Discrete Mathematics MCQ Set 1
1. Expansion of function f(x) is
a) f(0) + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)
b) 1 + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)
c) f(0) – x⁄1! f‘ (0) + x2⁄2! f” (0)…….+(-1)^n xn⁄n! fn (0)
d) f(1) + x⁄1! f‘ (1) + x2⁄2! f” (1)…….+xn⁄n! fn (1)
Answer
Answer: a [Reason:] By Maclaurin’s series, f(0) + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)
2. The necessary condition for the maclaurin expansion to be true for function f(x) is
a) f(x) should be continuous
b) f(x) should be differentiable
c) f(x) should exists at every point
d) f(x) should be continuous and differentiable
Answer
Answer: d [Reason:] By Maclaurin’s series, f(0) + x⁄1! f‘ (0) + x2⁄2! f” (0)…….+xn⁄n! fn (0)
Where, f(x) should be continuous and differentiable upto nth derivative.
3. The expansion of f(a+h) is
Answer
Answer: a [Reason:] By taylor expansion,
f(a+h) = f(a) + h⁄1! f’ (a) + h2⁄2! f” (a)…….
4. The expansion of eSin(x) is
a) 1 + x + x2⁄2 + x4⁄8 +….
b) 1 + x + x2⁄2 – x4⁄8 +….
c) 1 + x – x2⁄2 + x4⁄8 +….
d) 1 + x + x3⁄6 – x5⁄10 +….
Answer
Answer: b [Reason:] Now f(x) = eSin(x), f(0) = 1
Hence, f‘ (x)=f(x)Cos(x), f‘ (0) = 1
f” (x)=f‘ (x)Cos(x) – f(x)Sin(x),f” (0)=1
f”’ (x)=f” (x)Cos(x) – 2f‘ (x)Sin(x) – f(x) cos(x),f”’ (x) = 0
f”” (x)=f”’ (x)Cos(x) – 3f” (x)Sin(x) – 3f‘ (x) cos(x) + f(x) sin(x), f”” (x) = -3
Hence,
f(x) = eSin(x) = 1 + x + x2⁄2 – x4⁄8 +…. (By mclaurin’ sexpansion)
5. Expansion of y = Sin-1(x) is
Answer
Answer: a [Reason:] Given, y = Sin-1(x), hence at x = 0, y = 0
Now, differentiating it, we get
6. Find the expansion of f(x) = ln(1+ex)
Answer
Answer: a [Reason:] Given, f(x) = ln(1+ex), f(0) = ln(2)
Differentiating it we get
7. Find the expansion of exSin(x)
Answer
Answer: b [Reason:] Given, f(x) = exSin(x), f(0) = 1
8. Given f(x)= ln(cos(x) ),calculate the value of ln(cos(π⁄2)).
a) -1.741
b) 1.741
c) 1.563
d) -1.563
Answer
Answer: a [Reason:] Given f(x) = ln(Cos(x)), f(0) = 0
Differentiating it f'(x) = – tan(x), f'(0) = 0
9. Find the expansion of cos(xsin(t)).
Answer
Answer: b [Reason:]
10. Find the expansion of Sin(lSin-1 (x)).
Answer
Answer: b [Reason:] Given, y = f(x) = Sin(lSin-1 (x))
Now, differentiating,
11. Expand (1 + x)1⁄x, gives
a) e [1 + x⁄2 + 11x2⁄24 -…..].
b) e [1 – x⁄2 + 11x2⁄24 -…..].
c) e [x⁄2 – 11x2⁄24 -…..].
d) e [x⁄2 + 11x2⁄24 -…..].
