Continuity & DifferentiabilityhardFree

Continuity & Differentiability: Let Cases Sgn 2mm 2mm Cases Sgn Denote

JEE Maths question with a full step-by-step solution.

Question
Let
f(x)={(15−3b){x}−(b2−4b−5)sgn⁡(x+1),−π2<x<0k([x]+[−x]),0≤x≤π(a+2cos⁡x)(1+tan⁡x)ln⁡(1+π2−2πx+x2),π<x<3π2f(x) = \begin{cases} \left(15-3b\right)\{x\}-\left(b^2-4b-5\right)\operatorname{sgn}(x+1), & -\dfrac{\pi}{2} < x < 0 \\[2mm] k\left(\left[x\right]+\left[-x\right]\right), & 0 \le x \le \pi \\[2mm] \dfrac{\left(a+2\cos x\right)\left(1+\tan x\right)}{\ln\left(1+\pi^2-2\pi x+x^2\right)}, & \pi < x < \dfrac{3\pi}{2} \end{cases}
 where [y][y], {y}\{y\} and sgn⁡(y)\operatorname{sgn}(y) denote greatest integer function, fractional part function and signum function of yy respectively.
List-I
IIf ff is continuous in (−π2,0)\left(-\dfrac{\pi}{2},0\right), then the value of bb is
IIIf ff is continuous at x=πx = \pi, then value of (a+k)(a+k) is
IIIIf ff is continuous in (−π2,π)\left(-\dfrac{\pi}{2},\pi\right), then value of (b+k)(b+k) is
IVIf ff has exactly four points of discontinuity in (−π2,3π2)\left(-\dfrac{\pi}{2},\dfrac{3\pi}{2}\right), then (a+b+k)(a+b+k) is equal to
List-II
P00
Q11
R 55
S66
AI →\to (R); II →\to (Q); III →\to (R); IV →\to (S)correct
BI →\to (P); II →\to (Q); III →\to (R); IV →\to (S)
CI →\to (S); II →\to (P); III →\to (Q); IV →\to (P)
DI →\to (Q); II →\to (P); III →\to (Q); IV →\to (P)
Solution
PART (I) - continuity on (−π2,0)\left(-\tfrac{\pi}{2},0\right) Step 1: For −π2<x<0-\tfrac{\pi}{2} < x < 0, x+1∈(1−π2, 1)x+1 \in \left(1-\tfrac{\pi}{2},\,1\right) and 1−π2≈−0.571-\tfrac{\pi}{2} \approx -0.57, so sgn⁡(x+1)\operatorname{sgn}(x+1) changes sign at x=−1x = -1, which lies in the interval. Also {x}=x−[x]\{x\} = x-[x], so on (−π2,−1)\left(-\tfrac{\pi}{2},-1\right), [x]=−2[x] = -2 and {x}=x+2\{x\} = x+2, while on [−1,0)[-1,0), [x]=−1[x] = -1 and {x}=x+1\{x\} = x+1. Both terms break only at x=−1x = -1, so that is the only suspect point. Step 2:
left limit=(15−3b)(1)−(b2−4b−5)(−1)\text{left limit} = \left(15-3b\right)(1)-\left(b^2-4b-5\right)(-1)
right value=(15−3b)(0)−(b2−4b−5)(+1)\text{right value} = \left(15-3b\right)(0)-\left(b^2-4b-5\right)(+1)
 On equating the two,
15−3b+(b2−4b−5)=−(b2−4b−5)  ⟹  15−3b+2(b2−4b−5)=015-3b+\left(b^2-4b-5\right) = -\left(b^2-4b-5\right) \;\Longrightarrow\; 15-3b+2\left(b^2-4b-5\right) = 0
2b2−11b+5=0  ⟹  (2b−1)(b−5)=0  ⟹  b=5 or b=122b^2-11b+5 = 0 \;\Longrightarrow\; (2b-1)(b-5) = 0 \;\Longrightarrow\; b = 5 \ \text{or}\ b = \tfrac12
 Step 3: At x=−1x = -1 itself {−1}=0\{-1\} = 0 and sgn⁡(0)=0\operatorname{sgn}(0) = 0, so f(−1)=0f(-1) = 0, and continuity at x=−1x = -1 needs both one-sided limits to equal f(−1)f(-1):
15−3b+(b2−4b−5)=0,b2−4b−5=0  ⟹  15−3b=0  ⟹  b=515-3b+\left(b^2-4b-5\right) = 0,\qquad b^2-4b-5 = 0 \;\Longrightarrow\; 15-3b = 0 \;\Longrightarrow\; b = 5
 and b=5b = 5 satisfies both equations. For b=12b = \tfrac12 both limits equal 274≠0\tfrac{27}{4} \ne 0, which is not possible.
(I)→(R) 5\textbf{(I)} \to \textbf{(R)}\ 5
