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

2016 · 9 Apr · Shift 1 · Q46
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. /2016 · 9 Apr · Shift 1 · Q46

Limits Continuity and Differentiability question

2016 · 9 Apr · Shift 1 · Q46

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If lim⁡x→∞(1+ax−4x2)2x=e3,\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3},x→∞lim​(1+xa​−x24​)2x=e3, then 'a' is equal to :
  1. A
    2
  2. B
    32{3 \over 2}23​
  3. C
    23{2 \over 3}32​
  4. D
    12{1 \over 2}21​
View written solutionFree

Correct answer: B

  1. We need to evaluate
lim⁡x→∞(1+ax−4x2)2x=e3.\lim_{x\to\infty}\left(1+\frac{a}{x}-\frac{4}{x^2}\right)^{2x}=e^3.x→∞lim​(1+xa​−x24​)2x=e3.
  1. Use the standard exponential limit: If
lim⁡x→∞ux=0,\lim_{x\to\infty}u_x=0,x→∞lim​ux​=0,

then

lim⁡x→∞(1+ux)vx=elim⁡vxux\lim_{x\to\infty}(1+u_x)^{v_x}=e^{\lim v_x u_x}x→∞lim​(1+ux​)vx​=elimvx​ux​

provided the limit exists suitably.

Here,

ux=ax−4x2,vx=2x.u_x=\frac{a}{x}-\frac{4}{x^2}, \qquad v_x=2x.ux​=xa​−x24​,vx​=2x.

Since ux→0u_x\to 0ux​→0 as x→∞x\to\inftyx→∞, we compute

lim⁡x→∞(2x)(ax−4x2)=lim⁡x→∞(2a−8x)=2a.\lim_{x\to\infty} (2x)\left(\frac{a}{x}-\frac{4}{x^2}\right) =\lim_{x\to\infty}\left(2a-\frac{8}{x}\right)=2a.x→∞lim​(2x)(xa​−x24​)=x→∞lim​(2a−x8​)=2a.

Therefore,

lim⁡x→∞(1+ax−4x2)2x=e2a.\lim_{x\to\infty}\left(1+\frac{a}{x}-\frac{4}{x^2}\right)^{2x}=e^{2a}.x→∞lim​(1+xa​−x24​)2x=e2a.
  1. Given that this limit equals e3e^3e3, we get
e2a=e3.e^{2a}=e^3.e2a=e3.

So,

2a=3⇒a=32.2a=3 \quad \Rightarrow \quad a=\frac{3}{2}.2a=3⇒a=23​.
  1. Check options:
  • A: 222  incorrect
  • B: 32\frac{3}{2}23​  correct
  • C: 23\frac{2}{3}32​  incorrect
  • D: 12\frac{1}{2}21​  incorrect

Hence, the correct answer is

32.\boxed{\frac{3}{2}}.23​​.
PreviousNext

More from Limits Continuity and Differentiability

  • Let a, b ∈ R, (a e 0). If the function f defined as f(x)=⎩⎨⎧​a2x2​,a,x32b2−4b​,​0≤x<11≤x<2​2​≤x<∞​…2016 · MCQ
  • x→0lim​2xtanx−xtan2x(1−cos2x)2​ is :2016 · MCQ
  • For x∈R,f(x)=∣log2−sinx∣ and g(x)=f(f(x)), then :2016 · MCQ
  • Let p=x→0+lim​(1+tan2x​)2x1​ then logp is equal to :2016 · MCQ
  • If the function. g(x)={kx+1​,mx+2,​0≤x≤33<x≤5​ is differentiable, then the value of k+m is :2015 · MCQ
  • x→0lim​xtan4x(1−cos2x)(3+cosx)​ is equal to2015 · MCQ
  • x→0lim​x2sin(πcos2x)​ is equal to :2014 · MCQ
  • x→0lim​xtan4x(1−cos2x)(3+cosx)​ is equal to2013 · MCQ