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Limit fixing a = 5 and b = 10, then four checks on a, b | JEE Advanced

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

Question
Let aa be a natural number such that
lim⁡x→1(1x−1−1xa−4x+3)=b(b≠0),\lim_{x\to1}\left(\frac1{x-1}-\frac1{x^a-4x+3}\right) = b \qquad \left(b \ne 0\right),
 then
A∫ab[x2][x2]+[x2−30x+225]dx=52\displaystyle\int_a^b\frac{\left[x^2\right]}{\left[x^2\right]+\left[x^2-30x+225\right]}dx = \frac52 (where [ ⋅ ]\left[\,\cdot\,\right] represents the greatest integer function)correct
Bthe least value of nn for which (n−2)x2−8x+n+4>k ∀ x∈R\left(n-2\right)x^2-8x+n+4>k \ \forall\,x \in \mathbb R, where k=tan⁡−1(tan⁡(a+b))+sin⁡−1(sin⁡(a+b))k = \tan^{-1}\left(\tan\left(a+b\right)\right)+\sin^{-1}\left(\sin\left(a+b\right)\right) and n∈Nn \in \mathbb N, is 55correct
Cthe number of points of discontinuity of f(x)=[asin⁡x]f\left(x\right) = \left[a\sin x\right], x∈[π,2π]x \in \left[\pi,2\pi\right], is equal to bb (where [ ⋅ ]\left[\,\cdot\,\right] represents the greatest integer function)correct
Dthe value of cc of the chord ax+by=cax+by = c, which subtends a right angle at the centre of the conic 2x2+3y2=12x^2+3y^2 = 1, is ±5\pm5correct
Solution
Step 1: Putting x=1+hx = 1+h,
xa=1+ah+a(a−1)2h2+…x^a = 1+ah+\frac{a\left(a-1\right)}2h^2+\ldots
xa−4x+3=(a−4)h+a(a−1)2h2+…x^a-4x+3 = \left(a-4\right)h+\frac{a\left(a-1\right)}2h^2+\ldots
 the constant terms 1−4+31-4+3 cancelling, and 1x−1=1h\dfrac1{x-1} = \dfrac1h, so
1x−1−1xa−4x+3=(a−5)h+a(a−1)2h2+…h[(a−4)h+a(a−1)2h2+…]\frac1{x-1}-\frac1{x^a-4x+3} = \frac{\left(a-5\right)h+\frac{a\left(a-1\right)}2h^2+\ldots} {h\left[\left(a-4\right)h+\frac{a\left(a-1\right)}2h^2+\ldots\right]}
 Step 2: Case-I: a=4a = 4. The bracket loses its hh term, so the denominator is 6h3+…6h^3+\ldots while the numerator is −h+6h2+…-h+6h^2+\ldots, and
−h+6h2+…6h3+…⟶−∞\frac{-h+6h^2+\ldots}{6h^3+\ldots} \longrightarrow -\infty
 which is not possible for a finite bb. Step 3: Case-II: a≠4a \ne 4. The denominator is now (a−4)h2+…\left(a-4\right)h^2+\ldots, of order h2h^2, so the numerator's hh term must vanish:
a−5=0  ⟹  a=5a-5 = 0 \implies a = 5
 Step 4: With a=5a = 5, the numerator is 5⋅42h2+…=10h2+…\tfrac{5\cdot4}2h^2+\ldots = 10h^2+\ldots and the denominator is h[h+10h2+…]=h2+…h\left[h+10h^2+\ldots\right] = h^2+\ldots, so
b=lim⁡h→010h2h2=10b = \lim_{h\to0}\frac{10h^2}{h^2} = 10
 ∴ a=5\therefore\ a = 5, b=10b = 10. The limit is infinite for every other natural aa from 11 to 99. Step 5: (A) x2−30x+225=(x−15)2x^2-30x+225 = \left(x-15\right)^2, so
I=∫510[x2][x2]+[(x−15)2]dxI = \int_5^{10}\frac{\left[x^2\right]}{\left[x^2\right]+\left[\left(x-15\right)^2\right]}dx
