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

Limits Continuity and Differentiability question

2020 · 2 Sep · Shift 1 · Q28
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. /Limits Continuity and Differentiability
  5. /2020 · 2 Sep · Shift 1 · Q28

Limits Continuity and Differentiability question

2020 · 2 Sep · Shift 1 · Q28

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1
If lim⁡x→1x+x2+x3+...+xn−nx−1\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}x→1lim​x−1x+x2+x3+...+xn−n​= 820, (n ∈\in∈ N) then the value of n is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 40

  1. Let S(x)=x+x2+x3+⋯+xn.S(x)=x+x^2+x^3+\cdots+x^n.S(x)=x+x2+x3+⋯+xn. Then the given limit is lim⁡x→1S(x)−nx−1=820.\lim_{x\to 1}\frac{S(x)-n}{x-1}=820.limx→1​x−1S(x)−n​=820.

  2. Observe that S(1)=1+1+⋯+1=n,S(1)=1+1+\cdots+1=n,S(1)=1+1+⋯+1=n, so the numerator becomes 000 at x=1x=1x=1. Hence the limit is of the form lim⁡x→1S(x)−S(1)x−1=S′(1).\lim_{x\to 1}\frac{S(x)-S(1)}{x-1}=S'(1).limx→1​x−1S(x)−S(1)​=S′(1).

  3. Differentiate S(x)S(x)S(x): S′(x)=1+2x+3x2+⋯+nxn−1.S'(x)=1+2x+3x^2+\cdots+n x^{n-1}.S′(x)=1+2x+3x2+⋯+nxn−1. Therefore, S′(1)=1+2+3+⋯+n=n(n+1)2.S'(1)=1+2+3+\cdots+n=\frac{n(n+1)}{2}.S′(1)=1+2+3+⋯+n=2n(n+1)​.

  4. Given that the limit equals 820820820, we get n(n+1)2=820.\frac{n(n+1)}{2}=820.2n(n+1)​=820. So, n(n+1)=1640.n(n+1)=1640.n(n+1)=1640.

  5. Solve the quadratic: n2+n−1640=0.n^2+n-1640=0.n2+n−1640=0. Factorizing, n2+n−1640=(n−40)(n+41)=0.n^2+n-1640=(n-40)(n+41)=0.n2+n−1640=(n−40)(n+41)=0. Thus, n=40orn=−41.n=40 \quad \text{or} \quad n=-41.n=40orn=−41. Since n∈Nn\in\mathbb{N}n∈N, we take n=40.n=40.n=40.

  6. Final answer: 40\boxed{40}40​

PreviousNext

More from Limits Continuity and Differentiability

  • x→0lim​(tan(4π​+x))x1​ is equal to :2020 · MCQ
  • Let [t] denote the greatest integer ≤ t. If for some λ∈ R - {1, 0}, x→0lim​​λ−x+[x]1−x+∣x∣​​ = L, then L is equal to :2020 · MCQ
  • If x→0lim​{x81​(1−cos2x2​−cos4x2​+cos2x2​cos4x2​)} = 2-k then the value of k is ​…2020 · Numerical
  • x→alim​(3a+x)31​−(4x)31​(a+2x)31​−(3x)31​​ (ae 0) is equal to :2020 · MCQ
  • Suppose a differentiable function f(x) satisfies the identity f(x+y) = f(x) + f(y) + xy2 + x2y, for all real x and y. x→0lim​xf(x)​=1, then f'(3) is equal to ​.2020 · Numerical
  • Let f:(0,∞)→(0,∞) be a differentiable function such that f(1) = e and t→xlim​t−xt2f2(x)−x2f2(t)​=0. If f(x) = 1, then x is equal…2020 · MCQ
  • The function f(x)={4π​+tan−1x,21​(∣x∣−1),​∣x∣≤1∣x∣>1​ is :2020 · MCQ
  • If α is positive root of the equation, p(x) = x2 - x - 2 = 0, then x→α+lim​x+α−41−cos(p(x))​​ is equal to :2020 · MCQ