Mathematics MCQ Quiz - Objective Question with Answer for Mathematics - Download Free PDF

Concept:

• \(\text{Related Rates}: \)  Problems where two or more quantities change with respect to time.
• \(\text{Volume of Cone } \) The formula is \(V = \frac{1}{3}\pi r^2 h\), where \(V\) is volume, \(r\) is radius, and \(h\) is height.
• \(\text{Constant Ratio Condition} : \) Given \(h = \frac{1}{6}r\), the radius and height are proportional, so express \(r\) in terms of \(h\).
• Differentiation: Differentiate \(V\) with respect to time \(t\) to find how fast the height changes as the volume increases.

Calculation:

Given,

\(\frac{dV}{dt} = 12 \, \text{cm}^3/\text{s}\)

Height and radius relation: \(h = \frac{1}{6}r\) so \(r = 6h\)

Volume of cone:

\(V = \frac{1}{3}\pi r^2 h\)

⇒ Substitute \(r = 6h\):

\(V = \frac{1}{3}\pi (6h)^2 h\)

⇒ \(V = \frac{1}{3}\pi \times 36h^2 \times h\)

⇒ \(V = 12\pi h^3\)

Differentiate both sides w.r.t \(t\):

\(\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}\)

Substitute \(\frac{dV}{dt} = 12\) and \(h = 4\):

⇒ \(12 = 36\pi \times (4)^2 \times \frac{dh}{dt}\)

⇒ \(12 = 36\pi \times 16 \times \frac{dh}{dt}\)

⇒ \(12 = 576\pi \frac{dh}{dt}\)

⇒ \(\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}\)

The height increases at a rate of \(\frac{1}{48\pi} \, \text{cm/s}\). 

⇒ 96πh = 2

Hence 2 is the correct answer. 

Calculation

Given \(f(0) = 0 \) & \(f'(1) = 0\) & \(f(1) = \frac{1}{2}\)

\(\therefore f(x) = \frac{2x - x^2}{2}\)

\(f^{-1}(x)\) is the image of \(f(x)\) on \(y = x.\)

Also, \(2x + 2y = 3\) passes through \(A(1, \frac{1}{2})\) & \(B(\frac{1}{2}, 1)\)

so bounded Area \(A\)

 \(= AreaOAB = 2[Area OCM + Area CMNA - Area ONA]\)

\(A = 2[\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} + \frac{1}{2} (\frac{3}{4} + \frac{1}{2}) \times \frac{1}{4} - \frac{1}{2} \int_0^1 (2x - x^2) , dx]\)

\(\Rightarrow A = 2[\frac{9}{32} + \frac{5}{32} - [x^2 - \frac{x^3}{3}]_0^1]\)

\(\Rightarrow A = 2[\frac{14}{32} -(1 - \frac{1}{3})] \)
\(​\Rightarrow A = \frac{28}{32} - \frac{2}{3} = \frac{7}{8} - \frac{2}{3} \)\(\Rightarrow A = \frac{21 - 16}{24} = \frac{5}{24} \)

⇒ 48A = 10

Concept:

Exponential Generating Function & Series Coefficients:

• The expression uses combinations ⁿC_r and factorials, hinting at exponential and binomial expansion identities.
• Function f(x) = 1 + (1 + x)/1! + (1 + x)²/2! + (1 + x)³/3! + ... is considered to evaluate the coefficient of x².
• This function can be transformed using the identity: e^1+x / (1 + x)
• The coefficient of x² in this expansion corresponds to the RHS of 'a' series.
• The value of b is derived using the identity: 1 + 2/1! + 2²/2! + 2³/3! + ... = e²

Calculation:

Let f(x) = 1 + (1 + x)/1! + (1 + x)²/2! + (1 + x)³/3! + ...

⇒ f(x) = e^(1+x) / (1 + x)

Expand RHS:

= (1 + x + x²/2! + ...) / (1 + x)

⇒ (1 + x + (1 + x)²/2! + (1 + x)³/3! + (1 + x)⁴/4! + ...)

So, coefficient of x² in RHS is:

²C₂/3! + ³C₂/4! + ⁴C₂/5! + ... = a - 1

coefficient of x2 in RHS:

e × (1 + x+ x2/2!) × (1 -x+ x2/2!)

is e- e+ e/2! =a 

Now, expand LHS expression:

e × (1 + x+ x²/2!) × (1 -x+ x²/2!)

⇒ e × (1 - (x⁴/4!)) = e 

So, coefficient of x² = e × e = e²

Thus, b = 1 + 2/1! + 2²/2! + 2³/3! + ... = e²

a = e\2!

