MATHEMATICS
The following questions given below consist is a point of extremum for y = f(x). If f(x) is
of an "Assertion" (A) and "Reason" (R) non-linear.
Type questions. Use the following Key to
Reason : xlim
k
g(x) does not exists but xlim
a
choose the appropriate answer.
(A) If both (A) and (R) are true, and (R) is the g(f(x)) exists, f(x) will approach k when x a
correct explanation of (A). through only one side.
Sol.[D] Because maximum and minima is also
(B) If both (A) and (R) are true but (R) is not the
correct explanation of (A). dependent of f(a).
(C) If (A) is true but (R) is false.
(D) If (A) is false but (R) is true. x b
Q.5 Assertion : lim does not exist, (where
x 0 a x
Lim [.] denotes the greatest integer function).
Q.1 Assertion : If f(x) is an odd function and x0
1
Reason : lim does not exist.
f(x) exists, then this limit must be zero. x 0 x
Reason : Odd function are symmetrical w.r.t. x b b b x b b
Sol.[D] lim = lim =
origin. [A] x 0 a x x x 0 a a x a
.
Q.2 Assertion : If [x] represents greatest integer
lim
x, then [ |x – 1| + | x |] doesn't exist. 1 x
x 1 Q.6 Assertion : lim sec does not exist.
x x 1
Reason : RHL and LHL both are 1. [D]
Reason : sec–1t is defined for those t, whose
modulus value is more than or equal to 1.
xn
Q.3 Assertion: nlim
= 0 for every n > 0.
n! Sol.[A] Reason is true and correct reasoning for
x
Reason : Every sequence whose nth term Assertion, because lim = 1–.
x x 1
contains n! in the denominator converges to
Hence, (A) is the correct answer.
zero. [C]
Sol. The assertion is true since for any x > 0 we can 1/ x
Q.7 Assertion : lim tan x =e
xn x 0 4
choose sufficiently larger n such that is
n!
Reason : xlim
lim f ( x ).g ( x )
a
(1 + f(x))g(x) = e xa
n ! 2 if xlim f(x) = 0 and xlim g(x) = .
small. The reason is false, since a a
n!
1/ x 1/ x
1 tan x 2 tan x
contains n! in the denominator but diverges to Sol.[D] lim lim 1
x 0 1 tan x x 0 1 tan x
. 1/ x
2 tan x
= xlim
0
exp
= e2
x (1 tan x )
Q.4 Assertion : xlim lim
a f(x) exists = k, but x k g(x)
Q.8 Assertion : The value of xlim
/ 2
(sin x)tan x is e.
does not exists. If xlim
a
g(f(x)) exists, then x = a
Reason : lim
lim g ( x ) f ( x )
x a
(1 + f(x))g(x) is e x a .
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If xlim
a
f(x) = 0 and xlim
a
g(x) = . Reason : If lim f (g ( x )) f lim g ( x ) =
x a x a
Sol. [D] f(m) provided ‘ f ’ is continuous at x = m.
Sol. [D]
1 cos(2 x 2)
Q.9 Assertion : xlim
1 exists
x 1
Questions Add (24–6-09)
Reason : xlim
a
f(x) exists if the left-hand limit
is equal to right-hand limit. n n 1
Q.13 Assertion(A) : lim x n x 1
= 0,
Sol. [D] x n
[x ]
nI
lim 1 Reason(R) : x – 1 < [x] x
Q.10 Assertion : x 0 x sin = 1.
x where [] represents greatest integer function.
1
Reason : ylim
y sin y = 1. Sol.[D] For all x, x – 1 < [x] x,
xn – 1 < [xn] xn
Sol. [D]
1 1 1
n n n
f (x) x [x ] x 1
Q.11 Assertion : For the existence of xlim
a g ( x )
,
x n nx n 1 1
xlim
xlim
xn
f ' (x)
It is necessary that xlim exists. x n nx n 1 1
a g ' ( x )
[x n ]
Reason : If f(0) = g(0) = 0 then
< xlim
lim f ( x ) = lim f ' ( x ) .Provided that
x 0 g ( x ) x 0 g ' ( x ) x n nx n 1 1
x n 1
f(x) and g(x) are continuous function.
n n 1
Sol. [D] lim x nx 1
= lim
x n x
x
1 cos (1 cos x ) x n nx n 1 1
Q.12 Assertion : lim 4 =1
x 0 x x n 1
hence by sandwich theorem xlim
1 cos x 2 x n nx n 1 1
1 cos .x =1
= x2 = [x n ]
lim 4
x 0 x hence (A) is false and (R) is clearly true.
