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Limits — JEE Maths practice question

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

Question
If lim⁡x→∞4x(π4−tan⁡−1x+1x+2)=y2+4y+5\displaystyle\lim_{x\to\infty}4x\left(\dfrac{\pi}{4}-\tan^{-1}\dfrac{x+1}{x+2}\right)=y^{2}+4y+5, then yy can be equal to
A11
B−1-1
C−4-4correct
D−3-3correct
Solution
Step 1: Use the difference of inverse tangents. Note π4=tan⁡−11\dfrac{\pi}{4}=\tan^{-1}1:
π4−tan⁡−1x+1x+2=tan⁡−11−tan⁡−1x+1x+2=tan⁡−11−x+1x+21+x+1x+2\dfrac{\pi}{4}-\tan^{-1}\dfrac{x+1}{x+2}=\tan^{-1}1-\tan^{-1}\dfrac{x+1}{x+2}=\tan^{-1}\dfrac{1-\dfrac{x+1}{x+2}}{1+\dfrac{x+1}{x+2}}
 Step 2: Simplify:
=tan⁡−11x+22x+3x+2=tan⁡−112x+3=\tan^{-1}\dfrac{\dfrac{1}{x+2}}{\dfrac{2x+3}{x+2}}=\tan^{-1}\dfrac{1}{2x+3}
 Step 3: As x→∞x\to\infty, 12x+3→0\dfrac{1}{2x+3}\to 0, so tan⁡−112x+3∼12x+3\tan^{-1}\dfrac{1}{2x+3}\sim\dfrac{1}{2x+3}. Step 4: Multiply by 4x4x and take the limit:
lim⁡x→∞4x2x+3=2\lim_{x\to\infty}\dfrac{4x}{2x+3}=2
 Step 5: Set equal to y2+4y+5y^{2}+4y+5:
y2+4y+5=2  ⇒  y2+4y+3=0  ⇒  (y+1)(y+3)=0y^{2}+4y+5=2\;\Rightarrow\;y^{2}+4y+3=0\;\Rightarrow\;(y+1)(y+3)=0
 So y=−1y=-1 or y=−3y=-3. Correct answers: (2) −1-1 and (4) −3-3
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