Answer
Answer: b [Reason:] Given, y = (1 + x)1⁄x
12. Find the solution of differential equation, dy⁄dx = xy + x2, if y = 1 at x = 0.
Answer
Answer: d [Reason:] Given dy⁄dx = xy + x2
hence, dy⁄dy (x=0) = 0
and, d2y⁄dx2 = xy1 + y + 2x
hence, y2 = xy1 + y + 2x
hence, d2y⁄dx2 (x=0)=1
Differentiating it n times we get,
Discrete Mathematics MCQ Set 2
1. A and B are two events such that P(A) = 0.4 and P(A ∩ B) = 0.2 Then P(A ∩ B) is equal to
a) 0.4
b) 0.2
c) 0.6
d) 0.8
Answer
Answer: a [Reason:] P(A ∩ B) = P(A – (A ∩ B))
= P(A) – P(A ∩ B)
= 0.6 – 0.2 Using P(A) = 1 – P(A)
= 0.4.
2. A problem in mathematics is given to three students A, B and C . If the probability of A solving the problem is 1⁄2 and B not solving it is 1⁄4 . The whole probability of the problem being solved is 63⁄64 then what is the probability of solving it ?
a) 1⁄8
b) 1⁄64
c) 7⁄8
d) 1⁄2
Answer
Answer: c [Reason:]
Let A be the event of A solving the problem
Let B be the event of B solving the problem
Let C be the event of C solving the problem
Given P(a) = 1⁄2, P(~B) = 1⁄4 and P(A ∪ B ∪ C) = 63/64
We know P(A ∪ B ∪ C) = 1 – P(A ∪ B ∪ C)
= 1 – P(A ∩ B ∩ C)
= 1 – P(A) P(B) P(C)
Let P(C) = p
ie 63⁄64 = 1 – (1⁄2)(1⁄4)(p)
= 1 – p⁄8
⇒ P =1/8 = P(C)
⇒P(C) = 1 – P = 1 – 1⁄8 = 7⁄8.
3. Let A and B be two events such that P(A) = 1⁄5 While P(A or B) = 1⁄2. Let P(B) = P. For what values of P are A and B independent?
a) 1⁄10 and 3⁄10
b) 3⁄10 and 4⁄5
c) 3⁄8 only
d) 3⁄10
Answer
Answer: c [Reason:]
For independent events,
P(A ∩ B) = P(A) P(B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) P(B)
= 1⁄5 + P (1⁄5)P
⇒ 1⁄2 = 1⁄5 + 4⁄5P
⇒ P= 3⁄8.
4. If A and B are two mutually exclusive events with P(~A) = 5⁄6 and P(b) = 1⁄3 then P(A /~B) is equal to
a) 1⁄4
b) 1⁄2
c) 0, since mutually exclusive
d) 5⁄18
Answer
Answer: a [Reason:] As A and B are mutually exclusive we have
.
5. If A and B are two events such that P(a) = 0.2, P(b) = 0.6 and P(A /B) = 0.2 then the value of P(A /~B) is
a) 0.2
b) 0.5
c) 0.8
d) 1⁄3
Answer
Answer: a [Reason:] For independent events,
P(A /~B) = P(a) = 0.2.
6. If A and B are two mutually exclusive events with P(a) > 0 and P(b) > 0 then it implies they are also independent
a) True
b) False
Answer
Answer: b [Reason:]
P(A ∩ B) = 0 as (A ∩ B) = ∅
But P(A ∩ B) ≠ 0 , as P(a) > 0 and P(b) > 0
P(A ∩ B) = P(A) P(B), for independent events.
7. Let A and B be two events such that the occurrence of A implies occurrence of B, But not vice-versa, then the correct relation between P(a) and P(b) is
a) P(A) < P(B)
b) P(B) ≥ P(A)
c) P(A) = P(B)
d) P(A) ≥ P(B)
Answer
Answer: b [Reason:] Here, according to the given statement A ⊆ B
In this case P(B) = P(A ∪ (A ∩ B))
= P(A) + P(A ∩ B)
Therefore, P(B) ≥ P(A)
8. In a sample space S, if P(a) = 0, then A is independent of any other event
a) True
b) False
Answer
Answer: a [Reason:] P(a) = 0 (impossible event)
Hence, A is not dependent on any other event.