⇒ 8b / a² = 2 × e² / (e/2!)² = 32

∴ 8b / a² = 32

Concept:

• Let f(x) = 3 + 2x, which is a linear function of the form f(x) = ax + b.
• Let g_n(x) be the function obtained by composing f with itself n times: g_n(x) = f ∘ f ∘ ... ∘ f (n times)(x).
• If g_n(x) passes through a fixed point (α, β) for all n ∈ ℕ, then the point satisfies the equation g_n(α) = β for all n.
• This implies that the point (α, β) is a fixed point of all functions g_n.
• We find this by solving for the fixed point that satisfies f(α) = α, which guarantees that the point stays fixed under repeated function compositions.

Calculation:

Let f(x) = 3 + 2x

We want a point (α, β) such that g_n(α) = β for all n ∈ ℕ

⇒ fⁿ(α) = β for all n

Let’s compute a few values:

g₁(x) = f(x) = 3 + 2x

g₂(x) = f(f(x)) = f(3 + 2x) = 3 + 2(3 + 2x) = 3 + 6 + 4x = 9 + 4x

g₃(x) = f(g₂(x)) = f(9 + 4x) = 3 + 2(9 + 4x) = 3 + 18 + 8x = 21 + 8x

So pattern is: g_n(x) = A_n + B_nx

Let’s deduce the recurrence relations:

Initial: A₁ = 3, B₁ = 2

⇒ A_n = 3 + 2A_n-1

⇒ B_n = 2B_n-1

B_n = 2ⁿ

Let us find fixed point: f(α) = α

⇒ 3 + 2α = α

⇒ α = -3

Let β = g_n(α) = g_n(-3)

g_n(x) = A_n + B_nx = A_n - 3 × B_n

So β = A_n - 3 × B_n

We want this constant for all n

Let’s check for small values:

g₁(x) = 3 + 2x ⇒ g₁(-3) = 3 - 6 = -3

g₂(x) = 9 + 4x ⇒ g₂(-3) = 9 - 12 = -3

g₃(x) = 21 + 8x ⇒ g₃(-3) = 21 - 24 = -3

So, β = -3 always

∴ 2α + 3β = -3× 2 + (-3)× 3 = -15

Concept:

• n^th Root of Unity: The complex numbers satisfying αⁿ = 1 are called the n^th roots of unity.
• For n = 5, the fifth roots of unity are: 1, α, α², α³, α⁴, where α = e^2πi/5.
• The sum of all five roots of unity is always zero: 1 + α + α² + α³ + α⁴ = 0
• We can use symmetry of these roots on the unit circle in the complex plane to evaluate modulus expressions involving them.

Calculation:

Given that α is a fifth root of unity,

⇒ α⁵ = 1

The 5 roots are: 1, α, α², α³, α⁴

Sum of all roots: 1 + α + α² + α³ + α⁴ = 0

⇒ 1 + α + α² + α³ = −α⁴

Take modulus on both sides:

|1 + α + α² + α³| = |−α⁴|

Since α⁴ lies on the unit circle, |α⁴| = 1

⇒ |1 + α + α² + α³| = 1

∴ The correct answer is Option (2): |1 + α + α² + α³| = 1

Concept:

sin (2nπ ± θ) = ±  sin θ

sin (90 + θ) = cos θ

Calculation:

Given: sin (1920°)

⇒ sin (1920°) = sin(360° × 5° + 120°) = sin (120°)

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

Given:

\({\rm{y}} = {\rm{x}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} + {\left( {\frac{{{\rm{dx}}}}{{{\rm{dy}}}}} \right)} \)

\({\rm{y}} = {\rm{x}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} + \frac{1}{{{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}}}} \)

\(\Rightarrow {\rm{y}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)} = {\rm{x}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^3} + 1\)

For the given differential equation the highest order derivative is 1.

Now, the power of the highest order derivative is 3.

We know that the degree of a differential equation is the power of the highest derivative

Hence, the degree of the differential equation is 3.

Mistake PointsNote that, there is a term (dx/dy) which needs to convert into the dy/dx form before calculating the degree or order. 

Given:

The given data is 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Concept used:

The mode is the value that appears most frequently in a data set

At the time of finding Median

First, arrange the given data in the ascending order and then find the term

Formula used:

Mean = Sum of all the terms/Total number of terms

Median = {(n + 1)/2}^th term when n is odd 

Median = 1/2[(n/2)^th term + {(n/2) + 1}^th] term when n is even

Range = Maximum value – Minimum value 

Calculation:

Arranging the given data in ascending order 

2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19

Here, Most frequent data is 4 so 

Mode = 4

Total terms in the given data, (n) = 15 (It is odd)

Median = {(n + 1)/2}th term when n is odd 

⇒ {(15 + 1)/2}^th term 

⇒ (8)^th term

⇒ 6 

Now, Range = Maximum value – Minimum value 

⇒ 19 – 2 = 17

Mean of Range, Mode and median = (Range + Mode + Median)/3

⇒ (17 + 4 + 6)/3 

⇒ 27/3 = 9

∴ The mean of the Range, Mode and Median is 9

Formula used:

