Class Xi Maths Limits And Derivatives 1 r6537

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CONTENTS

LIMITS AND DERIVATIVES Theory Solved Examples Exercises Level – I Level – II Level – III Questions asked in AIEEE and other Engineering Exams Answers

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Mathematics : Limits and Derivatives

INSTITUTE OF CORRESPONDENCE COURSES

LIMITS AND DERIV ATIVES DERIVATIVES AIEEE Syllabus Limits, Differentiation of the sum, difference, product and quotient of two functions, differentiation of trigonometric, inverse trigonometric, logarithmic, exponential, composite and implicit functions, derivatives of order up to two.

CONTENTS ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦

Definition of a limit Trigonometric limits Exponential and logarithmic limits Approximations Some useful expansions Indeterminate forms Limit of greatest integer function Sandwich Theorem Derivative of a function Some differentiation formulae Algebra of differentiation Differentiation of implicit functions Derivative of parametric functions Derivative of a function w.r.t. another function Use of finding derivatives of the function of type (f(x))g(x) Differentiation using trigonometrical substitutions Higher order differentiation Derivative of infinite series Differentiation of a determinant function

INTRODUCTION This chapter is an introduction to calculus. Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. In this chapter we define limit and some algebra of limits. Also we study derivative and algebra of derivatives and derivatives of certain standard functions.

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1.

DEFINITION If y = f(x) is any function which is defined in a neighbourhood of a then for some ‘ ∈ ’ greater than zero there exists a δ > 0 such that |f(x) – l| < ∈ ⇒ | x − a | < δ then l is said to be limit of the function when x-approches a. It is symbolically written as Lt f (x) = l x →a

2.

STANDARD FORMULA Lt

x →a

xn − an = na n −1 ; x ≠ a ; n is a rational number or integer. x−a

Remark :

3.

x m − a m m m−n = a xn − an n

Lt

x →a

TRIGONOMETRIC LIMITS

4.

(i)

lim

sin x =1 x

(ii)

lim cos x = 1

(iii)

lim

tan x =1 x

(iv)

lim

sin −1 x =1 x

(v)

lim

tan −1 x =1 x

(vi)

lim

sin x o π = x 180

x →0

x →0

x →0

x →0

x →0

x →0

EXPONENTIAL AND LOGARITHMIC LIMITS Lt

ex − 1 =1 x

(ii)

(iii)

lim

a x − bx a = log e   , a, b > 0 x b

(v)

 1 Lt (1 + x)1/ x = e = Lt  1 +  x →0 x →∞  x

(vii)

 a lim  1 +  = ea x →∞  x

(i)

x →0

x →0

Lt

ax −1 = log e a (a > 0) x

(iv)

lim

(1 + x) n − 1 =n x

(vi)

lim (1 + ax)1/ x = ea

x →0

x →0

x x →0

x

 1  (viii) lim 1 + f (x)  x →∞   (ix)

lim (1 + f (x) )

(xi)

lim

f (x)

1/ f (x)

x →a

x →0

= e , where f (x) → ∞ as x → ∞

=e

(x)

lim

x →∞

log x = 0 (m > 0) xm

log a (1 + x) = log a e (a > 0, a ± 1) x


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5.

APPROXIMATIONS sin ax ! ax

(ii) cos ax ! 1 −

(iii)

tan ax ! ax

(iv) eax ! 1 + ax

(v)

e−ax ! 1 − ax

(vi) log(1 + ax) ! ax

(vii) a ! 1 + (logea)x x

(ix)

(xi)

6.

a2x2 2

(i)

(viii)

(x) coshax ! 1 +

tanhax ! ax n

1± x !1±

sinhax ! ax

a2x2 2

x , |x| < 1 n

SOME USEFUL EXPANSIONS If x → 0 and there is at least one function in the given expansion which can be expanded, then we express numerator and denominator in the ascending powers of x and remove the common factor there, the following expansions of some standard functions should be ed. (a)

ex = 1 +

x x2 + + ...... 1 2

(c)

ax =1+

(log a)x (log a)2 x 2 + + ...... 1 2

(e)

  x 2 x3 log(1 − x) = −  x + + + ..... 2 3  

(g)

sin x = x −

(i)

tan x = x −

(k)

cos h x = 1 +

x2 x4 + + ....... 2 4

(m)

sin −1 x = x +

x 3 9x 5 + + ...... 3 5

(o)

(1 + x)n = 1 + nx +

(p)

 1  x 11 2   1 +  = e  1 − + x + ......   x  2 24 

x x 2 x3 + − + ..... 1 2 3

(b)

e− x = 1 −

(d)

log(1 + x) = x −

(f)

cos x = 1 −

x3 x5 + − ....... 3 5

(h)

sin h x = x +

x3 x5 + + ....... 3 5

x 3 2x 5 + − ....... 3 15

(j)

tan h x = x +

x 3 2x 5 + + ....... 3 15

(l)

cos −1 x =

(n)

tan −1 x = x −

x 2 x3 + − ..... 2 3

x2 x4 + − ....... 2 4

 π  x 3 9x 5 −x + + + ......  2  3 5  x3 x5 x7 + − + ...... 3 5 7

n(n − 1) 2 x + ....... where n ∈ z + 2

x


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7.

INDETERMINATE FORMS The forms which cannot be defined exactly are called indeterminate forms, they are

0 ∞ , , 0 × ∞ , ∞ − ∞ , 00, ∞0 and 1∞ 0 ∞

L’ HOSPITAL’S RULE f ′(x) f ′(x) 0 ∞ f (x) takes the form of or then the limit of the function is xLt , if xLt → a → a g′(x) itself x → a g(x) g′(x) ∞ 0

If Lt

takes the form again

till

1.

f ′′(x) 0 ∞ , then the limit of the function is xLt →a g ′′(x) and the process is continued 0 ∞

0 ∞ , is eliminated then limit is obtained. 0 ∞ If 0 × ∞ form is given, convert it in the form of

0 ∞ , by taking one term to the denominater 0 ∞

then apply L’Hospital’s Rule. 2.

If (∞ − ∞) form is given, take L.C.M convert it in the form of

0 ∞ or form, then take the 0 ∞

help of L’Hospital’s Rule. 3.

00 and ∞ 0 form is given take the help of logarithms convert the problem again in the form of

0 ∞ or form and then use L’Hospital’s Rule. 0 ∞ 4.

8.

If Lt [f (x)]

g(x)

x →a

takes the form of 1∞ then write it as Lt ( f (x) ) x →a

g( x )

Lt g( x )[f ( x ) −1]

= e x →a

LIMIT OF GREATEST INTEGER FUNCTION Greatest integer function is denoted by [.] Let a ∈ R then two cases arise. Case (1) if a ∈ integer then we have 1. 2. 3.

Lt [x] = a

x →a +

Lt [x] = a − 1

x →a−

Lt [x] does not exist

x →a

Case (2) If a ∉ integer then

Lt [x] = c

x →c


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sin[x] , [x] ≠ 0 [x]

If f (x) =

Example :

=0

[x] = 0

where [x] denotes the greatest integer less than or equal to x, then find Lt f (x) x →0

Lt f (0 − h) = Lt f (0 + h)

Solution :

h →0

⇒

9.

h →0

Lt

h →0

sin[ − h] sin[h] = Lt h → 0 [− h] [h]

⇒ sin1 ≠ 1

SANDWICH THEOREM Suppose that g(x) ≤ f(x) ≤ h(x) for all x in some open interval containing c, except possibly at ‘c’ itself. Suppose also that

y h(x) f(x) g(x)

!

Lt g(x) = Lt h(x) = ! then Lt f (x) = ! x →c x →c

x →c

This is called sandwich theorem.

10.

O

c

SPECIAL TYPES OF LIMITS 1.

Use of Leibnitz’s formula for evaluating the limit Consider the integral ψ(x)

g(x) =

∫

φ( x )

f (t) dt then

g ′(x) = f [ψ (x)] ψ ′(x) − f (φ(x)) φ′(x) 2.

