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 · Q45
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 · Q45

Limits Continuity and Differentiability question

2016 · 9 Apr · Shift 1 · Q45

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If the function f(x) = {−xx<1a+cos⁡−1(x+b),1≤x≤2\left\{ {\begin{matrix} { - x} & {x \lt 1} \\ {a + {{\cos }^{ - 1}}\left( {x + b} \right),} & {1 \le x \le 2} \\ \end{matrix} } \right.{−xa+cos−1(x+b),​x<11≤x≤2​ is differentiable at x = 1, then ab{a \over b}ba​ is equal to :
  1. A
    π−22{{\pi - 2} \over 2}2π−2​
  2. B
    −π−22{{ - \pi - 2} \over 2}2−π−2​
  3. C
    π+22{{\pi + 2} \over 2}2π+2​
  4. D
    −1−cos⁡−1(2)- 1 - {\cos ^{ - 1}}\left( 2 \right)−1−cos−1(2)
View written solutionFree

Correct answer: C

  1. Given piecewise function
f(x)={−x,x<1a+cos⁡−1(x+b),1≤x≤2 f(x)= \begin{cases} -x, & x<1 \\ a+\cos^{-1}(x+b), & 1\le x\le 2 \end{cases}f(x)={−x,a+cos−1(x+b),​x<11≤x≤2​

We are told that f(x)f(x)f(x) is differentiable at x=1x=1x=1.

For differentiability at x=1x=1x=1, two conditions are necessary:

  1. fff is continuous at x=1x=1x=1
  2. Left-hand derivative === right-hand derivative at x=1x=1x=1
  1. Continuity at x=1x=1x=1

From the left side:

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

From the definition at x=1x=1x=1 (right piece):

f(1)=a+cos⁡−1(1+b)f(1)=a+\cos^{-1}(1+b)f(1)=a+cos−1(1+b)

So continuity gives

a+cos⁡−1(1+b)=−1...(1)a+\cos^{-1}(1+b)=-1 \qquad ...(1)a+cos−1(1+b)=−1...(1)
  1. Equality of derivatives at x=1x=1x=1

Left derivative:

ddx(−x)=−1\frac{d}{dx}(-x)=-1dxd​(−x)=−1

So,

f−′(1)=−1f'_-(1)=-1f−′​(1)=−1

Now for the right side,

f(x)=a+cos⁡−1(x+b)f(x)=a+\cos^{-1}(x+b)f(x)=a+cos−1(x+b)

Using

ddx(cos⁡−1u)=−u′1−u2\frac{d}{dx}\big(\cos^{-1}u\big)=\frac{-u'}{\sqrt{1-u^2}}dxd​(cos−1u)=1−u2​−u′​

with u=x+bu=x+bu=x+b, u′=1u'=1u′=1, we get

f+′(x)=−11−(x+b)2f'_+(x)= -\frac{1}{\sqrt{1-(x+b)^2}}f+′​(x)=−1−(x+b)2​1​

Thus at x=1x=1x=1,

f+′(1)=−11−(1+b)2f'_+(1)= -\frac{1}{\sqrt{1-(1+b)^2}}f+′​(1)=−1−(1+b)2​1​

Differentiability requires

−1=−11−(1+b)2-1=-\frac{1}{\sqrt{1-(1+b)^2}}−1=−1−(1+b)2​1​

Cancelling minus sign,

1=11−(1+b)21=\frac{1}{\sqrt{1-(1+b)^2}}1=1−(1+b)2​1​

So,

1−(1+b)2=1\sqrt{1-(1+b)^2}=11−(1+b)2​=1

Squaring both sides:

1−(1+b)2=11-(1+b)^2=11−(1+b)2=1 (1+b)2=0(1+b)^2=0(1+b)2=0 1+b=01+b=01+b=0 b=−1b=-1b=−1
  1. Find aaa using continuity

Substitute b=−1b=-1b=−1 into equation (1):

a+cos⁡−1(1−1)=−1a+\cos^{-1}(1-1)=-1a+cos−1(1−1)=−1 a+cos⁡−1(0)=−1a+\cos^{-1}(0)=-1a+cos−1(0)=−1

Since

cos⁡−1(0)=π2\cos^{-1}(0)=\frac{\pi}{2}cos−1(0)=2π​

we get

a+π2=−1a+\frac{\pi}{2}=-1a+2π​=−1 a=−1−π2a=-1-\frac{\pi}{2}a=−1−2π​
  1. Compute ab\dfrac{a}{b}ba​

Since b=−1b=-1b=−1,

ab=−1−π2−1=1+π2\frac{a}{b}=\frac{-1-\frac{\pi}{2}}{-1}=1+\frac{\pi}{2}ba​=−1−1−2π​​=1+2π​ ab=π+22\frac{a}{b}=\frac{\pi+2}{2}ba​=2π+2​
  1. Match with options
ab=π+22\frac{a}{b}=\frac{\pi+2}{2}ba​=2π+2​

This is Option C.

  1. Comparison with stored answer

Stored correct answer: C

Derived answer: C

So they agree.

PreviousNext

More from Limits Continuity and Differentiability

  • If x→∞lim​(1+xa​−x24​)2x=e3, then 'a' is equal to :2016 · MCQ
  • 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