 With b=5b = 5, 15−3b=015-3b = 0 and b2−4b−5=0b^2-4b-5 = 0, so f≡0f \equiv 0 on (−π2,0)\left(-\tfrac{\pi}{2},0\right). PART (II) - continuity at x=πx = \pi Step 1: [x]+[−x][x]+[-x] equals 00 when xx is an integer and −1-1 otherwise, and π\pi is not an integer, so
f(π)=−kf(\pi) = -k
 and the left-hand limit as x→π−x\to\pi^- is also −k-k. Step 2: Put h=x−π→0+h = x-\pi \to 0^+.
cos⁡x=cos⁡(π+h)=−cos⁡h,tan⁡x=tan⁡(π+h)=tan⁡h\cos x = \cos(\pi+h) = -\cos h ,\qquad \tan x = \tan(\pi+h) = \tan h
1+π2−2πx+x2=1+(x−π)2=1+h21+\pi^2-2\pi x+x^2 = 1+\left(x-\pi\right)^2 = 1+h^2
 so the branch becomes
(a−2cos⁡h)(1+tan⁡h)ln⁡(1+h2)\frac{\left(a-2\cos h\right)\left(1+\tan h\right)}{\ln\left(1+h^2\right)}
 Step 3: The denominator ln⁡(1+h2)→0\ln\left(1+h^2\right) \to 0, so a finite limit needs the numerator to tend to 00:
(a−2)(1)=0  ⟹  a=2\left(a-2\right)(1) = 0 \;\Longrightarrow\; a = 2
 With a=2a = 2,
2(1−cos⁡h)(1+tan⁡h)≈2⋅h22⋅(1+h)≈h22\left(1-\cos h\right)\left(1+\tan h\right) \approx 2\cdot\frac{h^2}{2}\cdot(1+h) \approx h^2
ln⁡(1+h2)≈h2\ln\left(1+h^2\right) \approx h^2
 so the right-hand limit is 11. Step 4:
−k=1  ⟹  k=−1,a+k=2+(−1)=1-k = 1 \;\Longrightarrow\; k = -1 ,\qquad a+k = 2+(-1) = 1
(II)→(Q) 1\textbf{(II)} \to \textbf{(Q)}\ 1
 PART (III) - continuity on (−π2,π)\left(-\tfrac{\pi}{2},\pi\right) Step 1: From Part (I), b=5b = 5 and f≡0f \equiv 0 on (−π2,0)\left(-\tfrac{\pi}{2},0\right). At x=0x = 0 the left limit is 00 and f(0)=k([0]+[0])=0f(0) = k\left([0]+[0]\right) = 0, so the left side matches. Step 2: On (0,π)(0,\pi), f(x)=−kf(x) = -k at the non-integers and f=0f = 0 at x=1,2,3x = 1,2,3, so continuity gives
−k=0  ⟹  k=0-k = 0 \;\Longrightarrow\; k = 0
 and then the right-hand limit at x=0x = 0 is 0=f(0)0 = f(0) as well.
b+k=5+0=5b+k = 5+0 = 5
(III)→(R) 5\textbf{(III)} \to \textbf{(R)}\ 5
 PART (IV) - exactly four points of discontinuity on (−π2,3π2)\left(-\tfrac{\pi}{2},\tfrac{3\pi}{2}\right) Step 1: The only candidates are
x=−1 (the {x}, sgn⁡ jump),x=0, 1, 2, 3,x=πx = -1 \ \text{(the } \{x\},\ \operatorname{sgn} \text{ jump)},\quad x = 0,\ 1,\ 2,\ 3,\quad x = \pi
 Step 2: If k=0k = 0, the middle branch is identically 00, so x=1,2,3x = 1,2,3 are points of continuity and only x=−1, 0, πx = -1,\ 0,\ \pi can break: at most three, never four. So k≠0k \ne 0. Step 3: With k≠0k \ne 0, at each of x=1,2,3x = 1,2,3 the value 00 differs from the limit −k-k, and at x=0x = 0 the right-hand limit −k-k differs from f(0)=0f(0) = 0. These are already four, so the remaining two candidates must be continuous: x=−1x = -1 gives b=5b = 5 by Part (I), and x=πx = \pi gives a=2a = 2, k=−1k = -1 by Part (II), with k=−1≠0k = -1 \ne 0 as assumed. Step 4:
a+b+k=2+5+(−1)=6a+b+k = 2+5+(-1) = 6
(IV)→(S) 6\textbf{(IV)} \to \textbf{(S)}\ 6
 MATCHING UP
I→R,II→Q,III→R,IV→S\text{I}\to\text{R},\qquad \text{II}\to\text{Q},\qquad \text{III}\to\text{R},\qquad \text{IV}\to\text{S}
 which is code (1). Answer: (1)
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