JEE Advanced Level Test: Limit & Derivatives- 3 - JEE MCQ

# JEE Advanced Level Test: Limit & Derivatives- 3 - JEE MCQ

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## 10 Questions MCQ Test - JEE Advanced Level Test: Limit & Derivatives- 3

JEE Advanced Level Test: Limit & Derivatives- 3 for JEE 2024 is part of JEE preparation. The JEE Advanced Level Test: Limit & Derivatives- 3 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Limit & Derivatives- 3 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Limit & Derivatives- 3 below.
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JEE Advanced Level Test: Limit & Derivatives- 3 - Question 1

### If α and β be the roots of ax2 + bx +c = 0 , then lim (1 + ax2 + bx + c)

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 1

If lim(x → a) f(x) = lim(x → a) g(x) = 0
then, lim(x → a) [1 + f(x)]1/g(x) = elim(x a) f(x)/g(x)
= lim(x → α) (1 + ax^2 + bx + c)
= elt(x → α) [a(x-α) (x-β)]/(x-α)
= e[a(x-β)]

JEE Advanced Level Test: Limit & Derivatives- 3 - Question 2

### If  f(x) exist and is finite & non zero and if  then the value of  f(x) is

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 2

lim(x→∞) (f(x) + ((3f(x) − 1)/f2(x)) = 3
Let lim(x→∞) f(x) = y
or lim (x→∞) f(x) + [3lim(x→∞)f(x) − 1)]/(lim(x→∞) f(x))2 = 3
or (y + (3y−1)/y2) = 3
or y3 − 3y2 + 3y − 1 = 0
or (y−1)3 = 0
or y = 1

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JEE Advanced Level Test: Limit & Derivatives- 3 - Question 3

### n ∈ N is equal to

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 3

JEE Advanced Level Test: Limit & Derivatives- 3 - Question 4

Let α and β be the distinct roots of ax2 + bx + c = 0, then is equal to

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 4

As α is root of ax2 + bx + c = 0
∴ aα2 + bα + c = 0.
Now,

JEE Advanced Level Test: Limit & Derivatives- 3 - Question 5

has the value

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 5

S = 1n + 2(n-1) + 3(n-2) +.................n(1)
S = (k → 1 to n)  k(n-k+1) = (k-->1 to n) (n+1)k - k2
= (n+1)[n(n+1)/2 - n(n+1)(2n+1)/6]
= n(n+1)/2 [(n+1) - (2n+1)/3]
= n(n+1)/2 * (n+2)/3
= [n(n+1)(n+2)]/6
S/[12 + 22 + .....n2] = {[n(n+1)(n+2)]/6}/{[n(n+1)(2n+1)]/6}
= (n+2)/(2n+1)
= 1/2(1 + 3/(2n+1))
= lim(n → ∞) 1/2(1+3/(2n+1))
= 1/2

JEE Advanced Level Test: Limit & Derivatives- 3 - Question 6

Let (tan α) x + (sin α) y = α and (α cosec α) x + (cos α) y = 1 be two variable straight lines, α being the parameter. Let P be the point of intersection of the lines. In the limiting position when α→ 0, the coordinates of P are

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 6

Solving tanα.x + sinα.y = α and α cosecα.x + cosα.y = 1, we get
x = (αcosα−sinα)/(sinα−α)
and y = (α−xtanα)/sinα
​lim x(α→0) = lim(α→0) (cosα − αsinα − cosα)/(cosα−1)
​= lim (α→0) αsinα/(2sin2(α/2))
= lim (α→0) [4(α/2)2(sinα/α)]/[(sinα/2)2]/2 = 2
lim y(α→0) = lim(α→α) (α−xtanα)/(sinα)
= lim(α→0) (α/sinα − x/cosα)
= 1 − 2
= − 1
Hence P = (2,−1)

JEE Advanced Level Test: Limit & Derivatives- 3 - Question 7

If Ajj = 1, 2, ....n and a1 < a2 < a3 < ..... < an  (A1 . A2. ...An), 1 < m < n

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 7

lim x→am (x−a1)/|x−a1| * (x−a2)/|x−a2| * (x−a3)/|x−a3|.......* (x−an)/|x−an|
lim x→am (x−am)/|x−am| doesnot exist because
lim x→am- (x−am)/|x−am| = -1
lim x→am+ (x−am)/|x−am| = 1
Since, LHL is not equal to RHL , hence the limit does not exist.
Therefore, the limit of the entire function does not exist

JEE Advanced Level Test: Limit & Derivatives- 3 - Question 8

Let a = min {x2 + 2x + 3, x ∈ R) & b =  The value of  ar bn - r is

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 8

Let a = min{x2+2x+3:x∈R} and b = lim(θ→0) (1−cosθ)/θ2
​Now, x2+2x+3
= x2 + 2x + 1 + 2
=(x+1)2 + 2
∴ min{x2 + 2x + 3 : x ∈ R} = min{(x+1)2 + 2 : x ∈ R} = 2
Also, b = lim(θ→0) (1−cosθ)/θ2
lim(θ→0) (sinθ/2θ) = 1/2 ...... {lim (θ→0) {sinθ/θ = 1}}
∑(r = 0 to n)​     ar ⋅bn-r
= ∑(r = 0 to n) 2r(1/2)n-r
= 2-n ∑(r = 0 to n) 22r
= 2-n{1 + 22 + 24 +.....22n}
= 2-n . [1(4n+1 − 1)]/(4−1)
= [4n+1 − 1]/[3.2n]

JEE Advanced Level Test: Limit & Derivatives- 3 - Question 9

If  then the constants 'a' and 'b' are (where a > 0)

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 9

JEE Advanced Level Test: Limit & Derivatives- 3 - Question 10

If f(x) =   then  f(0) is

Detailed Solution for JEE Advanced Level Test: Limit & Derivatives- 3 - Question 10