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

2025 · 4 Apr · Shift 1 · Q42
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. /2025 · 4 Apr · Shift 1 · Q42

Limits Continuity and Differentiability question

2025 · 4 Apr · Shift 1 · Q42

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If lim⁡x→1+(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\lim_{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1limx→1+​(x−1)3(x−1)(6+λcos(x−1))+μsin(1−x)​=−1, where λ,μ∈R\lambda, \mu \in \mathbb{R}λ,μ∈R, then λ+μ\lambda+\muλ+μ is equal to
  1. A
    20
  2. B
    19
  3. C
    18
  4. D
    17
View written solutionFree

Correct answer: C

  1. Let h=x−1,h=x-1,h=x−1, so that as x→1+x\to 1^+x→1+, we have h→0+h\to 0^+h→0+.

Then the given limit becomes lim⁡h→0+h(6+λcos⁡h)+μsin⁡(−h)h3=−1.\lim_{h\to 0^+} \frac{h(6+\lambda \cos h)+\mu \sin(-h)}{h^3}=-1.limh→0+​h3h(6+λcosh)+μsin(−h)​=−1. Since sin⁡(−h)=−sin⁡h\sin(-h)=-\sin hsin(−h)=−sinh, this is lim⁡h→0+6h+λhcos⁡h−μsin⁡hh3=−1.\lim_{h\to 0^+} \frac{6h+\lambda h\cos h-\mu \sin h}{h^3}=-1.limh→0+​h36h+λhcosh−μsinh​=−1.

  1. Use Taylor expansions near h=0h=0h=0: cos⁡h=1−h22+O(h4),\cos h = 1-\frac{h^2}{2}+O(h^4),cosh=1−2h2​+O(h4), sin⁡h=h−h36+O(h5).\sin h = h-\frac{h^3}{6}+O(h^5).sinh=h−6h3​+O(h5).

So,

=\lambda h-\frac{\lambda}{2}h^3+O(h^5),$$ and $$-\mu \sin h=-\mu\left(h-\frac{h^3}{6}+O(h^5)\right) =-\mu h+\frac{\mu}{6}h^3+O(h^5).$$ 3. Substitute into the numerator: $$6h+\lambda h\cos h-\mu \sin h$$ $$=6h+\left(\lambda h-\frac{\lambda}{2}h^3\right)+\left(-\mu h+\frac{\mu}{6}h^3\right)+O(h^5).$$ Thus, $$=(6+\lambda-\mu)h+\left(-\frac{\lambda}{2}+\frac{\mu}{6}\right)h^3+O(h^5).$$ 4. For the limit after division by $h^3$ to be finite, the coefficient of $h$ must vanish: $$6+\lambda-\mu=0$$ $$\mu=\lambda+6.$$ Then the limit equals the coefficient of $h^3$: $$-\frac{\lambda}{2}+\frac{\mu}{6}=-1.$$ Substitute $\mu=\lambda+6$: $$-\frac{\lambda}{2}+\frac{\lambda+6}{6}=-1.$$ Multiply by $6$: $$-3\lambda+\lambda+6=-6,$$ $$-2\lambda=-12,$$ $$\lambda=6.$$ Then $$\mu=\lambda+6=12.$$ 5. Therefore, $$\lambda+\mu=6+12=18.$$ So the correct option is **C**.
PreviousNext

More from Limits Continuity and Differentiability

  • Let m and n be the number of points at which the function f(x)=max{x,x3,x5,…x21},x∈R, is not differentiable and not continuous, respectively. Then m+n is equal to…2025 · Numerical
  • Let f be a differentiable function on R such that f(2)=1,f′(2)=4. Let x→0lim​(f(2+x))3/x=eα. Then the number of times the curve y=4x3−4x2−4(α−7)x−α meets…2025 · MCQ
  • limx→0+​(tan−13x​)2(e5(x)34​−1)tan(5(x)31​)loge​(1+3x2)​ is equal to2025 · MCQ
  • The number of points of discontinuity of the function f(x)=[2x2​]−[x​],x∈[0,4], where [⋅] denotes the greatest integer function, is ​.2025 · Numerical
  • If the function f(x)=tanx−sinxtan(tanx)−sin(sinx)​ is continuous at x=0, then f(0) is equal to ​.2025 · Numerical
  • For t>−1, let αt​ and βt​ be the roots of the equation ((t+2)1/7−1)x2+((t+2)1/6−1)x+((t+2)1/21−1)=0. If t→−1+lim​αt​=a and t→−1+lim​βt​=b, …2025 · Numerical
  • Given below are two statements: Statement I: x→0lim​(x5tan−1x+loge​1−x1+x​​−2x​)=52​ Statement II: x→1lim​(x1−x2​)=e21​…2025 · MCQ
  • If ∑r=1n​Tr​=64(2n−1)(2n+1)(2n+3)(2n+5)​, then limn→∞​∑r=1n​(Tr​1​) is equal to :2025 · MCQ