Summation of series using definite integral as the limit of a sum. It is used in the expression of the form

1 n ∑ n →∞ n r =1 Lt

r f   = ∫ f (x) dx n a b

To Evaluate such limits we note the following (a)

∑

(b)

r → x (r = x , n = 1) n

(c)

1 → dx n

(d)

Lower limit is always zero.

(e)

Upper limit is Coefficient of n in the upper limit of Σ

is replaced by sign of integration


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DERIVATIVES 11. DERIVATIVE OF A FUNCTION Let y = f(x) be a function defined on an interval [a, b]. Let for a small increment δx in x, the corresponding increment in the value of y be δy. Then y = f(x) and y + δy = f(x + δx) On subtraction, we get δy = f(x + δx) – f(x) or

δy f (x + δx) − f (x) = δx δx

Taking limit on both sides when δx → 0 we have,

lim

δx → 0

δy f (x + δx) − f (x) = lim δ → x 0 δx δx

if this limit exists, is called the derivative or differential coefficient of y with respect to x and is written as

∴

dy or f ′(x) . Thus dx

δy dy f (x + δx) − f (x) = lim = lim . This is called Differentiation from first principle. δ → δ → x 0 x 0 δx δx dx

Derivative at a point: The value of f ′(x) obtained by putting x = a, is called the derivative of f(x) at x = a and it is denoted

 dy    dx x = a

by f ′(a) or 

Note :

dy d d ( y ) in which is is simply a symbol of operation and not 'd' divided by dx. dx dx dx

12. SOME DIFFERENTIATION FORMULAE (i)

d (constant) = 0 dx

(ii)

d n n–1 (x ) = nx dx

(iii)

d x x (e ) = e dx

(iv)

d x x (a ) = a loge a dx

(v)

d 1 (logex) = dx x

(vi)

1 d (loga x) = x log a dx e

(vii)

d (sin x) = cos x dx

(viii)

d (cos x) = – sinx dx


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(ix)

d 2 (tan x) = sec x dx

(x)

d 2 (cot x) = – cosec x dx

(xi)

d (sec x) = sec x tan x dx

(xii)

d (cosec x) = – cosecx cotx dx

(xiii)

d –1 (sin x) = dx

(xiv)

d –1 (cos x) = – dx

(xv)

1 d –1 (tan x) = dx 1+ x2

(xvi)

d 1 –1 (cot x) = – dx 1+ x2

(xviii)

d −1 –1 (cosec x) = dx | x | x2 −1

(xvii)

1 1− x

2

1 d –1 (sec x) = dx | x | x2 −1

1 1− x 2

(xix)

d ax ax (e sin bx) = e (a sin bx + b cos bx) = dx

a 2 + b 2 e sin (bx + tan

(xx)

d ax ax (e cos bx) = e (a cos bx – b sin bx) = dx

a 2 + b 2 e cos (bx + tan

(xxi)

x d |x| |x| = | x | or : x≠0 dx x

(xxii)

ax

–1

ax

–1

b ) a b ) a

d 1 log |x| = dx x

13. ALGEBRA OF DIFFERENTIATION (i)

Sum and difference rule d d d [f1(x) ± f2 (x)] = [f1(x)] ± [f2 (x)] dx dx dx

(ii)

Scalar multiple rule d d [k f(x)] = k [f(x)] , where k is any constant dx dx

(iii) Product rule d d d [f1(x).f2 (x)] = [f1(x)] [f2 (x)] + [f2 (x)] [f1 (x)] dx dx dx

(iv) Quotient rule

d  f1(x)    = d x  f2 (x)  (v)

f2 (x)

d d [f1(x)] − f1(x) [f2 (x)] dx dx [f2 (x)]2

Chain rule d y d y du dv if y = f1(u), u = f2(v) and v = f3(x) then d x = d u . d v . d x 7

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14. DERIVATIVE OF PARAMETRIC FUNCTIONS Let x and y are two functions of variable 't' (parameter) such that x = f(t) and y = g(t). then

 dy  dy  dt  g′(t) = = dx  dx  f ′(t)    dt  Example - 1

If x = a(cosθ + θsinθ), y = a(sinθ – θcosθ) then find

Solution :

dx = a[ − sin θ + sin θ + θ cos θ] = aθ cos θ ; dθ ∴

dy . dx

dy = a(cos θ − cos θ + θ sin θ) = aθ sin θ dθ

dy = tan θ dx

15. DIFFERENTIATION OF IMPLICIT FUNCTIONS If in an equation, x and y both occur together. i.e. f(x, y) = 0 or f(x, y) = c and this function can not be solved either for 'y' or 'x' then f(x, y) is called the implicit function of x (or y). ∂f − dy x −(yx y −1 + y x log y) ∂ y x b = = If x + y = a , then ∂f dx (x y log x + xy x −1 ) ∂y

15.1. WORKING RULE FOR FINDING THE DERIVATIVE Method – 1 (i) Differentiate every term of f(x, y) = 0 with respect to 'x'. (ii)

Collect the coefficients of

dy dy and obtain the value of . dx dx

Method – 2 ∂f − ∂f ∂f f dy ∂x = = − x where If f(x, y) = constant, then and ∂x ∂ y are partial differential coefficients of ∂f dx fy ∂y f(x, y) with respect to x and y respectively.

Note :

An implicit function can be differentiated either with respect to 'x' or with respect to 'y'

16. DERIVATIVE OF A FUNCTION WITH RESPECT TO ANOTHER FUNCTION Let y = f(x) and z = g(x) be two functions of 'x' then the derivative of f(x) w.r.t g(x) or derivative of y is denoted by

i.e.

dy dz

 dy  dy  dx  f ′(x) = = dz  dz  g′(x)  dx   


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Example - 2

 1− x2   2x  Find the derivative of sin−1  with respect to cos −1  2  2   1+ x   1+ x 

Solution :

−1  2x  = 2 tan−1 x ∴ Let f(x) = sin  2   1+ x 

f ′(x) =

2 1+ x2

2 2 −1  1 − x  = 2 tan−1 x ∴ g′(x) = Let g(x) = cos  2  + 1 x 1 + x2   Hence the derivative of f(x) with respect to g(x) is

f ′(x) 2 /1 + x 2 = =1 g′(x) 2 /1 + x 2

17. USE OF FINDING DERIVATIVES OF THE FUNCTION OF TYPE (f(x) )g(x) Let y = [f(x)]

g( x )

Taking log on both sides we get log y = g(x) . log f(x) Differentiating we get dy f ′(x)  g( x )  = [f(x)] ⋅ g′(x) log f(x) + g(x) dx f(x)  

d tan x tan x   = x tan x  sec 2 x log x + x dx x  

{

Example - 3

}

18. DIFFERENTIATION USING TRIGONOMETRICAL SUBSTITUTIONS –1

(i)

sin

(ii)

cos

(iii) tan

(v)

–1

–1

y = sin

–1

x ± cos

–1

x ± tan

2 cos

(vii) 2 tan

(ix)

x ± sin

–1

–1

–1

y = cos

y = tan

x = cos

x = sin

–1

–1

–1

[x 1− y 2 ± y 1− x 2 ] –1

–1

[xy ∓

x±y   1 ∓ x y 

(1 − x 2 )(1 − y 2 )] –1

(iv)

2 sin

(2x – 1)

(vi)

2 tan

 2x     1 + x2   

(viii)

2 tan

(x)

3 sin

2

π –1 –1 – tan x = tan 4

 1− x    1+ x 

–1

–1

–1

x = sin

–1

( 2x 1 − x 2 )  2x     1− x2 

–1

x = tan

x = cos

x = sin

–1

–1

 1− x2     1 + x2    3

(3x – 4x )


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(xi) 3 cos

–1

x = cos

–1

–1

3

–1

(4x – 3x)

(xii)

3 tan

π 2

(xiv)

tan x + cot x =

–1

(xiii) sin x + cos x = –1

–1

x = tan

–1

–1

 3x − x 3   1 − 3x 2 

   

