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Now put x=x-2 → limits also change from 0 to 2
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This question can be easily solved using LH rule i.e L.Hospital’s rule which says:
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Find ‘n’, if
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Now cosx=1-2sin2 (x/2)
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Given,
Differentiate each of the functions w.r. to x in
Let y=
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Differentiate each of the functions w.r. to x in
Let
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= 3x2 – 3x-4 + 3 – 3x-2
Differentiate each of the functions w.r. to x in
(3x + 5) (1 + tan x)
Given
Applying product rule of differentiation i.e
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Differentiate each of the functions w.r. to x in
(sec x – 1) (sec x + 1)
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Differentiate each of the functions w.r. to x in
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Applying division rule of differentiation i.e
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Differentiate each of the functions w.r. to x in
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Applying division rule of differentiation i.e
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Differentiate each of the functions w.r. to x in
Given
Applying division rule of differentiation i.e
Differentiate each of the functions w.r. to x in
(ax2 + cotx) (p + q cosx)
Given
Applying product rule of differentiation i.e
Differentiate each of the functions w.r. to x in
Given
Applying division rule of differentiation i.e
Differentiate each of the functions w.r. to x in
(sin x + cosx)2
Given
Applying the concept of chain rule
Differentiate each of the functions w.r. to x in
(2x – 7)2 (3x + 5)3
This question will involve the concept of both chain rule and product rule.
Given
Applying product rule of differentiation i.e
Differentiate each of the functions w.r. to x in
x2 sinx + cos2x
This question will involve the concept of both chain rule and product rule.
Given
Applying product rule of differentiation i.e
Differentiate each of the functions w.r. to x in
sin3x cos3x
This question will involve the concept of chain rule .
Given
Differentiate each of the functions w.r. to x in
This question will involve the concept of chain rule .
Given
Differentiate each of the functions with respect to ‘x’
Differentiate using first principle cos (x2 + 1)
Let f(x) = Cos (x2 + 1) ------------(i)
f(x + ∆x) = Cos [(x + ∆x)2 + 1] -------(ii)
Subtracting eq. (i) from eq. (ii),
f(x + ∆x) - f(x) = Cos [(x + ∆x)2 + 1] - Cos (x2 + 1)
Dividing both sides by ∆x,
As per the definition of differentiations,
Taking limits,
= -2 Sin (x2 + 1).1.(x) = -2x Sin (x2 + 1)
As
Hence, the required answer is -2x Sin (x2 + 1).
Differentiate each of the functions with respect to ‘x’
Differentiate using first principle
Let ----- (i)
------- (ii)
Subtracting eq. (i) from eq. (ii),
Dividing both sides by ∆x and taking the limit,
Using differentiation,
Taking limits, we have,
Hence the answer is.
Differentiate each of the functions with respect to ‘x’
Differentiate using first principle
Let f(x) = x2/3 ------ (i)
f(x + ∆x) =(x + ∆x)2/3 ------ (ii)
Subtracting eq. (i) from eq. (ii),
f(x + ∆x) - f(x) = (x + ∆x)2/3 - x2/3
Dividing both sides by ∆x and taking the limit,
By differentiation,
Expanding by binomial theorem, and rejecting the higher powers of ∆x as ∆x → 0
Hence, the answer is .
Differentiate each of the functions with respect to ‘x’
Differentiate using first principle x cos x.
Let y = x Cosx ----- (i)
y + ∆y = (x + ∆x) Cos (x + ∆x) ----- (ii)
Subtracting eq. (i) from eq. (ii),
y + ∆y – y = (x + ∆x) Cos (x + ∆x) - x Cosx
∆y = xCos (x + ∆x) + ∆x Cos (x + ∆x) – x Cosx
Dividing both sides by ∆x and take the limits,
Taking the limits,
=x(-Sin x)+Cos x
= -x Sinx + Cos x
Hence the answer is -x Sinx + Cos x.
Evaluate each of the following limits
Taking the limits,
= x Sec x tanx + Secx
= Secx (x tan x + 1)
Hence, the answer is Secx (x tan x + 1).
Evaluate each of the following limits
Hence, the answer is .
Evaluate each of the following limits
{Taking limits}
Hence, the answer is -4.
Evaluate each of the following limits
Hence, the answer is .
Evaluate each of the following limits
Show that does not exists
{|x - 4| = -(x - 4) if x < 4}
{|x - 4| = (x - 4) if x > 4}
LHL ≠ RHL
Hence, the limit does not exist.
Let and if find the value of k.
and if
k = 6
Hence the answer is 6.
Evaluate each of the following limits
Let find ‘c’ if exists.
f(x) = x + 2, x ≤ -1
f(x) = cx2, x > -1
=1
= c
As the limits exist,
LHL = RHL
c = 1
Hence the answer is 1.
Choose the correct answer out of 4 options given against each Question
is
A. 1
B. 2
C. –1
D. –2
= -1
Hence, the answer is Option (C).
Choose the correct answer out of 4 options given against each Question
is
A. 2
B.
C.
D. 1
= 2 Cos0 = 2× 1 = 2
Hence Option (A) is the correct answer.
Choose the correct answer out of 4 options given against each Question
is
A. n
B. 1
C. –n
D. 0
= n(1)(n-1)
= n
Hence Option (A) is the correct answer.