9. If then,
a) P(a) > P(b)
b) P(a) > P(b)
c) P(a) = P(b)
d) P(~A) less than P(~B)
Answer
Answer: c [Reason:] A ⊂ B, B ⊂ A
That means A = B
Hence P(a) = P(b).
10. If A is a subset of B then,
a) P(a) is greater than P(b)
b) P(~A) is greater than or equal to P(~B)
c) P(b) is equal to P(a)
d) P(b) is equal to P(~B)
Answer
Answer: b [Reason:] A is a subset of B which means (~B) is a subset of (~A)
Therefore P(~A) is greater than or equal to P(~B).
11. If A is a perfect subset of B and P(a < Pb), then P(B – A) is equal to
a) P(a) / P(b)
b) P(a)P(b)
c) P(a) + P(b)
d) P(b) – P(a)
Answer
Answer: d [Reason:]
From Basic Theorem of probability,
P(B – A) = P(b) – P(a), this is true only if the condition given in the question is true.
12. What is the probability of an impossible event?
a) 0
b) 1
c) Not defined
d) Insufficient data
Answer
Answer: a [Reason:] If the probability of an event is 0, then it is called as an impossible event.
13. If A = A1 ∪ A2……..∪ An, where A1…An are mutually exclusive events then
Answer
Answer: b [Reason:]
A = A1 ∪ A2……..∪ An, where A1…An
Since A1…An are mutually exclusive
P(a) = P(A1) + P(A2) + … + P(An)
Therefore 
.
Discrete Mathematics MCQ Set 3
1. If P(B⁄A) = p(b), then P( A and B) =
a) p(b)
b) p(a)
c) p(b)p(a)
d) p(a) + p(b)
Answer
Answer: c [Reason:] P(B /A) = p(b) implies A and B are independent events
Therefore, P(A and B) = p(a)p(b).
2. Two unbiased coins are tossed. What is the probability of getting at most one head?
a)1⁄2
b)1⁄3
c)1⁄6
d)3⁄4
Answer
Answer: d [Reason:] Total outcomes = (HH, HT, TH, TT)
Favorable outcomes = (TT, HT, TH)
At most one head refers to maximum one head,
Therefore, Probability = 3⁄4.
3. If A and B are two events such that p(a) > 0 and p(b) is not a sure event, then
P(~A /~B) =
a) 1 – P(A /B)
b) P(~A)/P(~B)
c) Not Defined
d) (1 – P(A or B) ) /P(~B)
Answer
Answer: d [Reason:] From definition of conditional probability we have
4. If A and B are two events, then the probability of exactly one of them occurs is given by
a) P(A ∩ B) + P( A ∩ B)
b) P(A) + P(B) – 2P(A) P(B)
c) P(A) + P(B) – 2P(A) P(B)
d) P(A) + P(B) – P(A ∩ B)
Answer
Answer: a [Reason:] The set corresponding to the required outcome is
(A ∩ B) ∪ ( A ∩ B)
Hence the required probability is
P(A ∩ B) + P( A ∩ B).
5. The probability that at least one of the events M and N occur is 0.6. If M and N have probability of occurring together as 0.2, then P(~M) + P(~N) is
a) 0.4
b) 1.2
c) 0.8
d) Indeterminate
Answer
Answer: b [Reason:] Given : P(M or N) = 0.6, P(M and N) = 0.2
P(M or N) + P(M and N) = P(M) + P(N)
2 – ( P(M or N) + P(M and N) ) = 2 – ( P(M) + P(N) )
= ( 1 – P(M) ) + ( 1 – P(N) )
2 – (0.6 + 0.2) = P(~M) + P(~N) = 1.2.