π 2

π 2

–1

(xv) sec x + cosec x =

19. SUITABLE SUBSTITUTION (i)

If the function involve the term

a 2 − x 2 , then put x = a sin θ, or x = a cos θ

(ii)

If the function involve the term

x 2 + a 2 , then put x = a tan θ

(iii) If the function involve the term

x 2 − a 2 , then put x = a sec θ

(iv) th

20. n

 a−x  If the function involve the term  a + x 

   , then put x = a cos θ 

DIFFERENTIATION OF SUITABLE FUNCTION n

m

n

(1)

D (ax + b) = m(m – 1) (m – 2) ........... (m – n + 1) a (ax + b)

(2)

If m ∈ N and m > n, then

m–n

m! n m n m–n D (ax + b) = (m − n) ! a (ax + b) Dn ( x m ) =

(3)

n

m! x m −n (m − n) ! n

n

D (ax + b) = n ! a n

n

D (x ) = n ! (4)

n n  1  ( −1) n ! a = Dn   n +1  ax + b  (ax + b) n  1  ( −1) n ! Dn   = x n +1 x

(5)

n

D {log (ax +b)} =

( −1)n −1(n − 1) ! xn

n

D (log x) = n

ax

( −1)n −1(n − 1) ! n a (ax + b)n

n

ax

(6)

D (e ) = a e

(7)

D (a ) = (log a) a

n

mx

n

mx

.m

n


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(8)

n

n

D (sin x) = sin (x + n (9)

π ) 2

n

D {sin (ax + b)} = a sin (ax + b + n

n

π ) 2 π ) 2

n

D {cos (ax + b)} = a cos (ax + b + n π ) 2

n

D (cos x) = cos (x + n n

ax

n

ax

(10) D {e (11) D {e

n

(12) D (tan

2

2

n

2 n/2

cos (bx + c)} = (a + b )

–1

–1

e

ax

e

ax

sin (bx + c + n tan

–1

cos (bx + c + n tan

–1

b ) a b ) a

x ( −1)n−1(n − 1) ! sinn θ sin n θ ) = a an

Where θ = tan (13) D (tan

2 n/2

sin (bx + c)} = (a + b )

–1

(

a ) x

n–1

x) = (–1)

Where θ = tan

–1

(

n

(n–1) ! sin θ sin nθ

1 ) x

Example - 4

dy 3 3 If x = a cos θ, y = a sin θ, then find d x .

Solution :

Since x = a cos θ

3

∴

dx d (cos3 θ) d (cos θ) =a . dθ (cos θ) dθ 2

(Using chain rule) 2

= 3a cos θ (–sin θ) = –3a cos θ sin θ 3 and y = a sin θ ∴

dy d (sin3 θ) d (sin θ) 2 =a . = 3a sin θ . cos θ dθ d (sin θ) dθ

Now,

dy   d y  d θ  3a sin2 θ cos θ = = = − tan θ d x  d x  −3a cos2 θ sin θ   dθ

21. HIGHER ORDER DIFFERENTIATION th

If y = f(x) be a differentiable function of x, such that whose second, third............, n derivatives exist. th

The first, second, third ........., n derivatives of y = f(x) are denoted respectively by

d y d2 y dn y , 2 , ......., d x dx dx n 11


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Also denoted by y', y"........y

(n)

y1, y2, y3..............,yn n

f', f".............f 2

3

n

dy, d y, d y,.........d y

Example - 5

d2 y ln x If y = then find d x2 x

Solution :

We have y =

ln x x

xy = ln x

.....(1)

Differentiating both sides w.r.t. x, we get dy 1 x d x + y .1= x

⇒

x

2

dy d x + xy = 1

⇒

x

dy d x + ln x = 1

2

[From (1)]

.....(2)

Again differentiating both sides w.r.t. x, we get

x

d2 y

2

dx

⇒

x3

⇒

x

Hence

x3

2

d2 y dx

2

d2 y

3

dx 2 d2 y dx

2

+

dy 1 . 2x + = 0 dx x

+ 2x 2

dy +1= 0 dx

+ 2 (1 – ln x) + 1 = 0

= (2 ln x − 3)

or

[from (2)]

d2 y dx

2

=

2 ln x − 3 x3

22. DERIVATIVE OF INFINITE SERIES If taking out one or more than one form an infinite series.

f ( x ) + f ( x ) + f ( x ) + .........∞

(A) If y = ⇒

then y =

f ( x) + y

2

(y – y) = f(x)

dy Differentiating both sides w.r.t. x, we get (2y – 1) d x = f '(x) {f ( x )}....... ∞

(B) If y = {f(x)}{f ( x )} ⇒

y = e

then y = {f(x)}

y

y !n f(x)


NARAYANA



Differentiating both sides w.r.t. x, we get dy y dx = e

ln f(x)

 1 dy  ⋅ f '(x) + ln f(x) ⋅  y ⋅ dx   f(x)

⇒

dy dy  y  y ⋅ f '(x) + ln f(x) ⋅   d x = {f(x)}  f(x) dx 

⇒

dy y y–1 {1 – {f(x)} ln f(x)} d x = y {f(x)} . f '(x)

(C) y = f(x)

f(x)

then

dy = f(x)f ( x ) ⋅ f ′(x)[1 + log f(x)] dx

d x (x ) = x x (1 + log x) dx d(sin x sin x ) = (sin x)sin x .cos x[1 + log sin x]

(D) y = f(x)

g(x)

then

 dy f ′(x)  = f(x)g( x ) g′(x)logf(x) + g(x) ⋅ dx f(x)  

d 1  (sin x)log x = (sin x)log x  log(sin x) + log x.cot x  dx x 

{

}

dy y 2 f ′(x) 1 = 2 then dx y +1 1 f(x) + .....∞ f(x) +

(E)

y = f(x) +

(F)

dy 2f ′(x) 1 + f(x)  y = log  then dx =  1 − f 2 (x)  1 − f(x) 

 1 + tan x  dy 2sec 2 x y = log  = then  dx 1 − tan2 x  1 − tan x  Example - 6

dy If y = x + y + x + y + .......∞ then find d x

Solution :

We have y =

x + y + x + y + ......∞

⇒

y =

⇒

y – x =

⇒

(y –x) = 2y

x+ y+y

2

2

2y

2


NARAYANA



Differentiating both sides w.r.t. x, we get  dy  dy 2 2 (y – x)  2 y d x − 1 = 2 d x  

⇒

dy 2 2 (2y (y – x) –1) d x = (y – x)

Hence

dy ( y 2 − x) = d x (2y 3 − 2xy − 1)

23. DIFFERENTIATION OF A DETERMINANT FUNCTION

If F(x) =

f g h ! m n u v w

Where f, g, h, !, m, n, u, v, w are functions of x and differentiable then

F'(x) =

or

F'(x) =

f ' g' h' ! m n u v w

f' g h !' m n u' v w

+

f g h !' m' n' u v w

+

f g' h ! m' n u v' w

+

+

f g h ! m n u' v' w '

f g h' ! m n' u v w'


NARAYANA



SOLVED EXAMPLES Example - 1

Solution :

Lt

a x − bx = tan x

(1)

log(a/b)

(2)

log(b/a)

(3)

logab

(4)

a/b

x →0

Ans. (1)

Lt

(a x − 1) − (b x − 1)  x   tan x  x

Lt

a x − 1 bx − 1 − x x

x →0

x →0

(∵ Lt

x →0

tan x = 1) x

= loga – logb = log(a/b) Example - 2

Solution :

Lt (cos x + a sin bx)1/ x =

x →0

(1)

eab

(2)

ea/b

(3)

log(a/b)

(4)

logab

Ans. (1) ∞

Given limit is in the form 1 Lt

e x→0

=e Example - 3

Solution :

2  ∵ cos x " 1 − x   2 

1 [cos x + a sin bx −1] x

 x2  1− + abx −1 2   Lt x →0 x

= e ab

( sin bx " bx )

