2xf(x)x4=3x4          (Divide both sides by x4)

 ddx(f(x)x2)=3x4

Using integration on both sides,

 f(x)x2=3x4dx

 f(x)x2=3×13x3+C

 f(x)x2=1x3+C

 f(x)=1x+Cx2

Since f(1) = 4

 4=1+C  C=4+1=5

Now, we get f(x) =1x+5x2

 2f(2)=2·(12+5×4)

                     =2×392=39.



Q 14 :    

If f(x)=1x1/4(1+x1/4)dx, f(0) = – 6, then f(1) is equal to :          [2025]

  • loge2+2

     

  • 4(loge22)

     

  • 4(loge2+2)

     

  • 2loge2

     

(2)

We have, f(x)=1x1/4(1+x1/4)dx

Putting x=t4  dx=4t3dt

f(t4)=4t3t(1+t)dt=4t21+tdt

=4(t21)+11+t dt

=4[(t1)dt+1t+1 dt]

=4[t22t+loge (t+1)]+C

  f(x)=2x1/24x1/4+4 loge (x1/4+1)+C

          f(0)=6  C=6

  f(1)=24+4 loge26=4 loge28=4(loge22)



Q 15 :    

If (1+x2+x)10(1+x2x)9dx=1m((1+x2+x)n(n1+x2x))+C where C is the constant of integration and m, n  N, then m + n is equal to __________.          [2025]



(379)

Let I=(1+x2+x)10(1+x2x)9dx=(1+x2+x)191dx           (On rationalising)

Let 1+x2+x=t

 (x1+x2+1)dx=dt

 dx=dtt(1+x2)

          =dtt·(t2+12t)=t2+12t2·dt

So, I=t19(t2+12t2)dt

              =12(t19+t17)dt=12[t2020+t1818]+C

              =t19360[9t+10t]+C=t19360[9(t+1t)+1t]

=(1+x2+x)19360[9(21+x2)+(1+x2x)]+C

=(1+x2+x)19360[191+x2x]+C

Then, m = 360, n = 19

  m + n = 360 + 19 = 379.



Q 16 :    

If (1x+1x3)(3x24+x2623)dx=α3(α+1)(3xβ+xγ)α+1α+C, x > 0, (α,β,γZ), where c is the constant of integration, then α+β+γ is equal to __________.          [2025]



(19)

We have, (1x2+1x4)(3x+1x3)123dx

Put t=3x+1x3  dt=3(1x2+1x4)dx

Now, t1/23dt3=t24/23(2423)(3)+C

On comparing, we get α=23, β=1, γ=3

  α+β+γ=19.



Q 17 :    

If 2x2+5x+9x2+x+1dx=xx2+x+1+αx2+x+1+β loge |x+12+x2+x+1|+C, where C is the constant of integration, then α+2β is equal to __________.          [2025]



(16)

Given integral is 2x2+5x+9x2+x+1dx

Using partial fraction decomposition, we get

2x2+5x+9=A(x2+x+1)+B(2x+1)+C

                              =Ax2+(A+2B)x+A+B+C

Comparing terms, we get A=2, B=32 and C=112

  2x2+5x+9x2+x+1dx=2x2+x+1x2+x+1dx+322x+1x2+x+1dx+1121x2+x+1dx

=2(x+12)2+(32)2dx+32×2x2+x+1+1121(x+12)2+(32)2dx

=2×((x+122)x2+x+1+32×4 ln |x+12+x2+x+1|)+3x2+x+1+112 ln |x+12+x2+x+1|+C

On comparing, we get α=72 and β=254

Hence, α+2β=72+2×254=16.