6. A jar contains ‘y’ blue colored balls and ‘x’ red colored balls. Two balls are pulled from the jar without replacing. What is the probability that the first ball Is blue and second one is red
Answer
Answer: b [Reason:] Number of blue balls = y
Number of Red balls = x
Total number of balls = x + y
Probability of Blue ball first = y / x + y
No. of balls remaining after removing one = x + y – 1
7. A survey determines that in a locality, 33% go to work by Bike, 42% go by Car, and 12% use both. The probability that a random person selected uses neither of them is
a) 0.29
b) 0.37
c) 0.61
d) 0.75
Answer
Answer: b [Reason:] Given: p(b) = 0.33, Pc) = 0.42
P(B and C) = 0.12
P(~B and ~C) = ?
P(~B and ~C) = 1 – P(B or C)
= 1 – p(b) – Pc) + P(B and C)
= 1 – 0.22 – 0.42 + 0.12
= 0.37.
8. A coin is biased so that its chances of landing Head is 2⁄3 . If the coin is flipped 3 times, the probability that the first 2 flips are heads and the 3rd flip is a tail is
a) 4⁄27
b) 8⁄27
c) 4⁄9
d) 2⁄9
Answer
Answer: a [Reason:] Required probability = 2⁄3 x 2⁄3 x 1⁄3 = 4⁄27.
9. Husband and wife apply for two vacant spots in a company. If the probability of wife getting selected and husband getting selected are 3/7 and 2/3 respectively, what is the probability that neither of them will be selected?
a) 2⁄7
b) 5⁄7
c) 4⁄21
d) 17⁄21
Answer
Answer: c [Reason:]Let H be the event of husband getting selected
W be the event of wife getting selected
Then, the event of neither of them getting selected is = (H ∩ W)
P (H ∩ W) = P (H) x P (W)
= ( 1 – P (H) ) x ( 1 – P (W) )
= ( 1 – 2⁄3 ) x (1 – 3⁄7 )
= 4⁄21.
10. For two events A and B, if P (B) = 0.5 and P (A ∪ B) = 0.5, then
P (A|B) =
a) 0.5
b) 0
c) 0.25
d) 1
Answer
Answer: d [Reason:] We know that,
P (A│B) = P(A ∩ B) /P(B)
= P((A ∪B) /P(B) )
= ( 1 – P(A ∪ B) ) /P(B)
= (1 – 0.5) /0.5
= 1.
11. A fair coin is tossed thrice, what is the probability of getting all 3 same outcomes?
a) 3⁄4
b) 1⁄4
c) 1⁄2
d) 1⁄6
Answer
Answer: b [Reason:]S= {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
All 3 same outcomes mean either all head or all tail
Total outcomes = 8
Favourable outcomes = {HHH, TTT} = 2
∴ required probability = 2⁄8 = 1⁄4.
12. A bag contains 5 red and 3 yellow balls. Two balls are picked at random. What is the probability that both are of the same colour?
Answer
Answer: a [Reason:]Total no.of balls = 5R+3Y = 8
No.of ways in which 2 balls can be picked = 8C2
Probability of picking both balls as red = 5C2 /8C2
Probability of picking both balls as yellow = 3C2 /8C2
∴ required probability 
.
Discrete Mathematics MCQ Set 4
1. nth derivative of Sinh(x) is
a) 0.5(ex – e-x)
b) 0.5(e-x – ex)
c) 0.5(ex – (-1)n e-x)
d) 0.5((-1)-n e-x -ex)
Answer
Answer: c [Reason:] Y = Sinh(x)
Y = 0.5[ex – e-x].
y1 = 0.5 [ex – (-1)e-x].
y2 = 0.5 [ex – (-1)2 e-x].
Similarly,
yn = 0.5 [ex – (-1)n e-x].
2. If y=log(x⁄(x2 – 1)) , then nth derivative of y is ?
a) (-1)(n-1) (n-1)!(x(-n) + (x-1)(-n) + (x+1)(-n) )
b) (-1)n (n)! (x(-n-1) + (x-1)(-n-1) + (x+1)(-n-1) )
c) (-1)(n+1) (n+1)!(x(-n) + (x-1)(-n) + (x+1)(-n) )
d) (-1)n(n)! (x(-n-1) + (x-1)(-n+1) + (x+1)(-n+1) )
Answer
Answer: a [Reason:] Y=log(x) – log(x2 – 1)
y1 = x(-1)-2x/(x2-1)
y1 = x(-1)-(x-1)(-1) + (x+1)(-1)
yn = (-1)(n-1) (n-1)!(x(-n)-(x-1)(-n) + (x+1)(-n)).