 1  1 1 + + .... + Lt  = (2n + 1)(2n + 3)   3.5 5.7

n →∞

(1)

1 2

(2)

1 3

(3)

1 6

(4)

1 4

Ans. (3)

Lt

n →∞

1 1 1 1 1 1 1  −  − + − + .... +  2 3 5 5 7 2n + 1 2n + 3 

1 1 1 = × = 2 3 6 15 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

NARAYANA



Example - 4

Solution :

(x + 1)10 + (x + 4)10 + (x + 9)10 + ... + (x + 100)10 = x →∞ x10 + 10010 Lt

(1)

100

(2)

1000

(3)

10

(4)

1

Ans. (3) 10 10  1 10  4  100   x10  1 +  +  1 +  + .... +  1 +   x    x   x  Lt x →∞  10010  x10  1 + 10  x  

= 1 + 1 + ..... + 10 times = 10 Note : If the degree of numerater and denomenater are equal, then the ratio of constant is the limit when x → 0 and the ratio of coefficients of highest degree is the limit when x → ∞ . x2

Example - 5

Solution :

Lt

∫ sin

t dt

0

x3

x →0

=

(1)

3 2

(2)

1 3

(3)

1 2

(4)

2 3

Ans. (4) x d    ∫ sin t dt  dx  0  2

Lt

d 3 (x ) dx

x →0

sin x (2x) x →0 3x 2

= Lt

= Lt

x →0

Example - 6

Solution :

(Use Leibnitz’s rule and sinx " x)

2x ⋅ x 2 = 3x 2 3

[x] + [2x] + .... + [nx] is, where [.] denotes the greatest integer function x →∞ n2 Lt

(1)

x 3

(2)

x 6

(3)

does not exist

(4)

x 2

Ans. (4) Using the fact nx − 1 < [nx] ≤ nx


NARAYANA



∴

∑ (nx − 1) < ∑ [nx] ≤ nx

⇒

x

∴

n x.n(n + 1)  x n(n + 1) n  Lt  ⋅ Lt [rx] ≤ Lt − < ∑  2 2 n →∞ n →∞ n n  n →∞ r =1 2n 2 2

⇒

x ≤ Lt 2 n →∞

∴

Lt

n n(n + 1) n(n + 1) − n < ∑ [rn] ≤ x ⋅ 2 2 r =1

n

Example - 7

Solution :

n

x

∑[rx] ≤ 2 r =1

x

∑ [rx] = 2

n →∞

r =1

(Using sandwich theorem)

1   1 Lt  2 − = 2  sin x sinh x 

x →0

(1)

2 3

(2)

0

(3)

1 3

(4)

−

2 3

Ans. (1)

Lt

x →0

sinh 2 x − sin 2 x sinh 2 x − sin 2 x = Lt x →0 sinh 2 x sin 2 x x4

= Lt

x →0

2sinh x cosh x − 2sin x cos x (L.H.R) 4x 3

sinh 2x − sin 2x = Lt = Lt x →0 x →0 4x 3

e 2x − e −2x − sin 2x 2 4x 3

23 ⋅ e2 x − (−2)3 e −2x  3π  − 23 sin  + 2x  2  2  = Lt x →0 4.3! ( ∵ the degree of denominator is 3, we take the 3rd order derivative)

8+8 +8 16 2 = 2 == = 4.3.2 8.3 3 Example - 8

1 − cos(ax 2 + bx + c) is equal to x →α (x − α) 2

Let a, b be the distinct roots of ax2 + bx + c = 0, then Lt

(1)

(α − β)2 2

(2)

−

(3)

0

(4)

a2 (α − β) 2 2

a2 (α − β) 2 2


NARAYANA



Solution :

Ans. (4)

1 − cos a(x − α)(x − β) x →α (x − α) 2 Lt

(∵ ax 2 + bx + c ≡ a(x − α)(x − β) )

 a2x2  a 2 (x − α ) 2 (x − β) 2 a 2 (α − β)2 cos ax 1 − " Lt = =   2  x →α 2(x − α) 2 2  Example - 9

Lt x→

π 2

(1) (3) Solution :

(1 − tan x / 2)(1 − sin x) = (1 + tan x / 2) (π − 2x)3

1 8 1 32

(2)

0

(4)

∞

Ans. (3)

lim

π x→ 2

(1 − tan x / 2)(1 − sin x) (1 + tan x / 2)(π − 2x)3

If x →

π π then h → 0 put x = − h 2 2

π h 1 − tan  −   4 2  ⋅ 1 − cosh = lim tan  π + π + h  ⋅ 1 − cosh = lim   3 h →0 h →0  π h  (2h)3  4 4 2  8h 1 + tan  −  4 2

2  cosh " 1 − h   2 

h h2 h h2 ⋅ tan ⋅ 2 2 = lim 2 2 = 1 = lim h →0 h → 0 8h 3 8h 3 32 Example - 10

Solution :

Lt

x →0

tan[e 2 ]x 4 − tan[ −e2 ]x 4 = sin 4 x

(1) 0 (3) 8 Ans. (2) e2 = (2.718)2 = 7.3875

(2) (4)

( ∵ tanax " ax , sin x " x )

[e2] = 7, [–e2] = –8

tan 7x + tan 8x 15x = 4 = 15 sin 4 x x 4

∴ Example - 11

Lt

x →0

d  2  −1 1 + x sin  cot d x  1− x 

(1)

−

(3)

1 2

1 2

15 7

4

4

   = 

(2)

0

(4)

–1


NARAYANA



Solution :

Example - 12

Solution :

Ans. (1)

Let

 −1 1 + x 2  y = sin  cot 1− x 

∴

 −1 2 y = sin cot 

∴

y = sin θ =

∴

dy 1 =− . dx 2

2

   . Put x = cos 2θ. 

 1 + cos 2θ      = sin2 cot–1 (cot θ)  1 − cos 2θ  

1 − cos 2θ 1− x 1 x = = − 2 2 2 2

If y = (a − x )( x − b ) - (a – b) tan

–1

dy a−x  then find  dx.  x −b

(1)

x−a    x −b

(2)

a−x   x+b

(3)

a−x    x −b

(4)

x+a x+b

Ans. (3) 2

2

Let

x = a cos θ + b sin θ

∴

a – x = a – a cos θ – b sin θ = (a – b) sin θ

2

2

2

2

2

.....(1)

2

and x – b = a cos θ + b sin θ – b = (a – b) cos θ ∴

.....(2)

–1

y = (a – b) sin θ cos θ – (a – b) tan tanθ =

(a − b ) sin2θ – (a – b) θ 2

d y   d y  d θ  (a − b) cos 2θ − (a − b) 1 − cos 2θ = = = = tan θ Then (b − a) sin 2θ sin 2θ d x d x   dθ

=

Example - 13

a−x    x −b –1

Derivative of sec

 1   2  w.r.t  2x − 1 

[From (1) and (2)] −1 1+ 3 x at x = 3 is-

(1)

0

(2)

1 2

(3)

1 3

(4)

1 6 19


NARAYANA



Solution :

Example - 14

Solution :

Ans. (1)

 1   2 −  and  2x 1 

y = sec

∴

dy dy dx = = dx dz dx

∴

 d y     d x  x

=

1 θ

If x = θ –

−1 3

z=

1 + 3x

−4x 2 −4x × 1 + 3x (2x 2 − 1)2 3 ⋅ = 2 2 3  1   1  − 1  2   2  −1 2 1 + 3x  2x − 1   2x − 1  (2x 2 − 1) ⋅ 1

=0.

and y = θ +

dy 1 , then d x = θ

(1)

x y

(2)

y x

(3)

−x y

(4)

−x y

Ans. (1) x=θ–

1 θ

dx 1 ⇒ dθ =1+ 2 , θ

y=θ+

1 θ

⇒

dy dx

dy dθ dx dθ

∴

Example - 15

–1

Let

If y =

=

1 dy =1– 2 dθ θ 1−

1

θ 1 1+ 2 θ

=

2

1 θ = x = 1 y θ+ θ θ−

  a − b  . tan −1      a + b  (a 2 − b 2 ) 2

d2 y  x  tan   then find . d x2  2 

b cos x

(1)

b sin x

(a + b cos x )

2

(2)

2

(4)

b sin x

(3) Solution :

(b + a cos x )

(a + b cos x )2 −

b cos x

(a + b cos x )

2

Ans. (2) We have y =

2

–1

(a − b ) 2

2

. tan

  a − b   x   tan      2    a + b 


NARAYANA



Let

a−b x u=   tan   a+b 2

∴

y=

∴

dy = du

2 (a − b ) 2

2

. tan −1 u

2

.

