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

2021 · 20 Jul · Shift 1 · Q34
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. /2021 · 20 Jul · Shift 1 · Q34

Limits Continuity and Differentiability question

2021 · 20 Jul · Shift 1 · Q34

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let a function f : R →\to→ R be defined as f(x)={sin⁡x−exifx≤0a+[−x]if0<x<12x−bifx≥1f(x) = \left\{ {\begin{matrix} {\sin x - {e^x}} & {if} & {x \le 0} \\ {a + [ - x]} & {if} & {0 \lt x \lt 1} \\ {2x - b} & {if} & {x \ge 1} \\ \end{matrix} } \right.f(x)=⎩⎨⎧​sinx−exa+[−x]2x−b​ififif​x≤00<x<1x≥1​ where [ x ] is the greatest integer less than or equal to x. If f is continuous on R, then (a + b) is equal to:
  1. A
    4
  2. B
    3
  3. C
    2
  4. D
    5
View written solutionFree

Correct answer: B

  1. Given piecewise function
f(x)={sin⁡x−ex,x≤0a+[−x],0<x<12x−b,x≥1f(x)= \begin{cases} \sin x-e^x, & x\le 0 \\ a+[-x], & 0<x<1 \\ 2x-b, & x\ge 1 \end{cases}f(x)=⎩⎨⎧​sinx−ex,a+[−x],2x−b,​x≤00<x<1x≥1​

We need fff to be continuous on R\mathbb RR.

Since each piece is continuous in its own interval, we only need to check continuity at the junction points:

  • x=0x=0x=0
  • x=1x=1x=1
  1. Continuity at x=0x=0x=0

For x≤0x\le 0x≤0,

f(0)=sin⁡0−e0=0−1=−1f(0)=\sin 0-e^0=0-1=-1f(0)=sin0−e0=0−1=−1

Now for 0<x<10<x<10<x<1, we have −1<−x<0-1<-x<0−1<−x<0, so

[−x]=−1[-x]=-1[−x]=−1

Hence in (0,1)(0,1)(0,1),

f(x)=a+[−x]=a−1f(x)=a+[-x]=a-1f(x)=a+[−x]=a−1

So the right-hand limit at x=0x=0x=0 is

lim⁡x→0+f(x)=a−1\lim_{x\to 0^+} f(x)=a-1x→0+lim​f(x)=a−1

Continuity at x=0x=0x=0 gives

a−1=−1a-1=-1a−1=−1 a=0a=0a=0
  1. Continuity at x=1x=1x=1

For 0<x<10<x<10<x<1, still [−x]=−1[-x]=-1[−x]=−1, so

f(x)=a−1=−1f(x)=a-1=-1f(x)=a−1=−1

(using a=0a=0a=0)

Thus,

lim⁡x→1−f(x)=−1\lim_{x\to 1^-} f(x)=-1x→1−lim​f(x)=−1

For x≥1x\ge 1x≥1,

f(1)=2(1)−b=2−bf(1)=2(1)-b=2-bf(1)=2(1)−b=2−b

Continuity at x=1x=1x=1 gives

2−b=−12-b=-12−b=−1 b=3b=3b=3
  1. Compute a+ba+ba+b
a+b=0+3=3a+b=0+3=3a+b=0+3=3
  1. Option check

The correct option is:

B: 3\boxed{\text{B: }3}B: 3​

PreviousNext

More from Limits Continuity and Differentiability

  • If the value of x→0lim​(2−cosxcos2x​)(x2x+2​) is equal to ea, then a is equal to ​.2021 · Numerical
  • If f:R→R is given by f(x)=x+1, then the value of n→∞lim​n1​[f(0)+f(n5​)+f(n10​)+......+f(n5(n−1)​)]…2021 · MCQ
  • Let a function g:[0,4]→R be defined as g(x)={0≤t≤xmax​{t3−6t2+9t−3}4−x​0≤x≤33<x≤4​ then the number of…2021 · Numerical
  • If x→0lim​xsin2xαxex−βloge​(1+x)+γx2e−x​=10,α,β,γ∈R, then the value of α+β+γ is ​.2021 · Numerical
  • Let f : R → R be defined as f(x)={(1−cos2x)2x3​loge​((1−xe−x)21+2xe−2x​),α,​xe0x=0​…2021 · MCQ
  • Let f : R → R be a function defined as f(x)={3(1−2∣x∣​)0​ifif​∣x∣≤2∣x∣>2​ Let g : R → R be given…2021 · Numerical
  • If f : R → R is a function defined by f(x)= [x - 1] cos(22x−1​)π, where [.] denotes the greatest integer function, then f is :2021 · MCQ
  • n→∞lim​tan{r=1∑n​tan−1(1+r+r21​)} is equal to ​.2021 · Numerical