1 cos 2 x
Q. 14 Assertion (A) : xlim
0 =1
x2 2x
1 cos
1 1 sin x
2 Reason (R) : xlim
0 x = 0
lim
x 0 x2 4 8
| sin x |
2 Sol.[D] lim = 1 or –1 as x 0+ or x 0¯
x 0 x
Because
Q.15 Statement-1 : The graph of the function
y = f(x) has a unique tangent at the point (a, 0)
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through which the graph passes then log e (3 x ) x 2 n sin x
log e (1 6(f ( x )) Q.19 Statement - I : nlim
Lt =2 1 x 2n
x a 3f ( x ) = log (3 + x) x R
Statement-2 : Since the graph passes through
Statement - II : For –1 < x < 1, nlim 2n
x = 0
(a, 0). Therefore f(a) = 0, when f(a) = 0 given
limit is zero by zero form. So that it can be
log(3 x ) x 2n sin x
evaluate by using L'Hospital's rule. [B] Sol.[D] I. lim
n 1 x 2n
Lt 1 cos 2 x f(1 + h) = lim lim
Q.16 Statement-1 : does not h 0 n
x 0
x
exists. log(3 1 h ) (1 h ) 2 n sin(1 h )
Statement-2 : | sin x | = 1 (1 h ) 2 n
= – sin 1
sin x, 0x lim lim
2 f(1 – h) = h 0 n
sin x, x 0 log(3 1 h ) (1 h ) 2 n sin(1 h )
2
1 (1 h ) 2 n
[B]
Q.17 Statement – I : when | x | < 1 then log 3 0
= log 3 False
2n
1 0
lim log(x 2) x cos x log(x 2) II. – 1 < x < 1 0 x2 < 1
n 2n
x 1 lim x2n = lim (x2)n = 0 True
n n
Statement – II : –1 < x < 1, then n x2n 0
Q.20 Statement 1 : when |x| < 1,
Sol.[A] statement II is a fact so true
log ( x 2) x 2 n cos x
2n Lt = log (x + 2).
log(x 2) x cos x n x 2n 1
statement I lim
n x 2n 1 Statement 2 : For – 1 < x < 1, as x , x2n 0.
log( x 2) [B]
= lim log( x 2)
n 1
Ans (both correct & correct explaination) Q.21 Statement 1 : When | x | < 1,
log(x 2) x 2n cos x
3
5x 3x 1 5 Lt = log (x + 2)
Q.18 Statement: I – lim
3 2
n x 2n 1
x 2 x 5x 2 2
Statement: II – If P(x) and Q(x) are Statement 2 : For – 1 < x < 1, as x ,
polynomials of same degree, then x2n 0. [B]
P( x )
lim is a non-zero finite real number.
x Q( x )
5x 3 3x 1
Sol. [A] I. Lim
x 2 x 3 5x 2 2 1 cos 2 x does
Q.22 Statement-1 : xlim
0
not
x
3 1
x35 3 exists.
x x 500 5
Lim Statement-2 :
x 3 5 2 200 2
x 2 3
x x
II. is a explanation of I
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If xlim f(x) = 0 and xlim g(x) = [B]
sin x , 0x a a
| sin x | = 2
sin x , x 0 Q.
2
[B]
Q.23 Statement-1 : The graph of the function
y = f(x) has a unique tangent at the point
(a, 0) through which the graph passes then
lim log e (1 6 (f ( x )) = 2
x a 3f ( x )
Statement-2 : Since the graph passes through
(a, 0). Therefore f(a) = 0, when f(a) = 0 given
limit is zero by zero form. So that it can be
evaluate by using L’Hospital’s rule. [B]
Q. 24 Statement-1 : when | x | < 1, nlim
log ( x 2) x 2 n cos x = log(x + 2)
x 2n 1
Statement-2 : For –1 < x < 1,
as n , x2n 0. [B]
Q.25 Statement 1 : lim x sin 1 = 1
x 0 x
1
Statement 2 : ylim = 1
y sin
y
[A]
1 cos 2 x
Q.26 Statement 1 : xlim
0 2 exist's.
x
Statement 2 : xlim
a
f(x) exists if the left hand
limit is equal to right hand limit. [A]
Q.27 lim (sinx)tanx is 1.
Statement 1 : Value of x / 2
Statement 2: xlim
a
(1 + f(x))g(x) is
lim f ( x ) g ( x )
e x a ,
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