1

(a − b ) 1 + u 2

.............(1)

2

2

2

=

.

1

2 (a − b ) 1 + u 2

2

  1   . (a 2 − b 2 )  1 +  a − b tan 2 x    2   a+b  2

=

2

=

1

.

(a − b ) 1 +  a − b      a+b 2

2

[From (1)]

 1 − cos x     1 + cos x 

=

(a + b) (1 + cos x) (a 2 − b 2 ) {(a + b) (1 + cos x) + (a − b) (1 − cos x)}

=

(a + b) (1 + cos x ) (a − b ) (2a + 2b cos x )

2

.

2

2

2

.

.....(2)

and

du 1 1 a−b x a −b sec 2   =  =   .  . dx 2  a + b  (1 + cos x ) 2 a+b

∴

dy dy du = . dx du dx (1 + cos x ) a−b a+b .    .  (a + b cos x )  a + b  a−b

=

........(3)

[from (2) and (3)]

1 a + b cos x

=

d2 y b sin x = dx 2 (a + b cos x)2

Example - 16

Solution :

If x = e

tan−1

 y − x2  dy  2  , then d x equals x  2

2

(1)

x [1 + tan (log x) + sec x]

(2)

2x [1 + tan (log x)] + sec x

(3)

2x [1 + tan (log x)] + sec x

(4)

none of these

Ans. (4) 21


NARAYANA



Taking log on both sides, we get –1

log x = tan

 y − x2     x2    2

2

⇒

tan (log x) = (y – x ) / x

⇒

y = x + x tan (log x)

∴

dy 2 d x = 2x + 2x tan (log x) + x sec (log x)

2

2

2

= 2x [1 + tan (log x)] + x sec (log x) Example - 17

Solution :

dy 2 y x If x e + 2xye + 13 = 0, then d x equals

(1)

−

(3)

−

2xe y − x + 2y ( x + 1)

(2)

x( xe y − x + 2) 2 xe x − y + 2y ( x + 1)

(4)

x ( xe y − x + 2)

2xe x − y + 2y ( x + 1) x( xe y − x + 2)

none of these

Ans. (1) Using partial derivatives, we have

 2xe y + 2ye x + 2xye x  dy = −  dx x 2 e y + 2xe x   2xe y − x + 2y + 2xy   x 2 e y − x + 2x  

=– 

 2xe y − x + 2y (x + 1)  −  =  y−x  x (xe + 2)  Example - 18

Solution :

dy If y = sin x + sin x + ........ , then d x equals-

(1)

sin x 2y + 1

(2)

cos x 2y − 1

(3)

cos x 2y + 1

(4)

none of these

Ans. (2) y = sin x + y 2

2

⇒

y = sin x + y

⇒ y – y – sin x = 0

∴

− cos x cos x dy =− = dx 2y − 1 2y − 1


NARAYANA



Example - 19

2

dy 1 1 4 4 2 , x + y = t + 2 , then d x equalst t

2

If x + y = t – 1

(1)

(2)

x y3

1 x3y

(4)

−

x y

(3) Solution :

1

2

1 x y3

Ans. (3) Squaring the first equation, we have 4

4

2 2

2

x + y + 2x y = t +

Example - 20

Solution :

2

1

2 2

⇒

t +

⇒

x y = –1

∴

dy 2 2y d x = 3 x

t

2

2

+ 2x y = t +

2 2

2 ⇒ y = −

1 t2 1 t2

–2

–2

(from second equation)

1 x2

1 dy ⇒ dx = 3 x y

2

If y = p(x) is a polynomial of degree 3, then 2

d  3 d2 y  y is equal tod x  dx 2 

(1)

p′′′ (x) p′(x)

(2)

p′′(x) p′′′(x)

(3)

p(x) p′′′(x)

(4)

none of these

Ans. (3) 2

Giveny = p(x)

...(i)

p′(x) = 2yy′ p′′(x) = 2yy′′ + 2y′

2

p′′′(x) = 2yy′′′ + 6y′y′′ Also 2

...(ii)

d 3 d  3 d2 y  = 2 ( y y' ' ) y dx d x  d x 2  3

2

= 2 [y y’’’ + 3y y’y’’] 2

= y [2yy’’’ + 6y’y’’] = p(x) p’’’(x) Example - 21

from (i) and (ii)

 x + y  f(x) + f(y) If f  ∀ x,y ∈ R and f ′(0) = −1 , f(0) = 1, then f(2) = = 2  2  (1)

1 2

(2)

1

(3)

–1

(4)

−

1 2 23


NARAYANA



Solution :

Ans. (3)

 2x + 0  f(2x) + f(0) f(x) = f  = 2  2 

∴

 2x + 2h  f − f(x)  2  f(x + h) − f(x) = Lt f ′(x) = Lt h→ 0 h→ 0 h h  f(2x) + f(2h) f(2x) + f(0)  = Lt  − /h h→ 0 2 2   = Lt

2h→ 0

Example - 22

f(2h) − f(0) = f ′(0) 2h

∴

f ′(x) = f ′(0) = −1 ⇒ f(x) = − x + c

⇒

f(0) = c

∴

f(2) = –2 + 1 = –1

⇒ c=1

If f(x + y) = f(x) f(y) and f(x) = 1 + xg(x) H(x) where Lt g(x) = 2 and Lt H(x) = 3 then x →0

x →0

f ′(x) =

Solution :

(1)

f(x)

(2)

2f(x)

(3)

3f(x)

(4)

6f(x)

Ans. (4) f(x + h) − f(x) f(x)f(h) − f(x) f(h) − 1 = Lt = f(x) Lt h → 0 h → 0 h h h

f ′(x) = Lt

h→ 0

= f(x) Lt

h →0

1 + hg(h)H(h) − 1 = f(x) Lt g(h)H(h) h→0 h

= f(x) (2 × 3) = 6f(x) Example - 23

Solution :

If f(a) = 2, f ′(a) = 1 , g(a) = – 1, g′(a) = 2 then the value of Lt g(x)f(a) − g(a)f(x) is x →a x−a (1)

–5

(2)

1 5

(3)

5

(4)

4

Ans. (3) Use L.H.R Lt

x →a

g′(x)f(a) − g(a)f ′(x) = g′(a)f(a) − g(a)f ′(a) 1

= 2(2) – (–1) (1) = 4 + 1 = 5 x

Example - 24

If g(x) =

1 {3t − 2g′(t)} dt then g′(2) = x ∫2

(1)

–2/3

(2)

–3/2

(3)

2/3

(4)

3/2


NARAYANA



Solution :

Ans. (4) x

Given g(x) =

1 {3t − 2g′(t)} dt x ∫2 x

⇒

x g(x) = ∫ {3t − 2g′(t)} dt 2

⇒

g(x) + x g′(x) = {3x − 2g′(x)} (1) − 0

⇒

g(2) + 2 g′(2) = 6 − 2g′(2) 4g′(2) = 6 − g(2) = 6 − 0 = 6

∴

g′(2) =

3 2 1/ x

Example - 25

Solution :

 f(1 + x)    Let f : R → R be such that f(1) = 3 and f ′(1) = 6 then xLt →0  f(1) 

(1)

1

(3)

e

2

(2)

e

(4)

e

=

1/2

3

Ans. (3) 1/ x

 f (1 + x)  Let y =    f (1) 

1 {log f (1 + x) − log f (1)} x

⇒

log y =

⇒

log Lt (y) = Lt

∴

x →0

(

x →0

)

x →0

f ′(1 + x) f ′(1) 6 = = =2 f (1 + x) f (1) 3

Lt (y) = e 2


NARAYANA



EXERCISES LEVEL - I sin x

1.

 sin x  x − sin x = Lt   x →0  x  (1)

1 e2

(2)

(3) e2

1 e

(4) e

1 1 1 + 2 + .... + n 2 2 2 = Lt n →∞ 1 1 1 1 + + 2 + ..... + n 3 3 3 1+

2.

