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NDA Exam Mathematics Study Material

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Calculus

Let x be a variable and a be a constant. If x takes values closer and closer to a, then we say that x approaches to a and write it as:

a) X → a
b) X → a?
c) X → a?
d) All of these

✔Answer

Correct Answer Is : X → a

Solution Is :

Let x be a variable and a be a constant. If x takes values less than a and approaches to a, then we say that x tends to a from left and write it as:

a) X → a
b) X → a?
c) X → a?
d) All of these

✔Answer

Correct Answer Is : X → a?

Solution Is :

Let x be a variable and a be a constant. If x takes values greater than a and approaches to a, then we say that x tends to a from right and write it as:

a) X → a
b) X → a?
c) X → a?
d) All of these

✔Answer

Correct Answer Is : X → a?

Solution Is :

True or false: if δ is small positive number such that a-δ < x < a+δ,that is x belongs to the open interval (a-δ,a+δ), this open interval is called δ neighbourhood of a

a) TRUE
b) FALSE
c) Maybe
d) None of these

✔Answer

Correct Answer Is : TRUE

Solution Is :

If δ is small positive number such that a-δ < x < a+δ,that is x belongs to to the open interval ___________ .

a) (a-δ,a+δ)
b) (a+δ,a-δ)
c) (-δ,+δ)
d) (a-δ,δ)

✔Answer

Correct Answer Is : (a-δ,a+δ)

Solution Is :

As x → a we have x ≠ a, it means ____________.

a) X ∊ Nδ(a) + {a}
b) X ∊ Nδ(a) - {a}
c) X ∊ Nδ(a) - {x}
d) X ∊ Nδ(x) - {a}

✔Answer

Correct Answer Is : X ∊ Nδ(a) - {a}

Solution Is :

A function f(x) is said to have a limit l to x tends to a if for every _________.

a) ∊ = 0
b) ∊ < 0
c) ∊ > 0
d) All of these

✔Answer

Correct Answer Is : ∊ > 0

Solution Is :

Evaluate: lim (x →a) (x⁷ - a⁷ / x - a )

a) A⁷
b) 7 a⁷
c) 7 a⁶
d) 6 a⁷

✔Answer

Correct Answer Is : 7 a⁶

Solution Is :

Find value of k, if lim(x→1)(x⁴ - 1 / x - 1 ) = lim(x→k)(x³- k³/ x² - k² )

a) 8
b) 3
c) ( 3 / 8)
d) ( 8 / 3 )

✔Answer

Correct Answer Is : ( 8 / 3 )

Solution Is :

Evaluate: lim (x →27) (x^(2/3) - 9 / x - 27 )

a) 1
b) 2
c) ( 1 / 2 )
d) ( 1 / 3 )

✔Answer

Correct Answer Is : ( 1 / 2 )

Solution Is :

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NDA Exam Mathematics Study Material

Calculus

Let x be a variable and a be a constant. If x takes values closer and closer to a, then we say that x approaches to a and write it as:

Correct Answer Is : X → a

Solution Is :

Let x be a variable and a be a constant. If x takes values less than a and approaches to a, then we say that x tends to a from left and write it as:

Correct Answer Is : X → a?

Solution Is :

Let x be a variable and a be a constant. If x takes values greater than a and approaches to a, then we say that x tends to a from right and write it as:

Correct Answer Is : X → a?

Solution Is :

True or false: if δ is small positive number such that a-δ < x < a+δ,that is x belongs to the open interval (a-δ,a+δ), this open interval is called δ neighbourhood of a

Correct Answer Is : TRUE

Solution Is :

If δ is small positive number such that a-δ < x < a+δ,that is x belongs to to the open interval ___________ .

Correct Answer Is : (a-δ,a+δ)

Solution Is :

As x → a we have x ≠ a, it means ____________.

Correct Answer Is : X ∊ Nδ(a) - {a}

Solution Is :

A function f(x) is said to have a limit l to x tends to a if for every _________.

Correct Answer Is : ∊ > 0

Solution Is :

Evaluate: lim (x →a) (x7 - a7 / x - a )

Correct Answer Is : 7 a6

Solution Is :

Find value of k, if lim(x→1)(x4 - 1 / x - 1 ) = lim(x→k)(x3 - k3 / x2 - k2 )

Correct Answer Is : ( 8 / 3 )

Solution Is :

Evaluate: lim (x →27) (x(2/3) - 9 / x - 27 )

Correct Answer Is : ( 1 / 2 )

Solution Is :

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Evaluate: lim (x →a) (x⁷ - a⁷ / x - a )

Correct Answer Is : 7 a⁶

Find value of k, if lim(x→1)(x⁴ - 1 / x - 1 ) = lim(x→k)(x³- k³/ x² - k² )

Evaluate: lim (x →27) (x^(2/3) - 9 / x - 27 )