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Limits Continuity and Differentiability question

2025 · 22 Jan · Shift 1 · Q45
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  5. /2025 · 22 Jan · Shift 1 · Q45

Limits Continuity and Differentiability question

2025 · 22 Jan · Shift 1 · Q45

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1
Let the function, f(x)={−3ax2−2,x<1a2+bx,x⩾1f(x)= \begin{cases}-3 \mathrm{ax}^2-2, & x\lt 1 \\ \mathrm{a}^2+\mathrm{b} x, & x \geqslant 1\end{cases}f(x)={−3ax2−2,a2+bx,​x<1x⩾1​ be differentiable for all x∈Rx \in \mathbf{R}x∈R, where a>1, b∈R\mathrm{a}\gt 1, \mathrm{~b} \in \mathbf{R}a>1, b∈R. If the area of the region enclosed by y=f(x)y=f(x)y=f(x) and the line y=−20y=-20y=−20 is α+β3,α,β∈Z\alpha+\beta \sqrt{3}, \alpha, \beta \in Zα+β3​,α,β∈Z, then the value of α+β\alpha+\betaα+β is ‾\underline{\hspace{2cm}}​ .
Numerical answer
View written solutionFree

Correct answer: 34

  1. Given function
f(x)={−3ax2−2,x<1a2+bx,x≥1f(x)= \begin{cases} -3ax^2-2, & x<1 \\ a^2+bx, & x\ge 1 \end{cases}f(x)={−3ax2−2,a2+bx,​x<1x≥1​

with a>1a>1a>1 and fff differentiable for all x∈Rx\in\mathbb Rx∈R.

For differentiability at x=1x=1x=1, we need:

  • continuity at x=1x=1x=1
  • equal left and right derivatives at x=1x=1x=1
  1. Continuity at x=1x=1x=1

Left value at x=1x=1x=1: −3a(1)2−2=−3a−2-3a(1)^2-2=-3a-2−3a(1)2−2=−3a−2

Right value at x=1x=1x=1: a2+b(1)=a2+ba^2+b(1)=a^2+ba2+b(1)=a2+b

So, −3a−2=a2+b(1)-3a-2=a^2+b \qquad (1)−3a−2=a2+b(1)

  1. Derivative matching at x=1x=1x=1

For x<1x<1x<1, f′(x)=−6axf'(x)=-6axf′(x)=−6ax so left derivative at x=1x=1x=1 is −6a-6a−6a

For x≥1x\ge 1x≥1, f′(x)=bf'(x)=bf′(x)=b so right derivative at x=1x=1x=1 is bbb

Thus, b=−6a(2)b=-6a \qquad (2)b=−6a(2)

Substitute (2) into (1): −3a−2=a2−6a-3a-2=a^2-6a−3a−2=a2−6a a2−3a+2=0a^2-3a+2=0a2−3a+2=0 (a−1)(a−2)=0(a-1)(a-2)=0(a−1)(a−2)=0

Given a>1a>1a>1, therefore a=2a=2a=2

Then b=−6a=−12b=-6a=-12b=−6a=−12

  1. Write the function explicitly
f(x)={−6x2−2,x<14−12x,x≥1f(x)= \begin{cases} -6x^2-2, & x<1 \\ 4-12x, & x\ge 1 \end{cases}f(x)={−6x2−2,4−12x,​x<1x≥1​

We need the area enclosed by y=f(x)y=f(x)y=f(x) and y=−20y=-20y=−20.

So find points of intersection with y=−20y=-20y=−20.

  1. Intersection points

For x<1x<1x<1:

−6x2−2=−20-6x^2-2=-20−6x2−2=−20 −6x2=−18-6x^2=-18−6x2=−18 x2=3x^2=3x2=3 x=±3x=\pm \sqrt{3}x=±3​

Since this branch is valid for x<1x<1x<1, both x=−3x=-\sqrt{3}x=−3​ and x=3x=\sqrt{3}x=3​ satisfy the equation, but only x=−3x=-\sqrt{3}x=−3​ is in the domain x<1x<1x<1.

