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Limits: Variable Line Always Passes Through Fixed Point Parameter

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

Question
If variable line x(3+λ)+2y(2−λ)−(7−λ)=0x(3+\lambda)+2y(2-\lambda)-(7-\lambda)=0 always passes through a fixed point (a,b)(a,b) where λ\lambda is a parameter and λ=lim⁡x→(a−b)−[sin⁡x−2]+{cos⁡x}x−[x]−1\lambda=\displaystyle\lim_{x\to(a-b)^{-}}\dfrac{[\sin x-2]+\{\cos x\}}{x-[x]-1} where [y][y] and {y}\{y\} denotes greatest integer ≤\le and fractional part of yy respectively, then
Aa+2b=3a+2b=3correct
Ba−b+2l=2a-b+2l=2correct
Cl=1l=1
Dll does not exist
Solution
Step 1: Find the fixed point. Rewrite the line equation:
3x+4y−7+λ(x−2y+1)=03x+4y-7+\lambda(x-2y+1)=0
 For this to hold for all λ\lambda:
3x+4y−7=0andx−2y+1=03x+4y-7=0\quad\text{and}\quad x-2y+1=0
 Step 2: Solve the system. From the second equation: x=2y−1x=2y-1. Substitute:
3(2y−1)+4y−7=0  ⇒  6y−3+4y−7=0  ⇒  10y=10  ⇒  y=13(2y-1)+4y-7=0\;\Rightarrow\;6y-3+4y-7=0\;\Rightarrow\;10y=10\;\Rightarrow\;y=1
 So x=2(1)−1=1x=2(1)-1=1. Fixed point is (a,b)=(1,1)(a,b)=(1,1). Step 3: Compute a−b=0a-b=0. So we need
λ=lim⁡x→0−[sin⁡x−2]+{cos⁡x}x−[x]−1\lambda=\lim_{x\to 0^{-}}\dfrac{[\sin x-2]+\{\cos x\}}{x-[x]-1}
 Step 4: Evaluate as x→0−x\to 0^{-}: - sin⁡x→0−\sin x\to 0^{-}, so sin⁡x−2→−2−\sin x-2\to -2^{-}, hence [sin⁡x−2]=−3[\sin x-2]=-3. - cos⁡x→1−\cos x\to 1^{-}, so {cos⁡x}→1−\{\cos x\}\to 1^{-} (close to but less than 11). - x→0−x\to 0^{-}, so [x]=−1[x]=-1, hence x−[x]−1=x−(−1)−1=x→0−x-[x]-1=x-(-1)-1=x\to 0^{-}. Step 5: Substitute:
λ=lim⁡x→0−−3+{cos⁡x}x\lambda=\lim_{x\to 0^{-}}\dfrac{-3+\{\cos x\}}{x}
 As x→0−x\to 0^{-}: numerator →−3+1=−2\to -3+1=-2, denominator →0−\to 0^{-}. So λ→−20−=+∞\lambda\to\dfrac{-2}{0^{-}}=+\infty. Step 6: This means λ\lambda does not exist (as a finite limit). So option (1) is correct (calling the limit ℓ\ell here). Step 7: Check (1): a+2b=1+2=3a+2b=1+2=3. True. Option (1) is correct. Correct answers: (1) and (4)
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