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

Binomial Theorem question

2024 · 31 Jan · Shift 1 · Q47
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. /Binomial Theorem
  5. /2024 · 31 Jan · Shift 1 · Q47

Binomial Theorem question

2024 · 31 Jan · Shift 1 · Q47

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
Let aaa be the sum of all coefficients in the expansion of (1−2x+2x2)2023(3−4x2+2x3)2024\left(1-2 x+2 x^2\right)^{2023}\left(3-4 x^2+2 x^3\right)^{2024}(1−2x+2x2)2023(3−4x2+2x3)2024 and b=lim⁡x→0(∫0xlog⁡(1+t)t2024+1dtx2)b=\lim_{x \rightarrow 0}\left(\frac{\int_0^x \frac{\log (1+t)}{t^{2024}+1} d t}{x^2}\right)b=x→0lim​(x2∫0x​t2024+1log(1+t)​dt​). If the equation cx2+dx+e=0c x^2+d x+e=0cx2+dx+e=0 and 2bx2+ax+4=02 b x^2+a x+4=02bx2+ax+4=0 have a common root, where c,d,e∈Rc, d, e \in \mathbb{R}c,d,e∈R, then d:c:\mathrm{d}: \mathrm{c}:d:c: e equals
  1. A
    2:1:42: 1: 42:1:4
  2. B
    1:1:41: 1: 41:1:4
  3. C
    1:2:41: 2: 41:2:4
  4. D
    4:1:44: 1: 44:1:4
View written solutionFree

Correct answer: B

  1. Find aaa = sum of all coefficients

For a polynomial P(x)P(x)P(x), the sum of all coefficients is P(1)P(1)P(1).

So,

a=(1−2(1)+2(1)2)2023(3−4(1)2+2(1)3)2024.a=\left(1-2(1)+2(1)^2\right)^{2023}\left(3-4(1)^2+2(1)^3\right)^{2024}.a=(1−2(1)+2(1)2)2023(3−4(1)2+2(1)3)2024.

Now simplify each factor:

1−2+2=1,1-2+2=1,1−2+2=1, 3−4+2=1.3-4+2=1.3−4+2=1.

Hence,

a=12023⋅12024=1.a=1^{2023}\cdot 1^{2024}=1.a=12023⋅12024=1.

So,

a=1.a=1.a=1.
  1. Find bbb

Given

b=lim⁡x→0(∫0xlog⁡(1+t)t2024+1 dtx2).b=\lim_{x\to 0}\left(\frac{\int_0^x \frac{\log(1+t)}{t^{2024}+1}\,dt}{x^2}\right).b=x→0lim​(x2∫0x​t2024+1log(1+t)​dt​).

We expand the integrand near t=0t=0t=0.

Since

log⁡(1+t)=t−t22+O(t3),\log(1+t)=t-\frac{t^2}{2}+O(t^3),log(1+t)=t−2t2​+O(t3),

and

11+t2024=1−t2024+O(t4048),\frac{1}{1+t^{2024}}=1-t^{2024}+O(t^{4048}),1+t20241​=1−t2024+O(t4048),

we get

log⁡(1+t)1+t2024=(t−t22+O(t3))(1−t2024+⋯ )=t−t22+O(t3).\frac{\log(1+t)}{1+t^{2024}}=\left(t-\frac{t^2}{2}+O(t^3)\right)(1-t^{2024}+\cdots)=t-\frac{t^2}{2}+O(t^3).1+t2024log(1+t)​=(t−2t2​+O(t3))(1−t2024+⋯)=t−2t2​+O(t3).

Therefore,

∫0xlog⁡(1+t)1+t2024dt=∫0x(t−t22+O(t3))dt=x22−x36+O(x4).\int_0^x \frac{\log(1+t)}{1+t^{2024}}dt =\int_0^x \left(t-\frac{t^2}{2}+O(t^3)\right)dt =\frac{x^2}{2}-\frac{x^3}{6}+O(x^4).∫0x​1+t2024log(1+t)​dt=∫0x​(t−2t2​+O(t3))dt=2x2​−6x3​+O(x4).

Hence,

b=lim⁡x→0x22−x36+O(x4)x2=12.b=\lim_{x\to 0}\frac{\frac{x^2}{2}-\frac{x^3}{6}+O(x^4)}{x^2}=\frac12.b=x→0lim​x22x2​−6x3​+O(x4)​=21​.

So,

b=12.b=\frac12.b=21​.
  1. Use the common root condition

The second equation is

2bx2+ax+4=0.2bx^2+ax+4=0.2bx2+ax+4=0.

Substitute a=1a=1a=1 and b=12b=\frac12b=21​:

2(12)x2+1⋅x+4=02\left(\frac12\right)x^2+1\cdot x+4=02(21​)x2+1⋅x+4=0 x2+x+4=0.x^2+x+4=0.x2+x+4=0.

Now the first equation is

cx2+dx+e=0.cx^2+dx+e=0.cx2+dx+e=0.

We are told these two quadratics have a common root.

Since the options ask for a fixed ratio d:c:ed:c:ed:c:e, the only consistent possibility is that the first quadratic is proportional to the second one, i.e.

cx2+dx+e=k(x2+x+4)cx^2+dx+e = k(x^2+x+4)cx2+dx+e=k(x2+x+4)

for some nonzero real constant kkk.

Thus,

c:d:e=1:1:4.c:d:e = 1:1:4.c:d:e=1:1:4.

So,

d:c:e=1:1:4.d:c:e = 1:1:4.d:c:e=1:1:4.
  1. Check options
  • A: 2:1:42:1:42:1:4 ❌
  • B: 1:1:41:1:41:1:4 ✅
  • C: 1:2:41:2:41:2:4 ❌
  • D: 4:1:44:1:44:1:4 ❌

Therefore, the correct option is B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

They agree.

PreviousNext

More from Binomial Theorem

  • In the expansion of (1+x)(1−x2)(1+x3​+x23​+x31​)5,xeq0, the sum of the coefficients of x3 and x−13 is equal to ​.2024 · Numerical
  • Let the coefficient of xr in the expansion of (x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+……….+(x+2)n−1 be αr​. If ∑r=0n​αr​=βn−γn,β,γ∈N, then the…2024 · Numerical
  • The remainder, when 19200+23200 is divided by 49 , is ​.2023 · Numerical
  • Let the sixth term in the binomial expansion of (2log2​(10−3x)​+52(x−2)log2​3​)m in the increasing powers of 2(x−2)log2​3, be 21 . If the…2023 · Numerical
  • If the term without x in the expansion of (x32​+x3α​)22 is 7315 , then ∣α∣ is equal to ​.2023 · Numerical
  • If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (42​+43​1​)n is 6​:1, then the third term from the beginning is :2023 · MCQ
  • If the coefficient of x7 in (ax2+2bx1​)11 and x−7 in (ax−3bx21​)11 are equal, then :2023 · MCQ
  • Among the statements : (S1) : 20232022−19992022 is divisible by 8 (S2) : 13(13)n−12n−13 is divisible by 144 for infinitely many n∈N2023 · MCQ