For x≥1x\ge 1x≥1:

4−12x=−204-12x=-204−12x=−20 −12x=−24-12x=-24−12x=−24 x=2x=2x=2

So intersections are at x=−3,  x=2x=-\sqrt{3},\; x=2x=−3​,x=2

Also note that at x=1x=1x=1, f(1)=4−12=−8f(1)=4-12=-8f(1)=4−12=−8 which is above −20-20−20, so the enclosed region lies between the curve and the line from x=−3x=-\sqrt{3}x=−3​ to x=2x=2x=2.

  1. Compute the area

Area A=∫−31(f(x)−(−20))dx+∫12(f(x)−(−20))dxA=\int_{-\sqrt{3}}^{1}\big(f(x)-(-20)\big)dx+\int_{1}^{2}\big(f(x)-(-20)\big)dxA=∫−3​1​(f(x)−(−20))dx+∫12​(f(x)−(−20))dx

That is, A=∫−31(−6x2−2+20)dx+∫12(4−12x+20)dxA=\int_{-\sqrt{3}}^{1}(-6x^2-2+20)dx+\int_{1}^{2}(4-12x+20)dxA=∫−3​1​(−6x2−2+20)dx+∫12​(4−12x+20)dx A=∫−31(18−6x2)dx+∫12(24−12x)dxA=\int_{-\sqrt{3}}^{1}(18-6x^2)dx+\int_{1}^{2}(24-12x)dxA=∫−3​1​(18−6x2)dx+∫12​(24−12x)dx

  1. First integral
I1=∫−31(18−6x2)dxI_1=\int_{-\sqrt{3}}^{1}(18-6x^2)dxI1​=∫−3​1​(18−6x2)dx

Antiderivative: 18x−2x318x-2x^318x−2x3

So,

I1=[18x−2x3]−31I_1=\left[18x-2x^3\right]_{-\sqrt{3}}^{1}I1​=[18x−2x3]−3​1​

At x=1x=1x=1: 18(1)−2(1)3=1618(1)-2(1)^3=1618(1)−2(1)3=16

At x=−3x=-\sqrt{3}x=−3​: 18(−3)−2(−3)318(-\sqrt{3})-2(-\sqrt{3})^318(−3​)−2(−3​)3 Now, (−3)3=−33(-\sqrt{3})^3=-3\sqrt{3}(−3​)3=−33​ so 18(−3)−2(−33)=−183+63=−12318(-\sqrt{3})-2(-3\sqrt{3})=-18\sqrt{3}+6\sqrt{3}=-12\sqrt{3}18(−3​)−2(−33​)=−183​+63​=−123​

Hence, I1=16−(−123)=16+123I_1=16-(-12\sqrt{3})=16+12\sqrt{3}I1​=16−(−123​)=16+123​

  1. Second integral
I2=∫12(24−12x)dxI_2=\int_{1}^{2}(24-12x)dxI2​=∫12​(24−12x)dx

Antiderivative: 24x−6x224x-6x^224x−6x2

Thus,

I2=[24x−6x2]12I_2=\left[24x-6x^2\right]_{1}^{2}I2​=[24x−6x2]12​

At x=2x=2x=2: 48−24=2448-24=2448−24=24

At x=1x=1x=1: 24−6=1824-6=1824−6=18

So, I2=24−18=6I_2=24-18=6I2​=24−18=6

  1. Total area

A=I1+I2=(16+123)+6=22+123A=I_1+I_2=(16+12\sqrt{3})+6=22+12\sqrt{3}A=I1​+I2​=(16+123​)+6=22+123​

Hence, α=22,β=12\alpha=22,\quad \beta=12α=22,β=12

Therefore, α+β=22+12=34\alpha+\beta=22+12=34α+β=22+12=34

  1. Comparison with stored answer

Derived answer is 343434, which matches the stored correct answer.

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