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 · 26 Feb · Shift 2 · Q36
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 · 26 Feb · Shift 2 · Q36

Limits Continuity and Differentiability question

2021 · 26 Feb · Shift 2 · Q36

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be defined as f(x)={2sin⁡(−πx2)if x<−1∣ax2+x+b∣if −1≤x≤1sin⁡(πx)if x>1f(x) = \left\{ \begin{array}{ll} 2\sin\left(-\dfrac{\pi x}{2}\right) & \text{if } x \lt -1 \\ |ax^2 + x + b| & \text{if } -1 \le x \le 1 \\ \sin(\pi x) & \text{if } x \gt 1 \end{array} \right.f(x)=⎩⎨⎧​2sin(−2πx​)∣ax2+x+b∣sin(πx)​if x<−1if −1≤x≤1if x>1​ If f(x)f(x)f(x) is continuous on R\mathbb{R}R, then a+ba + ba+b equals :
  1. A
    −-− 3
  2. B
    3
  3. C
    −-− 1
  4. D
    1
View written solutionFree

Correct answer: C

  1. Given piecewise function
f(x)={2sin⁡(−πx2),x<−1∣ax2+x+b∣,−1≤x≤1sin⁡(πx),x>1f(x)= \begin{cases} 2\sin\left(-\dfrac{\pi x}{2}\right), & x<-1 \\ |ax^2+x+b|, & -1\le x\le 1 \\ \sin(\pi x), & x>1 \end{cases}f(x)=⎩⎨⎧​2sin(−2πx​),∣ax2+x+b∣,sin(πx),​x<−1−1≤x≤1x>1​

For continuity on R\mathbb{R}R, we only need to check the junction points x=−1x=-1x=−1 and x=1x=1x=1, since each piece is continuous on its own interval.

  1. Continuity at x=−1x=-1x=−1

Left-hand limit:

lim⁡x→−1−2sin⁡(−πx2)=2sin⁡(−π(−1)2)=2sin⁡(π2)=2.\lim_{x\to -1^-} 2\sin\left(-\frac{\pi x}{2}\right) =2\sin\left(-\frac{\pi(-1)}{2}\right) =2\sin\left(\frac{\pi}{2}\right)=2.x→−1−lim​2sin(−2πx​)=2sin(−2π(−1)​)=2sin(2π​)=2.

Value from middle piece at x=−1x=-1x=−1:

f(−1)=∣a(−1)2+(−1)+b∣=∣a−1+b∣=∣a+b−1∣.f(-1)=|a(-1)^2+(-1)+b|=|a-1+b|=|a+b-1|.f(−1)=∣a(−1)2+(−1)+b∣=∣a−1+b∣=∣a+b−1∣.

So continuity at x=−1x=-1x=−1 gives

∣a+b−1∣=2.(1)|a+b-1|=2. \qquad (1)∣a+b−1∣=2.(1)
  1. Continuity at x=1x=1x=1

Right-hand limit:

lim⁡x→1+sin⁡(πx)=sin⁡(π)=0.\lim_{x\to 1^+} \sin(\pi x)=\sin(\pi)=0.x→1+lim​sin(πx)=sin(π)=0.

Value from middle piece at x=1x=1x=1:

f(1)=∣a(1)2+1+b∣=∣a+b+1∣.f(1)=|a(1)^2+1+b|=|a+b+1|.f(1)=∣a(1)2+1+b∣=∣a+b+1∣.

So continuity at x=1x=1x=1 gives

∣a+b+1∣=0.|a+b+1|=0.∣a+b+1∣=0.

Since absolute value is zero only when the inside is zero,

a+b+1=0  ⟹  a+b=−1.a+b+1=0 \implies a+b=-1.a+b+1=0⟹a+b=−1.
  1. Check with equation (1)

If a+b=−1a+b=-1a+b=−1, then

∣a+b−1∣=∣−1−1∣=∣−2∣=2,|a+b-1|=|-1-1|=|-2|=2,∣a+b−1∣=∣−1−1∣=∣−2∣=2,

which satisfies continuity at x=−1x=-1x=−1 as well.

So the required value is

−1.\boxed{-1}.−1​.
  1. Option check
  • A: −3-3−3 ❌
  • B: 333 ❌
  • C: −1-1−1 ✅
  • D: 111 ❌

Therefore, the correct option is C.

PreviousNext

More from Limits Continuity and Differentiability

  • If α, β are the distinct roots of x2 + bx + c = 0, then x→βlim​(x−β)2e2(x2+bx+c)−1−2(x2+bx+c)​ is equal to :2021 · MCQ
  • If x→∞lim​(x2−x+1​−ax)=b, then the ordered pair (a, b) is :2021 · MCQ
  • Let f:(−4π​,4π​)→R be defined as f(x)=⎩⎨⎧​(1+∣sinx∣)∣sinx∣3a​becot4x/cot2x​,,,​−4π​<x<0x=00<x<4π​​…2021 · MCQ
  • Let f : R → R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of x→2lim​x−2x2f(2)−4f(x)​ is equal to :2021 · MCQ
  • Let f:[0,3]→R be defined by f(x)=min{x−[x],1+[x]−x} where [x] is the greatest integer less than or equal to x. Let P denote the set containing all x ∈[0, 3] where f i discontinuous, and Q denote the set containing…2021 · Numerical
  • The value of x→0lim​(81−sinx​−81+sinx​x​) is equal to :2021 · MCQ
  • Let f:[0,∞)→[0,3] be a function defined by f(x)={max{sint:0≤t≤x},2+cosx,​0≤x≤πx>π​ Then which of the following…2021 · MCQ
  • The function f(x)=​x2−2x−3​.e∣9x2−12x+4∣ is not differentiable at exactly :2021 · MCQ