lim x→∞(x2−4 x2−1)x2+1

  

94 ^





1. lim

x→∞(^x2−4

x²−1)^x2+1.

Ans: Consider ln((^x_x²₂⁻⁴₋₁)^x2+1) = (x²+ 1)(ln(x²− 4) − ln(x² − 1))

x→∞lim(x²+ 1)(ln(x²− 4) − ln(x²− 1)) = lim

x→∞

ln(x²−4)−ln(x²−1)

x2+11

x→∞lim

x2−42x −_x2−1^2x

(x2+1)2−2x

= lim

x→∞

(x²+1)²(x²−1−x²+4)

−(x²−4)(x²−1) =−3 Since ln is a continuous function, lim

x→aln(x) = ln(lim

x→ax), and by the L’Hospital’s rule lim

x→∞(^x_x²₂⁻⁴₋₁)^x2+1 = e⁻³. 2. (a)
₁

0 x⁵e^x3dx.

(b)
^π

04 dθ

(tan²θ+4 tan θ+3) cos²θ

(c)
_8x2+4x−11 (x+3)(x−1)²dx

Ans: (a) Let u = x³. Then
₁

0 x⁵e^x3dx =
₁

0 u · ¹₃ · e^udu

=¹₃



ue^u|¹₀−
₁

0 e^udu



= ¹₃ (b) Let u = tan θ ⇒ du = sec²θdθ

=
₁

0 du

u²+4u+3 =
₁

0 1 2

 ₁

u+1 −_u+3¹ 

du = ¹₂ [ln(u + 1) − ln(u + 3)]|¹₀

= ¹₂[ln 3− ln 2]

(c) _(x+3)(x−1)^8x2+4x−11₂ = ^A

x+3 + ^B

x−1 + _(x−1)^C ₂ ⇒ A = ⁴⁹₁₆, B = ⁷⁹₁₆, C = ¹₄

=
⁴⁹₁₆

x+3 +_x−1⁷⁹¹⁶ +_(x−1)¹⁴ ₂dx = ⁴⁹₁₆ln|x + 3| + ⁷⁹₁₆ln|x − 1| −¹_4x−1¹ +C 3. (a)
₁

12

√ 1

2t−t²dt (b) lim

x→1 1 x−1(
_x

12

√ 1

2t−t²dt − ^π₆)

Ans: (a) Letting 1− t = sin θ, we have dt = − cos θdθ and for ¹₂ ≤ t ≤ 1, 0 ≤ sin θ ≤

12 ⇒ cos θ ≥ 0 and

1− (1 − t)² = cos θ.

Thus,
₁

12

√ 1

1−(1−t)²dt =
₀

π6

− cos θdθ

cos θ = −θ|^0π

6 = ^π₆ (b) Let f (x) =
_x

12

√ 1

2t−t²dt. Then lim

x→1 1 x−1(
_x

12

√ 1

2t−t²dt − ^π₆) = lim

x→1

f (x)−f (1) x−1 . Since√

2t − t² > 0 for 0 < t < 2, f (x) is differentiable at x = 1, and the 1

Fundamental Theorem of Calculus implies that

x→1lim

f (x)−f (1)

x−1 . = f^(1) = √ ¹ 2t−t²

t=1

= 1

4. ^{ }
_∞

0 e^x−2exdx
^? ^
,
^. Ans: Consider
_t

0 e^x

e^2exdx. (Let u = e^x)

=
_e^t

1 du e^2u =
_e^t

1 −¹₂(−2e^−2u)du

= −¹₂e^−2u^et

1 =−¹₂(e^−2et− e⁻²)

t→∞lim − ¹₂(e^−2et − e⁻²) = lim

t→∞

−12 ( ¹

e^2et − _e¹2) = _2e¹₂ 5.^R = {(x, y) ∈ R²|1 ≤ x ≤ 4, 1 < y ≤ 1 + x¹²}. :

(a) R x = 1
^
ëë.

(b) R x = 0
^
ëë.

Ans:

(a) V ol =
₄

1 2π(x − 1)[1 +√

x − 1]dx

=
₄

1 2π(x³² − x¹²)dx 2

= 2π(²₅x⁵² − ²₃x³²)⁴

= 2π(⁶²₅ −¹⁴₃) 1

: V ol = π · 3²+
₃

2 π{3²− [(y − 1)²− 1]²}dy

= π · 3²· 2 −
₃

2 π[(y − 1)²− 1]²dy (b) V ol =
₄

1 2πx[1 +√

x − 1]dx

=
₄

1 2πx³²dx

= ^4π₅ x⁵²⁴

1 = ^124π₅

: V ol = π(4²− 1²) +
₃

2 π[4²− (y − 1)⁴]dy

= π · 4²· 2 − π · 1²· 1 −
₃

2 π[(y − 1)²]²dy

6.

y = x¹² y = 0
^. ^1≤ x ≤ 4 ^
ë. Ans: Area =
₄

1 2π · x¹² ·

1 + (¹₂x⁻¹²)²dx =
₄

1 2π · x¹² ·

1 + ¹₄x⁻¹dx

= 2π
₄

1



x +¹₄dx = 2π(x + ¹₄)³² · ²₃⁴

1

= ^4π₃ (¹⁷₈³² − ⁵₈³²) = ^π₆(17³² − 5³²) 7. ^f (x) =√

x³+ x + 6, (f⁻¹)^(4),

f^{ }4^. Ans: 4 =√

x³+ x + 6 ⇒ x = 2 (^

x²+ 2x + 5
)

∴ (f⁻¹)^(4) = ¹

f
(2) = ₁₃¹

8 = ₁₃⁸

8.^f (x) =

3x + x²sin ¹

x ^x = 0

0 ^x = 0

 f^(x)^.^f^(x)^ ,
^. Ans: When x = 0, f^(x) = 3 + 2x sin¹_x + x²cos¹_x · (−_x¹2)

= 3 + 2x sin¹_x − cos¹_x At x = 0, lim

x→0

f (x)−f (0)

x−0 = lim

x→0

3x+x²sin¹_x x

= lim

x→0(3 + x sin_x¹)

And, since sin_x¹ ≤1, 0 ≤x sin ¹_x ≤ |x|

⇒ lim

x→00≤ lim

x→0

x sin ¹_x ≤lim

x→0|x| = 0

⇒ lim

x→0

x sin ¹_x= 0

⇒ lim

x→0x sin¹_x = 0

⇒ f^(0) = 3

3

9.^f (x) = _x2^x−1³ . ^f (x) ^,^
, ,
^
,
^, ^

y = f (x)^.

Ans:
^
; x = 1, x = −1, y = x.

local maximum: (−√

3, −^3√₂³).

local minimum: (√

3,^3√₂³).

point of reflection: (0, 0).

:

4