3.

(1)

4 3

(2)

2 3

(3)

1 3

(4)

1 2

Lt

h →0

(a + h)2 sin(a + h) − a2 sina = h

(1) a2 cosa + 2a sina

(2) a (cosa + 2 sina)

2

(3) a (cosa + 2 sina)

(4) 0 1

4.

If α, β be the roots of ax2 + bx + c = 0, then Lt (1 + ax 2 + bx + c) x −α is x →α

(1) a(α – β) a(α – β)

(4) ea(α + β)

(3) e 5.

6.

Lt

x →0

(2) log | a(α – β)|

(1 + a3 ) + 8e1/ x = 2 Then 1 + (1 − b3 )e1/ x

(1) a = 1, b = 2

(2) a = 1, b = –31/3

(3) a = 1, b = –1/2

(4) a = 2, b ∈ R

Lt

sinnθ

n→∞

n

=

(1) 0

(2) ∞

(3) 1

(4)

n


NARAYANA



 x x x  x Lt cos   ⋅ cos   ⋅ cos   ⋅ ⋅ ⋅ ⋅ cos  n 2 4 8 2 

7.

n →∞

  = 

(1) 1

(3)

(2)

x sin x

sin x x

(4) –1

2 n   1 + + ..... + Lt  = 2 2 1 − n2   1− n 1− n

8.

n →∞

(3)

1 2

10.

(2) – 2, –1

(3) 2, 1

(4) 2, –1

Lt

x →0

cos x − 3 cos x = sin2 x 1 6

(3) −

(2)

1 12

1 3

(4) −

1 14

1− x = x →1 (cos −1 x)2 Lt

(1)

1 4

(3) −

1 4

(2)

1 2

(4)

1 3

{[1 x] + [2 2

12.

1 3

(1) –2, 1

(1)

11.

(4)

sin2x + a sin x be finite, then the value of ‘a’ and the limit are given by x3

If xLt →0

9.

1 2

(2) −

(1) 0

If [x] denotes the greatest integer ≤ x , then Lt n →∞

2

}

x] + ....[n2 x]

n3

(1)

x 2

(2)

x 3

(3)

x 6

(4) 0


NARAYANA



 x − x −1  = cos −1  −1  d x x+x  d

13.

14.

15.

16.

1

(1)

1+ x2

(3)

1+ x2

2

If y = e x + e

x + ex +.....∞

−1

(2)

1+ x2

(4)

1+ x2

−2

dy , then d x =

y (1) 1 − y

1 (2) 1 − y

y (3) 1 + y

y (4) y − 1 dy x2 x3 + + ....... ∞, then d x = 2! 3!

If y = 1 + x + (1) y

(2) y – 1

(3) y + 1

(4) none of these

Differential coefficient of sec

1

–1

2x 2 − 1

w.r.t.

1 1− x 2 at x = 2 is-

(1) 2

(2) 4

(3) 6

(4) 1 x

17.

dy 1  If y = 1 +  , then d x = x  1  (1) 1 +  x 

x

1  (3)  x +  x 

18.

x

x

 1 1   log 1 +  −  + x 1 x   

1  (2) 1 +  x 

x   log (x − 1) − x + 1  

1  (4)  x +  x 

x

 1   log 1 +  x     1 1   log 1 +  +  x  1+ x   

dy  2x + 3  If f ′(x) = sin (log x) and y = f  =  , then dx  3 − 2x  9 cos (log x )

(1)

x (3 − 2 x ) 2

2x + 3   9 sin  log  3 − 2x 2   (3) (3 − 2x ) 2

2x + 3   9 cos  log  3 − 2x   (2) x (3 − 2 x )2

(4)

 12  2x + 3   sin  log   2 (3 − 2x)  3 − 2x   


NARAYANA



19.

dy π –1 1/2 If y = cot (cos 2x) , then the value of d x at x = will be6 2 (1)   3

1/ 2

 1 (2)   3

1/2

1/2

(3) (3) 20.

(3)

22.

(4) (6)

dy If y = logcos x sin x, then d x is equal to-

(1)

21.

cot x log cos x + tan x log sin x (log cos x )

(log sin x )

If y = sinn x cosnx , then

(4)

tan x log cos x + cot x log sin x (log cos x )2 cot x (logsin x)2

dy equals dx

(2) nsinn −1 x sin (n + 1) x

(3) nsinn −1 x cos (n − 1) x

(4) nsinn −1 x cosnx

π If f ( x ) = cos x cos 2x cos 4x cos8x cos16x then f '   is 4 (2)

2

1 2

(4) None of these

The value of the derivative of |x – 2| + |x – 3| at x = 3 is (1) 2

(2) –2

(3) 0

(4) 1

x3 Let f(x) = 6 p

sin x cos x d3 (f(x)) at x = 0 is −1 0 where p is constant, then dx 3 2 3 p p 2

(1) p

(2) p + p 3

(3) p + p 25.

2

(2)

(1) nsinn −1 x cos (n + 1) x

(3) 1

24.

2

cot x log cos x + tan x log sin x

(1)

23.

1/ 2

(4) Independent of p 2

2

If f ′(x) = g(x) and g′(x) = −f(x) for all x and f(2) = 4 = f ′(2) then f (19) + g (19) is (1) 16

(2) 32

(3) 64

(4) 8


NARAYANA



LEVEL - II x

1.

2.

3.

4.

Lt

x →a

x f (x) dx = x − a ∫a

(1) 2 f(a)

(2) f(a)

(3) a f(a)

(4) 0

 1 4 9 n2  Lt  3 + 3 + 3 + .... + 3 = n →∞ n + 1 n +1 n +1 n +1  (1) 1

(2) 2/3

(3) 1/3

(4) 0

Lt

x →0

log(1 + {x}) = (where {x} denotes the fractional part of x) {x}

(1) 1

(2) 0

(3) 2

(4) does not exist

a

x

− a1/

x

a (1) 4

x

+ a1/

x

Lt

x →0

, a > 1 is (2) 2

(3) –1 5.

6.

Lt

n →∞

(4) 0

a n + bn , where a > b > 1, is equal to a n − bn

(1) –1

(2) 1

(3) 0

(4) none of these

{

1/ sin 2 x

Lt 1

x →0

+2

1/ sin 2 x

+ .... + n

1/ sin 2 x

}

sin 2 x

is

(1) ∞ (3) 7.

(2) 0

n(n + 1) 2

(4) n

If G(x) = − 25 − x 2 , then Lt

x →1

(1)

1 24

(3) − 24

G(x) − G(1) is x −1 (2)

(4)

1 5 1 24


NARAYANA



If f(9) = 9, f ′(9) = 4 , then Lt

8.

x →9

(1) 3 (3) 6

Lt

9.

x →0

10.

12.

13.

x tan 2x − 2x tan x is (1 − cos 2x)2 (2) –2

(3) 1/2

(4) –1/2

Lt

π 2

sin x − (sin x)sin x = 1 − sin x + log(sin x)

(1) 1

(2) 2

(3) 3

(4) 4

2 Given f ′(2) = 6 , and f ′(1) = 4 , Lt f (2h + 2 + h ) − f (2) is h → 0 f (h − h 2 + 1) − f (1)

(1) 3/2

(2) 3

(3) 5/2

(4) –3

Let g(x) = sin2[π3]x2 – sin2[–π3]x2 , [.] represents greatest integer function then xLt →0 (1) –63

(2) 63

(3) 1

(4) –1

If y = x |x| then

If 2x + 2y = 2x + y then

|x| x (4) 2 |x|

(2)

dy = dx

(1) 2x – y (3) –2x – y If x3y2 = (x + y)5 then (1)

x y

(3)

−

(2) 2y – x (4) –2y – x dy = dx

(2) x y

g(x) = sin 4 x

dy = dx

(3) x|x|

15.