 x→15−xx \to 15-x maps [5,10]\left[5,10\right] onto itself and swaps the two brackets, so with gg the integrand, g(x)+g(15−x)=1g\left(x\right)+g\left(15-x\right) = 1 and
2I=∫5101 dx=5  ⟹  I=522I = \int_5^{10}1\,dx = 5 \implies I = \frac52
 Step 6: (B) a+b=15a+b = 15 radians and 15=5π−0.70796…15 = 5\pi-0.70796\ldots, so
tan⁡−1(tan⁡15)=15−5π,sin⁡−1(sin⁡15)=5π−15\tan^{-1}\left(\tan15\right) = 15-5\pi , \qquad \sin^{-1}\left(\sin15\right) = 5\pi-15
 both values lying in (−π2,π2)\left(-\tfrac{\pi}2,\tfrac{\pi}2\right), and being negatives of each other,
k=(15−5π)+(5π−15)=0k = \left(15-5\pi\right)+\left(5\pi-15\right) = 0
 Step 7: (B) (n−2)x2−8x+n+4>0\left(n-2\right)x^2-8x+n+4>0 for every real xx needs the expression to be a genuine upward parabola: n=1n = 1 gives a downward parabola and n=2n = 2 gives the line 6−8x6-8x, neither of which stays positive, so n−2>0n-2>0. Then
64−4(n−2)(n+4)<0  ⟹  n2+2n−24>0  ⟹  (n+6)(n−4)>0  ⟹  n>464-4\left(n-2\right)\left(n+4\right)<0 \implies n^2+2n-24>0 \implies \left(n+6\right)\left(n-4\right)>0 \implies n>4
 n=4n = 4 gives the minimum value 00, so the inequality is not strict there, and n=5n = 5 gives minimum 113>0\tfrac{11}3>0. The least natural number is n=5n = 5. Step 8: (C) On [π,2π]\left[\pi,2\pi\right], 5sin⁡x5\sin x falls from 00 to −5-5 at x=3π2x = \tfrac{3\pi}2 and rises back to 00. Question attachment Case-I: interior crossings. 5sin⁡x5\sin x crosses −1,−2,−3,−4-1,-2,-3,-4 once on the descending arc and once on the ascending arc, giving 88 jumps. At x=3π2x = \tfrac{3\pi}2, 5sin⁡x=−55\sin x = -5 is a minimum and [5sin⁡x]=−5\left[5\sin x\right] = -5 on both sides, so there is no jump there. Case-II: end points. f(π)=[0]=0f\left(\pi\right) = \left[0\right] = 0 while f→−1f \to -1 as x→π+x \to \pi^+, and f→−1f \to -1 as x→2π−x \to 2\pi^- while f(2π)=[0]=0f\left(2\pi\right) = \left[0\right] = 0, giving 22 more.
8+2=10=b8+2 = 10 = b
 Step 9: (D) The chord is 5x+10y=c5x+10y = c, i.e. 5x+10yc=1\dfrac{5x+10y}c = 1. Homogenising 2x2+3y2=12x^2+3y^2 = 1,
2x2+3y2=(5x+10yc)2  ⟹  (2c2−25)x2−100xy+(3c2−100)y2=02x^2+3y^2 = \left(\frac{5x+10y}c\right)^2 \implies \left(2c^2-25\right)x^2-100xy+\left(3c^2-100\right)y^2 = 0
 which is the pair of lines from the centre to the ends of the chord. They are perpendicular when the coefficients of x2x^2 and y2y^2 add to zero:
2c2−25+3c2−100=0  ⟹  5c2=125  ⟹  c=±52c^2-25+3c^2-100 = 0 \implies 5c^2 = 125 \implies c = \pm5
 Putting x=±1−2yx = \pm1-2y in 2x2+3y2=12x^2+3y^2 = 1 gives 11y2∓8y+1=011y^2 \mp 8y+1 = 0 with discriminant 64−44=20>064-44 = 20>0, so the chord really does cut the conic in two points in each case. Answer: (1),(2),(3),(4)\left(1\right),\left(2\right),\left(3\right),\left(4\right)
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