3. If x = a(Cos(t) + t2) and y = a(Sin(t) + t2 + t3) then dy/dx equals to
a) (Cos(t) + 3t2 + 2t) / (-Sin(t) + 2t)
b) (Sin(t) + 3t2 + 2t) / (-Cos(t) + 2t)
c) (Sin(t) + 3t2 + 2t) / (Cos(t) + 2t)
d) (Cos(t) + 3t2 + 2t) / (Sin(t) + 2t)
Answer
Answer: a [Reason:] dx/dt = a(-Sin(t) + 2t)
dy/dt = a(Cos(t) + 2t + 3t2)
Then,
dy/dx = (Cos(t) + 3t2+2t)/(-Sin(t) + 2t).
4. If y=tan(-1)(x) , then which one is correct ?
a) y3 + y12 + 4xy2 y1=0
b) y3 + y12 + xy2 y1=0
c) y3 + 2y12 + xy2 y1=0
d) y2 + 2y12 + 4xy2 y1=0
Answer
Answer: d
Explanation:
y = tan(-1)(x)
5. What is the value of dn (xm)⁄dxn for m<n, m=n, m>n ?
a) 0, n!, mPn x(m-n)
b) mPn x(m-n), n!, 0
c) 0, n!, mCn x(m-n)
d) mCn x(m-n), n!, 0
Answer
Answer: a [Reason:] For, m > n
6. Which of the following is true
a) Value of dm (Sin(nx))⁄dxm is always positive for m=0, 1, 4, 5, 8, 9… for 0 < nx < π⁄2 and n<0.
b) Value of dm (Sin(nx))⁄dxm is always positive for m=2, 3, 6, 7, 10, 11… for 0 < nx < π⁄2 and n>0.
c) Value of dm (Sin(nx))⁄dxm is always positive for m=0, 1, 4, 5, 8, 9… for 0 < nx < π⁄2 and n>0.
d) Value of dm (Sin(nx))⁄dxm is always positive for m=2, 3, 6, 7, 10, 11… for 0 < nx < π⁄2 and n<0.
Answer
Answer: c [Reason:]
7. If nth derivative of eax sin(bx+c) cos(bx+c) is, eax rn sin(bx+c+nα⁄2) cos(bx+c+nα⁄2) then,
Answer
Answer: b [Reason:] y = eax sin(bx+c) cos(bx+c)
y = eax sin2(bx+c)/2
yn = eax rn sin(2(bx+c+nα/2))/2
yn = eax rn sin(bx+c+nα/2) cos(bx+c+nα/2)
where
r = √(a2+4b2 ) , α = tan(-1)2b/a.
8. If y=x4⁄x2-1 , Then,
a) 0.5*(-1)n (n-1)! [(x-1)-n-1 + (x+1)-n-1 ].
b) 0.5*(-1)n (n-1)! [x– n-1 + (x-1)-n-1 + (x+1)-n-1].
c) 0.5*(-1)n (n-1)! [(x-1)-n + (x+1)-n) ].
d) 0.5*(-1)n (n-1)! [x-n + (x-1)-n + (x+1)-n].