=

(2) 4 (4) 8

(1) |x|

14.

x −3

(1) 2

x→

11.

f (x) − 3

y x

(4) logx


NARAYANA



16.

{

}

d (x + a)(x 2 + a2 )(x 4 + a4 )(x 8 + a8 ) = dx

(1)

15x16 − 16x15 a + a16 (x − a)2

(2)

(3)

x16 − a16 x−a

(4) 16x15 – a16

x

17.

If

x2

x3

f(x) = 1 2x 3x 2 then f ′(x) = 0 2 6x

(1) 0 (3) x2 18.

19.

20.

21.

x16 − x15 a + a16 (x − a)2

(2) 1 (4) 6x2

π π   If f(x) = cos2x + cos3  + x  − sin x sin  x +  and γ (5/4) = 3 then (gof)(x) = 3 3   (1) 0

(2) 1

π  (3) cosx + cos  + x  3 

(4) 3

Let f(x) be a polynomial function of second degree if f(1) = f(–1) and a, b, c are in A.P then f ′(a), f ′(b), f ′(c) are in (1) A.P (2) G.P (3) H.P (4) A.G.P

x  x If cos   ⋅ cos  2 2   2

  x  ⋅ cos  3  2

sin x   .........to ∞ = x then 

(1)

1 x 1  x tan   + 2 tan  2 2 2 2 2

 1  x  + 3 tan  3  2 2

1   + ......∞ = − cot x + x 

(2)

1 1 x 1  x  1  x  tan   + 2 tan  2  + 3 tan  3  + ......∞ = cot x − 2 2 x 2 2 2 2      

(3)

1 x 1  x sec 2   + 4 sec 2  2 2 2 2 2 2

(4)

1 1 1 x 1  x  sec 2   + 4 sec 2  2  + ......∞ = cosec 2 x + 2 + 3 2 22 2 2 x x    

1  2  + ...... + ∞ = cos ec x − 2 x 

A triangle has two of its vertices at P(a, 0), Q(0, b) and the third vertex R(x, y) is moving along the st.line y = x, if A be the area of the ∆, then

dA = dx

(1)

a−b 2

(2)

a−b 4

(3)

a+b 2

(4)

a+b 4


NARAYANA



22.

If f ′(x) = φ(x) and φ′(x) = f(x) ∀x , also f(3) = 5 and f ′(3) = 4 , then the value of [f(10)]2 – [φ(10)]2 = (1) 0 (3) 41

23.

(2) 9 (4) 25

π 5  If f(x) = sin  [x] − x  , 1 < x < 2 and [x] denotes the greatest integer less than or equal to x, then 2   π f ′ 5  =    2 (1)

π 5  2

4/5

(3) 0

(2)

π −5   2

(4)

π 3  2

4/5

4/5

dy −1  a sin x + bcos x  = If y = tan  then  dx  acos x − b sin x 

24

25.

(1) 1

(2) –1

(3) 0

(4)

a a cos x − b sin x

If y =

x a+

then

x b+

dy = dx

x a+

x b + ........

(1)

b a(b + 2y)

(2)

b b + 2y

(3)

a b(b + 2y)

(4)

ab a + b2y


NARAYANA



LEVEL - III 1.

  π   π  Lt  x cos   sin    =  4x   4x   

x →∞

1 π

(1)

(2)

(4) π

(3) 1 2.

3.

Lt

x →∞

π 4

[x] = x

(1) 1

(2) 3

(3) –1

(4) does not exist

π   − x 2  = Lt  π sec x − tan x x→ 2

4.

5.

6.

7.

(1) 1

(2) –1

(3) 2

(4) 3

xe3x − x

Lt

1 + x2 −1

x →0

=

(1) 3

(2) 6

(3) 4

(4) 1

Lt

x →0

e|x| − 1 = x

(1) 1

(2) –1

(3) does not exist

(4) 0

11/ x + 21/ x + .... + n1/ x  Lt   x →∞ n  

nx

is

(1) n

(2) n

(3) n − 1

(4) 0

  Lt  x + x + x − x  =  

x →∞

(1)

1 2

(3) 0

(2) 1 (4) does not exist


NARAYANA



Lt

8.

x →2

2 x + 23 − x − 6 = 2− x / 2 − 21− x

(1) 8 (3) 2

(2) 4 (4) 6

4 + 3a n an = If a n +1 = 3 + 2a then nLt →∞

9.

n

(1) 0

(2) –2

(3) 10.

(4) − 2

2

 1  1 1 + + .... + + .....n  x > 0, then Lt f (x) = If f (x) = x  (1 + 2x)(1 + 3x) n →∞ 1 + x (1 + x)(1 + 2x) 

1 1− x (3) 1

1 1+ x (4) 0

(1)

11.

(2)

 x 4 sin(1/ x) + x 2  lim  is The value of x →−∞  1+ | x |3   (1) 1 (3) 0

(2) –1 (4) ∞

11/ cos x + 21/ cos x + .... + n1/ cos x  The value of xlim π   → 2

12.

2

2

cos2 x

is

2

13.

14.

(1) 0

(2) n

(3) ∞

(4)

If siny = xcos(a + y) then

n(n + 1) 2

dy = dx

(1)

sin2 (a + y) sina

(2)

cos2 (a + y) cosa

(3)

cos2 (a − y) cosa

(4)

cos2 (a + y) sina

If sin2mx + cos2ny = a2 then

dy = dx

(1)

m sin 2mx n sin 2ny

(2)

n sin 2ny m sin 2mx

(3)

n sin 2mx m sin2ny

(4)

−

m sin 2mx n sin2ny 35


NARAYANA



15.

16.

17.

18.

19.

20.

21.

 xx − x−x  If f(x) = cot −1   then f ′(1) = 2  

(1) –1

(2) 1

(3) log2

(4) –log2

4x  dy −1  = If y = logcot x tan x ⋅ logtan x cot x + tan  2  then − 4 x dx   (1)

1 4 + x2

(2)

4 4 + x2

(3)

1 4 − x2

(4)

4 4 − x2

If 3f(cosx) + 2f(sinx) = 5x then f ′(cos x) = (1)

−

5 cos x

(2)

5 cos x

(3)

−

5 sin x

(4)

5 sin x

If f(x) = (cosx + sinx) (cos3x + isin3x) .... (cos[(2n – 1)x] + isin(2n – 1)x] then f ′′(x) = (1) n2f(x)

(2) –n4f(x)

(3) –n2f(x)

(4) n3f(x)

Let φ(x) be the inverse of the function f(x) and f ′(x) =

d 1 [φ(x)] = 5 then dx 1+ x 1

(1)

1 1 + (φ(x))5

(2)

1 + ( f(x) )

(3)

1 + (φ(x) )

(4)

1 + ( f(x) )

If

5

x 2 + y 2 = ae tan

−1

(y / x)

5

5

,a > 0 then y ′′(0) is

(1)

a −π / 2 e 2

(2)

aeπ / 2

(3)

2 − e −π / 2 a

(4)

2 π/2 e a

2 If y = a cos(logx) + b sin(logx) then x

d2 y dy +x = dx dx 2

(1) 0

(2) y

(3) –y

(4) 1


NARAYANA



22.

23.

24.

25.