Choose the correct answer out of 4 options given against each Question
is
A. 1
B.
C.
D.
Hence Option (B) is the correct answer.
Choose the correct answer out of 4 options given against each Question
is
A.
B.
C.
D. –1
Hence Option (A) is the correct answer.
Choose the correct answer out of 4 options given against each Question
is
A.
B. 1
C.
D. –1
Sin 2x = 2 Sinx Cosx
Hence Option (C) is the correct answer.
Choose the correct answer out of 4 options given against each Question
is
A. 2
B. 0
C. 1
D. –1
Taking limits,
Hence Option (C) is the correct answer.
Choose the correct answer out of 4 options given against each Question
is
A. 3
B. 1
C. 0
D.
= 1 + 1
= 2
Hence Option (D) is the correct answer.
Choose the correct answer out of 4 options given against each Question
is
A.
B.
C. 1
D. None of these
Taking limits,
Hence Option (B) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If where [.] denotes the greatest integer function, then is equal to
A. 1
B. 0
C. –1
D. None of these
f(x) = 0, [x] = 0
= -1
= 1
LHL ≠ RHL
So, the limit does n’t exist.
Hence Option (D) is the correct answer.
Choose the correct answer out of 4 options given against each Question
is
A. 1
B. –1
C. does not exist
D. None of these
= -1
= 1
LHL ≠ RHL
So, the limit does n’t exist.
Hence Option (C) is the correct answer.
Choose the correct answer out of 4 options given against each Question
Let the quadratic equation whose roots are
A. x2 – 6x + 9 = 0
B. x2 – 7x + 8 = 0
C. x2 – 14x + 49 = 0
D. x2 – 10x + 21 = 0
f(x) = x2 – 1, 0 < x < 2
f(x) = 2x + 3, 2 ≤ x < 3
=[(2 - h)2 - 1] = (4 + h2 - 4h - 1)
=(h2 - 4h + 3) = 3
And, f(x) = 2x + 3
= [2(2 + h) + 3]
= 7
the quadratic equation whose roots are 3 & 7 is x2 – (3 + 7)x + 3(7) = 0, i.e.., x2 – 10x + 21 = 0.
Hence Option (D) is the correct answer.
Choose the correct answer out of 4 options given against each Question
is
A. 2
B.
C.
D.
2x → 0
Hence Option (B) is the correct answer.
Choose the correct answer out of 4 options given against each Question
Let f(x) = x – [x], is
A. 3/2
B. 1
C. 0
D. –1
f(x) = x – [x]
Checking for differentiability of f(x) at ,
= 1
= 1
LHD = RHD
Hence Option (B) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If at x = 1 is
A. 1
B.
C.
D. 0
= 0
Hence Option (D) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If then f’(1) is
A.
B.
C. 1
D. 0
f’(x) at x = 1 =
Hence Option (A) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If
A.
B.
C.
D.
Hence Option (A) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If at x = 0 is
A. –2
B. 0
C.
D. does not exist
= -2
Hence Option (A) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If at x = 0 is
A. cos 9
B. sin 9
C. 0
D. 1
= Cos 9
Hence Option (A) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If then f’(1) is equal to
A.
B. 100
C. does not exist
D. 0
f’(1) = 1 + 1 + 1 + ……. + 1 (100 times)
= 100
Hence Option (A) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If for some constant ‘a’, then f’(a) is
A. 1
B. 0
C. does not exist
D.
It does not exist.
Hence Option (C) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If f(x) = x100 + x99 + … x + 1, then f’(1) is equal to
A. 5050
B. 5049
C. 5051
D. 50051
f(x)=x100 + x99 + ………….x + 1
f'(x) = 100x99 + 99x98 +…… + 1
f'(1) = 100(1)99 + 99(1)98 + …… + 1
[Using Arithmetic Progression, where d = -1, a = 100 & n = 100]
= 50 (200 – 99)
= 50 (101)
= 5050
Hence Option (A) is the correct answer.
Choose the correct answer out of 4 options given against each Question
If f(x) = 1 – x + x2 – x3 … –x99 + x100, then f’(1) is equal to
A. 150
B. –50
C. –150
D. 50
f(x) = 1 – x + x2 – x3 + …….- x99 + x100
f’(x) = -1 + 2x – 3x2 + ……. – 99x98 + 100x99
f’(1) = -1 + 2(1) – 3(1)2 + ……. – 99(1)98 + 100(1)99
= (-1 – 3 – 5…… -99) + (2 + 4 + 6 +….. + 100)
=
[Using Arithmetic Progression, where d = -2 & 2, a = -1 & 2 & n = 50 respectively]
= 25 (-2 – 98) + 25 (4 + 98)
= 25 (-100) + 25 (102)
= 25 (-100 + 102)
= 25 (2)
= 50
Hence Option (D) is the correct answer.
Fill in the blanks
If then ________
= 1
Hence, the answer is 1.
Fill in the blanks
then m = ________
Rationalizing the denominator,
Hence, the answer is .
Fill in the blanks
If _______
Hence, the answer is y.
Fill in the blanks
___________
------(i)
By using equation (i)
= 1
Hence, the answer is 1.