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Limits Continuity and Differentiability question

2019 · 12 Apr · Shift 2 · Q31
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Limits Continuity and Differentiability question

2019 · 12 Apr · Shift 2 · Q31

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x ∈\in∈ R. If f(x) attains maximum value at α\alphaα and g(x) attains minimum value at β\betaβ, then lim⁡x→−αβ(x−1)(x2−5x+6)x2−6x+8\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}x→−αβlim​x2−6x+8(x−1)(x2−5x+6)​ is equal to :
  1. A
    12{1 \over 2}21​
  2. B
    −12-{1 \over 2}−21​
  3. C
    32{3 \over 2}23​
  4. D
    −32-{3 \over 2}−23​
View written solutionFree

Correct answer: A

  1. Find α\alphaα from f(x)=5−∣x−2∣f(x)=5-|x-2|f(x)=5−∣x−2∣

Since ∣x−2∣≥0|x-2|\ge 0∣x−2∣≥0 for all xxx, the expression 5−∣x−2∣5-|x-2|5−∣x−2∣ is maximum when ∣x−2∣|x-2|∣x−2∣ is minimum.

The minimum value of ∣x−2∣|x-2|∣x−2∣ is 000, attained at x=2.x=2.x=2. Hence, α=2.\alpha=2.α=2.

  1. Find β\betaβ from g(x)=∣x+1∣g(x)=|x+1|g(x)=∣x+1∣

Since ∣x+1∣≥0|x+1|\ge 0∣x+1∣≥0 for all xxx, the minimum value occurs when x+1=0  ⟹  x=−1.x+1=0 \implies x=-1.x+1=0⟹x=−1. Thus, β=−1.\beta=-1.β=−1.

  1. Compute −αβ-\alpha\beta−αβ

We have αβ=2⋅(−1)=−2.\alpha\beta = 2\cdot (-1)=-2.αβ=2⋅(−1)=−2. Therefore, −αβ=−(−2)=2.-\alpha\beta = -(-2)=2.−αβ=−(−2)=2.

So the limit becomes lim⁡x→2(x−1)(x2−5x+6)x2−6x+8.\lim_{x\to 2} \frac{(x-1)(x^2-5x+6)}{x^2-6x+8}.limx→2​x2−6x+8(x−1)(x2−5x+6)​.

  1. Factor numerator and denominator

Factorize: x2−5x+6=(x−2)(x−3),x^2-5x+6=(x-2)(x-3),x2−5x+6=(x−2)(x−3), x2−6x+8=(x−2)(x−4).x^2-6x+8=(x-2)(x-4).x2−6x+8=(x−2)(x−4).

Hence,

=\frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}.$$ For $x\ne 2$, cancel $(x-2)$: $$=\frac{(x-1)(x-3)}{x-4}.$$ 5. **Evaluate the limit** Now substitute $x=2$: $$\lim_{x\to 2} \frac{(x-1)(x-3)}{x-4} =\frac{(2-1)(2-3)}{2-4} =\frac{1\cdot (-1)}{-2} =\frac{1}{2}.$$ 6. **Match with options** Thus the value of the limit is $$\boxed{\frac{1}{2}}.$$ So the correct option is **A**.
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