Answer
Answer: a [Reason:]
9. If y=sin(-1)(x) then select the true statement
a) y2 = xy13
b) y3 = xy23
c) y2 = xy12
d) y3 = xy12
Answer
Answer: a [Reason:]
10. nth derivative of y = sin2x cos3x is
a)1⁄8 cos(x + nπ⁄2) –1⁄16 5n cos(x + nπ⁄2) – 1⁄16 3n cos(3x + nπ⁄2)
b)1⁄8 sin(x+nπ⁄2) –1⁄16 5n cos(x + nπ⁄2) – 1⁄16 3n cos(3x + nπ⁄2)
c)1⁄8 cos(x+nπ⁄2) –1⁄16 5n sin(x + nπ⁄2) – 1⁄16 3n sin(3x + nπ⁄2)
d)1⁄8 sin(x + nπ⁄2) –1⁄16 5n sin(x + nπ⁄2) – 1⁄16 3n sin(3x + nπ⁄2)
Answer
Answer: a [Reason:]
y = sin2x cos2x cos(x)
y = 1⁄4 sin22x cos2x cos(x)
y = 1⁄8 (2sin22x) cos(x)
y = 1⁄8 (1 – cos4x) cos(x)
y = 1⁄8 (1 – cos4x) cos(x)
y = 1⁄8 cos(x) – 1⁄8 cos4x cos(x)
y = 1⁄8 cos(x) – 1⁄16 (cos5x + cos(3x))
Now, nth derivative is
yn = 1⁄8 cos(x + nπ⁄2) – 1⁄16 5n cos(x + nπ⁄2) – 1⁄16 3n cos(3x + nπ⁄2).
11. If 
, Then value of ‘c’ equals to
a) 1
b) 2
c) 3
d) 4
Answer
Answer: a [Reason:]
12. If y = exSin-1(x) and 
, Then find the value of ‘c’ ?
a) -2
b) 2
c) -0.5
d) 0.5
Answer
Answer: a [Reason:]
13. Find nth derivative of y = Sin(x) Cos3(x)
Answer
Answer: a [Reason:]
14. If nth derivative of 
then find the value of a and b
a) -1, -2
b) 2, 1
c) 1, 2
d) -2, -1
Answer
Answer: c [Reason:]
Discrete Mathematics MCQ Set 5
1. Mean Value Theorem tells about the
a) Existence of point c in a curve where slope of a tangent to curve is equal to the slope of line joining two points in which curve is continuous and differentiable
b) Existence of point c in a curve where slope of a tangent to curve is equal to zero
c) Existence of point c in a curve where curve meets y axis
d) Existence of point c in a curve where curve meets x axis
Answer
Answer: a [Reason:] Statement of Mean Value Theorem is that, If function f(x) is continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = [f(b)-f(a)]/(b-a).
2. If f(a) is euquals to f(b) in Mean Value Theorem, then it becomes
a) Lebniz Theorem
b) Rolle’s Theorem
c) Taylor Series of a function
d) Leibnit’x Theorem
Answer
Answer: b [Reason:] According to Mean Value Theorem, If a function exist at pt. ‘a’, ‘b’ and continuous in closed interval [a, b] and differentiable in open interval (a, b) then there exists a point ‘c’, such that c∈(a,b), Where,
f’(c)= [f(b)-f(a)]/(b-a).
Hence, By putting f(b) = f(a) in the statement of Mean Value Theorem, we get
f’(c) = [f(b)-f(a)]/(b-a) = 0 . Which is a statement of Rolle’s Theorem.
3. Mean Value theorem is applicable to the
a) Functions differentiable in closed interval [a, b] and continuous in open interval (a, b).
b) Functions continuous in closed interval [a, b] only and having same value at point ‘a’ and ‘b’
c) Functions continuous in closed interval [a, b] and differentiable in open interval (a, b).
d) Functions differentiable in open interval (a, b) only and having same value at point ‘a’ and ‘b’
Answer
Answer: c [Reason:] Statement of Mean Value Theorem is that, If function f(x) is continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = [f(b)-f(a)]/(b-a).
4. Mean Value theorem is also known as
a) Rolle’s Theorem
b) Lagrange’s Theorem
c) Taylor Expansion
4) Leibnit’s Theorem
Answer
Answer: b [Reason:] Statement of Mean Value Theorem is that, If function f(x) is continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = [f(b)-f(a)]/(b-a).
It is also known as Lgrange’s Theorem.