The mean of grouped data is given by,

\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)

Where, \(u_i \ = \ \frac{X_i\ -\ a}{h}\)

Xi = mean of ith class

fi = frequency corresponding to ith class

Given:

Calculation:

Now, to calculate the mean of data will have to find ∑fiXi and ∑fi as below,

Then,

We know that, mean of grouped data is given by

\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)

= \(\frac{1535}{43}\)

= 35.7

Hence, the mean of the grouped data is 35.7

Concept:

a² - b² = (a - b) (a + b)

sec x = 1/cos x and cosec x = 1/sin x

a³ + b³ = (a + b) (a² + b² - ab)

Calculation:

 \(\frac{{\left( {1 - {\rm{sinAcosA}}} \right)\left( {{\rm{si}}{{\rm{n}}^2}{\rm{A}} - {\rm{co}}{{\rm{s}}^2}{\rm{A}}} \right)}}{{{\rm{cosA}}\left( {{\rm{secA}} - {\rm{cosecA}}} \right)\left( {{\rm{si}}{{\rm{n}}^3}{\rm{A}} + {\rm{co}}{{\rm{s}}^3}{\rm{A}}} \right)}}\) 

⇒ \( \frac{{\left( {{\rm{1}} - {\rm{sinAcosA}}} \right)\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{sinA}} - {\rm{cosA}}} \right)}}{{{\rm{cosA}}\left[ {\frac{1}{{{\rm{cosA}}}} - \frac{1}{{{\rm{sinA}}}}} \right]\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{si}}{{\rm{n}}^2}{\rm{A}} + {\rm{co}}{{\rm{s}}^2}{\rm{A}} - {\rm{sinAcosA}}} \right)}}\)

⇒ \( \frac{{\left( {1 - {\rm{sinAcosA}}} \right)\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{sinA}} - {\rm{cosA}}} \right)}}{{{\rm{cosA}}\left[ {\frac{{{\rm{sinA}} - {\rm{cosA}}}}{{{\rm{sinA}}.{\rm{cosA}}}}} \right]\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {1 - {\rm{sinAcosA}}} \right)}}\)

⇒ \(\frac{sinA - cosA}{cosA[\frac{sinA - cosA}{sinA.cosA}]}\)

⇒ \(\frac{(sinA - cosA)\times sinA.cosA}{cosA[sinA - cosA]}\)

⇒ \(\frac{ sinA.cosA}{cosA}\)

⇒ sin A

∴ The correct answer is option (1).

Concept:

• Rational numbers are those numbers that show the ratio of numbers or the number which we get after dividing it with any two integers.
• Irrational numbers are those numbers that we can not represent in the form of simple fractions a/b, and b is not equal to zero.
• When we add any two rational numbers then their sum will always remain rational.
• But if we add an irrational number with a rational number then the sum will always be an irrational number.

Explanation:

Case:1 Take two irrational numbers π and 1 - π

⇒ Sum =  π +1 - π = 1

Which is a rational number.

Case:2 Take two irrational numbers π and √2 

⇒ Sum =  π + √2

Which is an irrational number.

Hence, a sum of two irrational numbers may be a rational or an irrational number.

Given:

tan0° tan1° tan2° tan3° tan4° …… tan89°

Formula:

tan 0° = 0

Calculation:

tan0° × tan1° × tan2° × ……. × tan89°

⇒ 0 × tan1° × tan2° × ……. × tan89°

⇒ 0

Concept:

Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
• Conjugate of z =  = x – iy

Calculation:

Let z = (1 + i) ³

Using (a + b) ³ = a³ + b³ + 3a²b + 3ab²

⇒ z = 1³ + i³ + 3 × 1² × i + 3 × 1 × i²

= 1 – i + 3i – 3

= -2 + 2i

So, conjugate of (1 + i) ³ is -2 – 2i

NOTE:

The conjugate of a complex number is the other complex number having the same real part and opposite sign of the imaginary part.

Concept:

cosec² x – cot² x = 1

Calculation:

Given: p = cosec θ – cot θ and q = (cosec θ + cot θ)^-1

⇒ cosec θ + cot θ = 1/q

As we know that, cosec² x – cot² x = 1

⇒ (cosec θ + cot θ) × (cosec θ – cot θ) = 1

\(\frac1q \times p=1\)

Concept:

sin² x + cos² x = 1

Calculation:

Given: sin θ + cos θ = 7/5 

By, squaring both sides of the above equation we get,

⇒ (sin θ + cos θ)² = 49/25

⇒ sin² θ + cos² θ + 2sin θ.cos θ = 49/25

As we know that, sin2 x + cos2 x = 1

⇒ 1 + 2sin θcos θ = 49/25

⇒ 2sin θcos θ = 24/25