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Limits Continuity and Differentiability question

2019 · 10 Apr · Shift 2 · Q26
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  5. /2019 · 10 Apr · Shift 2 · Q26

Limits Continuity and Differentiability question

2019 · 10 Apr · Shift 2 · Q26

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If lim⁡x→1x2−ax+bx−1=5\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5x→1lim​x−1x2−ax+b​=5, then a + b is equal to :
  1. A
    1
  2. B
    - 4
  3. C
    - 7
  4. D
    5
View written solutionFree

Correct answer: C

  1. We are given

    limx→1x2−ax+bx−1=5.\\lim_{x \to 1} \frac{x^2-ax+b}{x-1}=5.limx→1​x−1x2−ax+b​=5.

  2. For the limit to be finite at x=1x=1x=1, the numerator must also vanish at x=1x=1x=1. So,

    12−a(1)+b=01^2-a(1)+b=012−a(1)+b=0 1−a+b=01-a+b=01−a+b=0 b=a−1.b=a-1.b=a−1.

  3. Substitute b=a−1b=a-1b=a−1 into the numerator:

    x2−ax+b=x2−ax+a−1.x^2-ax+b=x^2-ax+a-1.x2−ax+b=x2−ax+a−1.

    Factor it:

    x2−ax+a−1=(x−1)(x+1−a).x^2-ax+a-1=(x-1)(x+1-a).x2−ax+a−1=(x−1)(x+1−a).

  4. Then the limit becomes

    limx→1(x−1)(x+1−a)x−1=limx→1(x+1−a).\\lim_{x\to 1} \frac{(x-1)(x+1-a)}{x-1} = \\lim_{x\to 1}(x+1-a).limx→1​x−1(x−1)(x+1−a)​=limx→1​(x+1−a).

    Hence,

    1+1−a=51+1-a=51+1−a=5 2−a=52-a=52−a=5 a=−3.a=-3.a=−3.

  5. Now find bbb:

    b=a−1=−3−1=−4.b=a-1=-3-1=-4.b=a−1=−3−1=−4.

  6. Therefore,

    a+b=−3+(−4)=−7.a+b=-3+(-4)=-7.a+b=−3+(−4)=−7.

  7. Option check:

    • A: 111 ❌
    • B: −4-4−4 ❌
    • C: −7-7−7 ✅
    • D: 555 ❌

So the correct answer is Option C.

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