LimitshardFree

Limits — JEE Maths practice question

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

Question

If variable line

x(3+\lambda)+2y(2-\lambda)-(7-\lambda)=0

always passes through a fixed point

(a,b)

where

\lambda

is a parameter and

\lambda=\displaystyle\lim_{x\to(a-b)^{-}}\dfrac{[\sin x-2]+\{\cos x\}}{x-[x]-1}

where

[y]

and

\{y\}

denotes greatest integer

\le

and fractional part of

y

respectively, then

a+2b=3

correct

a-b+2l=2

correct

l=1

l

does not exist

Solution

Step 1: Find the fixed point. Rewrite the line equation:

3x+4y-7+\lambda(x-2y+1)=0

For this to hold for all

\lambda

3x+4y-7=0\quad\text{and}\quad x-2y+1=0

Step 2: Solve the system. From the second equation:

x=2y-1

. Substitute:

3(2y-1)+4y-7=0\;\Rightarrow\;6y-3+4y-7=0\;\Rightarrow\;10y=10\;\Rightarrow\;y=1

x=2(1)-1=1

. Fixed point is

(a,b)=(1,1)

. Step 3: Compute

a-b=0

. So we need

\lambda=\lim_{x\to 0^{-}}\dfrac{[\sin x-2]+\{\cos x\}}{x-[x]-1}

Step 4: Evaluate as

x\to 0^{-}

: -

\sin x\to 0^{-}

, so

\sin x-2\to -2^{-}

, hence

[\sin x-2]=-3

. -

\cos x\to 1^{-}

, so

\{\cos x\}\to 1^{-}

(close to but less than

1

). -

x\to 0^{-}

, so

[x]=-1

, hence

x-[x]-1=x-(-1)-1=x\to 0^{-}

. Step 5: Substitute:

\lambda=\lim_{x\to 0^{-}}\dfrac{-3+\{\cos x\}}{x}

x\to 0^{-}

: numerator

\to -3+1=-2

, denominator

\to 0^{-}

. So

\lambda\to\dfrac{-2}{0^{-}}=+\infty

. Step 6: This means

\lambda

does not exist (as a finite limit). So option (1) is correct (calling the limit

\ell

here). Step 7: Check (1):

a+2b=1+2=3

. True. Option (1) is correct. Correct answers: (1) and (4)

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