5. Find the point c in the curve f(x) = x3 + x2 + x + 1 in the interval [0, 1] where slope of a tangent to a curve is equals to the slope of a line joining (0,1)
a) 0.64
b) 0.54
c) 0.44
d) 0.34
Answer
Answer: b [Reason:] f(x) = x3 + x2 + x + 1
f(x) is continuous in given interval [0,1].
f’(x) = 3x2+2x+1
Since, value of f’(x) is always finite in interval (0, 1) it is differentiable in interval (0, 1).
f(0) = 1
f(1) = 4
By mean value theorem,
f’(c) = 3c2 + 2c + 1 = (4-1)/(1-0) = 3
⇒ c = 0.548,-1.215
Since c belongs to (0, 1) c = 0.54.
6. Find the equation of curve whose roots gives the point which lies in the curve f(x) = xSin(x) in the interval [0, π⁄2] where slope of a tangent to a curve is equals to the slope of a line joining (0, π⁄2)
a) c = -Sec(c) – Tan(c)
b) c = -Sec(c) – Tan(c)
c) c = Sec(c) +Tan(c)
d) c = Sec(c) – Tan(c)
Answer
Answer: d [Reason:] f(x) = xSin(x)
Since f1(x) = x and f2(x)=Sin(x) both are continuous in interval [0, π⁄2], the curve f(x)=f1(x)f2(x) is also continuous.
f’(x) = xCos(x) + Sin(x)
f’(x) always have finite value in interval [0, π⁄2] hence it is differentiable in interval (0, π⁄2).
f(0) = 0
f(π⁄2) = π⁄2
By mean value theorem,
f’(c) = cCos(c) + Sin(c) = (π⁄2 – 0)/(π⁄2 – 0)=1
Hence, c = Sec(c) – Tan(c) is the required curve.
7. Can Mean Value Theorem be applied in the curve
a) True
b) False
Answer
Answer: b [Reason:] Continuity Check
Function f(x) is not continuous hence mean value theorem cannot be applied.
8. Find point c between [2,9] where, the slope of tangent to the function f(x)=1+∛x-1 at point c is equals to the slope of a line joining point (2,f(2)) and (9,f(9)).
(Providing given function is continuous and differentiable in given interval).
a) -2.54
b) 4.56
c) 4.0
d) 4.9
Answer
Answer: b [Reason:] Since the given function is continuous and differentiable in a given interval,
f(2) = 2
f(9) = 3
Applying mean value theorem,
f’(c) = 1/3∛(c-1)2 = [f(9)-f(2)]/(9-2) = 1/7
c = 1 ± (7/3)(3/2)
c = 4.56,-2.54
Since c lies in (2,9), c = 4.56.
9. Find point c between [-1,6] where, the slope of tangent to the function f(x) = x2+3x+2 at point c is equals to the slope of a line joining point (-1,f(-1)) and (6,f(6)).
(Providing given function is continuous and differentiable in given interval).
a) 2.5
b) 0.5
c) -0.5
d) -2.5
Answer
Answer: a [Reason:] Since the given function is continuous and differentiable in a given interval,
f(-1) = 0
f(6) = 56
Applying mean value theorem,
f’(c) = 2c+3 = [f(6)-f(-1)]/[6-(-1)] = 56/7 = 8
c = 5/2
c = 2.5.
10. If f(x) = Sin(x)Cos(x) is continuous and differentiable in interval (0, x) then
Answer
Answer: b [Reason:] f(x) = sin(x)cos(x)
Given f(x) is continuous and differentiable in interval (0, x),
Applying mean value theorem in interval (0, x)
f’(c) = Cos(2c) = [f(x)-f(0)]/[x-0] = (Cos(x)Sin(x))⁄x ……………………. (1)
Now, Given
0 < c < x
Multiplying by 2 and taking Cos, We get
1 < Cos(2c) < Cos(2x)
1 < (Cos(x)Sin(x))⁄x < Cos(2x).