If siny = x sin(a + y) and

dy A = then the value of A is dx 1 + x 2 − 2x cos a

(1) 2

(2) cosa

(3) sina

(4) –2

If y =

ax + b , where a, b, c are constants then (2xy′ + y)y′′′ is equal to x2 + c

(1) 3(xy′′ + y′)y′′

(2) 3(xy′ + y′′)y′′

(3) 3(xy′′ + y′)y′

(4) none of these

dy is equal to dx

If

1 − x 6 + 1 − y 6 = a 3 (x 3 − y3 ) , then

(1)

x 2 1 − y6 y2 1 − x 6

(2)

(3)

x2 1 − x6 y 2 1 − y6

(4) none of these

y 2 1 − y6 x2 1 − x6

 u(x)  Let f(x) = log   , u′(2) = 4, ν ′(2) = 2 , u(2) = 2, v(2) = 1, then f ′(2) is  ν(x) 

(1) 0

(2) 1

(3) -1

(4) 2


NARAYANA



QUESTIONS ASKED IN AIEEE & OTHER ENGINEERING EXAMS 1.

If xm ⋅ yn = (x + y)m+n , then y x

(2)

x+y xy

(3) xy

(4)

x y

(2)

1 cos ec1 2

(4)

1 tan1 2

(1)

2.

dy is dx

1 2 4 1 1  lim  2 sec 2 2 + 2 sec 2 2 + .... + sec 2 1 is n n n n n 

n→∞

(1)

1 sec1 2

(3) tan1

3.

4.

5.

[AIEEE - 2006]

Let α and β be the distinct roots of ax + bx + c = 0 then xlim →α 2

(1)

a2 (α − β)2 2

(2) 0

(3)

−a2 (α − β)2 2

(4)

 a b If lim  1 + + 2  x→∞  x x 

1 − cos(ax 2 + bx + c) is equal to (x − a)2

1 ( α − β )2 2

= e2 , then the values of a and b, are (2) a = 1, b ∈ R

(3) a ∈ R, b ∈ R

(4) a = 1 and b = 2.

lim

[AIEEE - 2005]

2x

(1) a ∈ R, b = 2

x →π / 2

[AIEEE - 2005]

[AIEEE - 2004]

  x  1 − tan  2   [1 − sin x]   is   x  3 + π − 1 tan [ 2x]  2     

(1)

1 8

(2) 0

(3)

1 32

(4) ∞

[AIEEE - 2003]


NARAYANA



log(3 + x) − log(3 − x) = k , the value of k is x

If lim

6.

x →0

(1) 0

(2) −

1 3

2 3

(4) −

2 3

(3)

[AIEEE - 2003]

Let f(a) = g(a) = k and their nth derivatives fn(a), gn(a) exist and are not equal for some n. Further if

7.

lim

x →a

f(a)g(x) − f(a) − g(a)f(x) + g(a) = 4 , then the value of k is g(x) − f(x)

(1) 4

(2) 2

(3) 1

(4) 0

If f(x) = xn , then the value of f(1) −

8.

(1) 2n (3) 0

10.

11.

f ′(1) f ′′(1) f ′′′(1) ( −1)n f n (1) + − + .... + is 1! 2! 3! n! (2) 2n–1 (4) 1

If x = e y + e

9.

y +....to ∞

, x > 0 then

1− x x

(2)

1 x

(3)

x 1+ x

(4)

1+ x x

1 − cos 2x

x→ 0

2x

[AIEEE - 2003]

dy is dx

(1)

lim

[AIEEE - 2003]

[AIEEE - 2003]

is

(1) λ

(2) –1

(3) zero

(4) does not exist

 x 2 + 5x + 3  lim  2  x →∞  x + x + 2 

[AIEEE - 2002]

x

is

(1) e4

(2) e2

(3) e3

(4) e

[AIEEE - 2002]

x

12.

 x − 3 For x ∈ R , lim   is x→∞  x + 2  (1) e

(2) e–1

(3) e–5

(4) e5

[AIEEE - 2002]


NARAYANA



13.

14.

15.

Let f(2) = 4 and f ′(2) = 4 . Then lim

x→2

xf(2) − 2f(x) is given by x−2

(1) 2

(2) –2

(3) –4

(4) 3

2 If y = (x + 1 + x 2 )n , then (1 + x )

d2 y dy +x is 2 dx dx

(1) n2 y

(2) –n2y

(3) –y

(4) 2x2y

If sin y = x sin(a + y) , then

[AIEEE - 2002]

dy is dx

sina sin (a + y)

(2)

sin2 (a + y) sina

(3) sina sin2 (a + y)

(4)

sin2 (a − y) sina

(2)

1 − log x 1 + log x

sina

(1)

16.

[AIEEE - 2002]

2

[AIEEE - 2002]

dy is If x y = ex − y , then dx

1+ x 1 + log x

(1)

log x

(3) not defined

17.

(4)

19.

[AIEEE - 2002]

 e x − esin x  lim   is equal to x →0  x − sin x  (1) – 1 (3) 1

18.

(1 + log x)2

lim x →0

(2) 0 (4) none of these

[UPSEAT - 2004]

cos 2x 3 − 1 is equal to sin 6 2x

(1)

1 16

(2) −

1 16

(3)

1 32

(4) −

1 32

4   lim  1 −  x →∞  x −1 

[CEET (Haryana) - 2004]

3x −1

is equal to:

(1) e12

(2) e−12

(3) e 4

(4) e3

[CET (Karnataka) - 2004]


NARAYANA



n

20.

21.

a  lim  1 + sin  equals to: n →∞ n   (1) ea (2) e 2a (4) 0 (3) e The differential coefficient of f(sinx) with respect to x where f(x) = logx is (1) tanx (2) cotx (3) f(cosx)

22.

(4)

If x = A cos 4t + B sin4t, then

(2) 16x (4) –x

x If ∆1 = a a

b b x x b and ∆ 2 = a a x

24.

(1)

x log e x then

d ∆1 = 3∆ 2 dx

(4) ∆1 = 3(∆ 2 )3/ 2

[UPSEAT - 2000]

dy at x = e is dx 1

1 e

(3)

25.

(2)

d ∆1 = 3(∆ 2 ) 2 dx

If y =


b are given, then x

(1) ∆1 = 3(∆ 2 ) 2

(3)


d2x is equal to dt 2

(1) –16x (3) x

23.

1 x

[CEE (Delhi) - 2004]

(2)

e

(4) none of these

e −1

[BIT (Mesra) - 2000]

 log(e / x 2 )  d2 y −1  3 + 2log x  + tan , = then   2  dx 2  1 − 6log x   log(ex ) 

If y = tan  (1) 2 (3) 0

(2) 1 (4) –1

[CEE (Delhi) - 2004]


NARAYANA



ANSWERS EXERCISES LEVEL - I 1. (2)

2. (1)

3. (1)

4. (3)

5. (2)

6. (1)

7. (2)

8. (2)

9. (2)

10. (3)

11. (1)

12. (2)

13. (4)

14. (1)

15. (1)

16. (1)

17. (1)

18. (4)

19. (1)

20. (1)

21. (1)

22. (1)

23. (3)

24. (4)

25. (2)

LEVEL - II 1. (3)

2. (3)

3. (4)

4. (3)

5. (2)

6. (3)

7. (4)

8. (2)

9. (3)

10. (2)

11. (2)

12. (2)

13. (4)

14. (4)

15. (2)

16. (1)

17. (4)

18. (4)

19. (1)

20. (1)

21. (3)

22. (2)

23. (2)

24. (1)

25. (1)

LEVEL - III 1. (2)

2. (1)

3. (3)

4. (2)

5. (3)

6. (1)

7. (1)

8. (1)

9. (3)

10. (3)

11. (2)

12. (2)

13. (2)

14. (1)

15. (1)

16. (2)

17. (3)

18. (2)

19. (3)

20. (3)

21. (3)

22. (3)

23. (1)

24. (1)

25. (2)

QUESTIONS ASKED IN AIEEE & OTHER ENGINEERING EXAMS 1. (2)

2. (4)

3. (1)

4. (2)

5. (3)

6. (3)

7. (1)

8. (3)

9. (1)

10. (4)

11. (1)

12. (3)

13. (3)

14. (1)

15. (2)

16. (4)

17. (3)

18. (4)

19. (2)

20. (1)

21. (2)

22. (1)

23. (2)

24. (2)

25. (3)


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