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Limits Continuity and Differentiability question

2022 · 26 Jul · Shift 1 · Q32
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  5. /2022 · 26 Jul · Shift 1 · Q32

Limits Continuity and Differentiability question

2022 · 26 Jul · Shift 1 · Q32

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f(x)={x3−x2+10x−7,x≤1−2x+log⁡2(b2−4),x>1f(x) = \left\{ {\begin{matrix} {{x^3} - {x^2} + 10x - 7,} & {x \le 1} \\ { - 2x + {{\log }_2}({b^2} - 4),} & {x \gt 1} \\ \end{matrix} } \right.f(x)={x3−x2+10x−7,−2x+log2​(b2−4),​x≤1x>1​. Then the set of all values of b, for which f(x) has maximum value at x = 1, is :
  1. A
    (−-− 6, −-− 2)
  2. B
    (2, 6)
  3. C
    [−6,−2)∪(2,6][ - 6, - 2) \cup (2,6][−6,−2)∪(2,6]
  4. D
    [−6,−2)∪(2,6]\left[ {-\sqrt 6 , - 2} \right) \cup \left( {2,\sqrt 6 } \right][−6​,−2)∪(2,6​]
View written solutionFree

Correct answer: C

  1. Given piecewise function
f(x)={x3−x2+10x−7,x≤1−2x+log⁡2(b2−4),x>1 f(x)= \begin{cases} x^3-x^2+10x-7, & x\le 1 \\ -2x+\log_2(b^2-4), & x>1 \end{cases}f(x)={x3−x2+10x−7,−2x+log2​(b2−4),​x≤1x>1​

We need the set of all values of bbb for which f(x)f(x)f(x) has a maximum value at x=1x=1x=1.

  1. Find the value at x=1x=1x=1

Since 1≤11\le 11≤1, we use the first branch:

f(1)=13−12+10(1)−7=1−1+10−7=3f(1)=1^3-1^2+10(1)-7=1-1+10-7=3f(1)=13−12+10(1)−7=1−1+10−7=3

So for x=1x=1x=1 to be a maximum point, we need

  • for all x≤1x\le 1x≤1, f(x)≤3f(x)\le 3f(x)≤3 near and in fact globally on that side,
  • for all x>1x>1x>1, f(x)≤3f(x)\le 3f(x)≤3.
  1. Check the left side: x≤1x\le 1x≤1

Let

g(x)=x3−x2+10x−7g(x)=x^3-x^2+10x-7g(x)=x3−x2+10x−7

Differentiate:

g′(x)=3x2−2x+10g'(x)=3x^2-2x+10g′(x)=3x2−2x+10

Now,

3x2−2x+103x^2-2x+103x2−2x+10

has discriminant

(−2)2−4⋅3⋅10=4−120=−116<0(-2)^2-4\cdot 3\cdot 10=4-120=-116<0(−2)2−4⋅3⋅10=4−120=−116<0

and leading coefficient is positive, so

g′(x)>0for all x.g'(x)>0\quad \text{for all }x.g′(x)>0for all x.

Hence g(x)g(x)g(x) is strictly increasing on (−∞,∞)(-\infty,\infty)(−∞,∞). Therefore, on x≤1x\le 1x≤1, its maximum occurs at x=1x=1x=1. So the left part already satisfies the required condition.

  1. Check the right side: x>1x>1x>1

For x>1x>1x>1,

h(x)=−2x+log⁡2(b2−4)h(x)=-2x+\log_2(b^2-4)h(x)=−2x+log2​(b2−4)

This is a linear function with slope −2<0-2<0−2<0, so it is strictly decreasing for x>1x>1x>1. Hence its largest value on the interval x>1x>1x>1 is approached as x→1+x\to 1^+x→1+.

So in order that f(1)=3f(1)=3f(1)=3 be the maximum, we must have

lim⁡x→1+h(x)≤3.\lim_{x\to 1^+} h(x) \le 3.x→1+lim​h(x)≤3.

Now,

lim⁡x→1+h(x)=−2(1)+log⁡2(b2−4)=−2+log⁡2(b2−4)\lim_{x\to 1^+} h(x) = -2(1)+\log_2(b^2-4)= -2+\log_2(b^2-4)x→1+lim​h(x)=−2(1)+log2​(b2−4)=−2+log2​(b2−4)

Thus the condition is

−2+log⁡2(b2−4)≤3-2+\log_2(b^2-4)\le 3−2+log2​(b2−4)≤3 log⁡2(b2−4)≤5\log_2(b^2-4)\le 5log2​(b2−4)≤5 b2−4≤32b^2-4\le 32b2−4≤32 b2≤36b^2\le 36b2≤36 ∣b∣≤6|b|\le 6∣b∣≤6
  1. Domain condition for logarithm

For log⁡2(b2−4)\log_2(b^2-4)log2​(b2−4) to be defined,

b2−4>0b^2-4>0b2−4>0 b2>4b^2>4b2>4 ∣b∣>2|b|>2∣b∣>2
  1. Combine both conditions

We need both:

∣b∣>2and∣b∣≤6|b|>2 \quad \text{and} \quad |b|\le 6∣b∣>2and∣b∣≤6

Therefore,

b∈[−6,−2)∪(2,6]b\in [-6,-2)\cup(2,6]b∈[−6,−2)∪(2,6]
  1. Match with options

This corresponds to:

Option C

[−6,−2)∪(2,6][-6,-2)\cup(2,6][−6,−2)∪(2,6]
  1. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.

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