Quantitative Aptitude PYQs

Concept Used:

If two numbers are opposite in sign then their product is always less than zero.

Calculations:

Given function is f(x) = 3x² - 5x + p

Put x = 0 and x = 1 in f(x) to find f(0) and f(1).

Put x = 0

⇒ f(0) = p

Put x = 1

⇒ f(1) = 3(1)2 - 5(1) + p

⇒ f(1) = - 2 + p

Here, f(0) and f(1) are opposite in sign.

So, f(0) × f(1) < 0

⇒ p × (-2 + p) < 0

⇒ p (p - 2) < 0

∴ p ∈ (0, 2) [p ∈ (0, 1) lies inside the (0, 2)]

Hence, f(x) = 3x² - 5x + p and f(0) and f(1) are opposite in sign then 0 < p < 1 is correct statement.

Calculations:

Given, e^θϕ = c + 4θϕ, where c is an arbitrary constant and ϕ is a function of θ

Differentiate  w.r.to θ, we get

⇒ \(\rm \dfrac {d}{d\theta}(e^{θϕ} )= \dfrac {d}{d\theta} (c + 4θϕ)\)

⇒e^θϕ  \(\rm (\theta;\dfrac {d\phi}{d\theta} +\phi)\) =  4\(\rm (\theta;\dfrac {d\phi}{d\theta} +\phi)\)

⇒(e^θϕ - 4)\(\rm (\theta;\dfrac {d\phi}{d\theta} +\phi)\) = 0

⇒ \(\rm (\theta;\dfrac {d\phi}{d\theta} +\phi)\) = 0

⇒ θ dϕ + ϕ dθ = 0

∴ ϕ dθ = -θ dϕ

Concept:

\(\rm \displaystyle \int a^x dx = \frac{a^x}{\ln a} + c\)

Calculation:

Given: p(x) = (4e)^2x

To Find: \(\rm \displaystyle \int p(x) dx\)

\(\text{Let I}=\rm \displaystyle \int p(x) dx\ ⇒ \rm \displaystyle \int (4e)^{2x} dx\)

Let 2x = t

Differentiating with respect to x, we get

⇒ 2dx = dt

⇒ dx = \(\rm \frac{dt}{2}\)

Now,

\(\rm \displaystyle \int p(x) dx\⇒ \rm \frac{1}{2} \displaystyle \int (4e)^{t} dt\⇒\rm \frac{1}{2} \frac{(4e)^{t}}{\ln 4e}+c\)

\(⇒\rm \frac{1}{2} \frac{(4e)^{t}}{(\ln 4 + \ln e)}+c\)            (∵ log mn = log m + log n)

\(⇒ \rm \frac{1}{2} \frac{(4e)^{t}}{(1+2\ln 2 )}+c\)

∴ The value of the integral ∫p(x) is \(\rm \frac{1}{2} \frac{(4e)^{t}}{(1+2\ln 2 )}+c\).

Concept:

1 + tan² x = sec² x

\(\rm \displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+c\)

Calculation:

\(\text{Let I} = \rm \displaystyle \int_0^{π/4} (\tan^3 x + \tan x)dx\Rightarrow\rm \displaystyle \int_0^{π/4} \tan x(\tan^2 x + 1)dx\)

\(\Rightarrow \rm \displaystyle \int_0^{π/4} \tan x \sec^2 x dx\)             (∵ 1 + tan² x = sec² x)

Let tanx = t

Differentiating with respect to x, we get

⇒ sec2 x dx = dt

\(\Rightarrow\rm \displaystyle \int_0^1 tdt\Rightarrow[\frac{t^2}{2}]_0^1\Rightarrow\frac{1}{2}-0=\frac{1}{2}\)

∴ The value of the integral \(\rm \displaystyle\int_0^{\pi/4}(\tan^3 x + \tan x)dx \ \)is 1/2.

Concept:

Let small charge in x be Δx and the corresponding change in y is Δy.

Therefore \(\rm \Delta y = \rm \frac{dy}{dx}\Delta x\)

 

Calculation:

Given: y = 3x² + 2

To Find: total change in y = Δy

x changes from 10 to 10.1

So, Δx = 10.1 - 10 = 0.1

Now, y = 3x² + 2

Differentiating with respect to x, we get

\(\Rightarrow \rm \frac{dy}{dx} = 6x\)

As we know, \(\rm \Delta y = \rm \frac{dy}{dx}\Delta x\)

\(\Rightarrow \rm \Delta y = \rm 6x\Delta x\)

Put x = 10 and Δx = 0.1

\(\therefore \rm \Delta y = \rm 6\times 10\times 0.1 = 6\approx 6.03\)

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{;}}\frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{;}}\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = ;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Note:

We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}\) where l is a finite value.

A function f(x) is said to be continuous at a point x = a, in its domain if \(\rm \mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right) = f\left( a \right)\) exists 

 

Calculation:

Given: \(\rm f(x)=\dfrac{\sin x}{x}\)

To Find: f(0)

Function is continuous at x = 0

Therefore, f(0) = \( \rm \lim_{x\rightarrow 0} f(x)\)

\(= \rm \lim_{x\rightarrow 0} \frac{\sin x}{x}\)          (Form 0/0)

Apply L-Hospital Rule,

\(= \rm \lim_{x\rightarrow 0} \frac{\cos x}{1}\=\frac{\cos 0}{1} = 1\)

Concept:

\(\rm \displaystyle \int \frac{dx}{1+x^2} = \tan^{-1} x + c\)

Calculation:

Given: dy = (1 + y²) dx

\(\rm \Rightarrow \frac{dy}{1+y^2}=dx\)

Integrating both sides, we get

\(\rm \Rightarrow \displaystyle \int \frac{dy}{1+y^2}=\displaystyle \int dx\rm \Rightarrow \tan^{-1} y = x + c \)

⇒ y = tan (x + c)

∴ The solution of the given differential equation is y = tan (x + c).

Concept:

1. Integration by parts: 

Integration by parts is a method to find integrals of products

The formula for integrating by parts is given by;

\(⇒ \smallint {\rm{u;vdx}} = {\rm{u;}}\smallint {\rm{vdx}} - {\rm{;}}\smallint \left( {\frac{{{\rm{du}}}}{{{\rm{dx}}}}{\rm{;}}\smallint {\rm{vdx}}} \right){\rm{dx}}\)

Where u is the function u(x) and v is the function v(x)

2. ILATE Rule: 

Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

 

Calculation:

Let I = ∫ (e^{log x} + sin x) cos x dx

\(⇒\rm \int (x \cos x + \sin x \cos x) dx\)        (∵ e^{log x} = x)

\(⇒ \rm \int x\cos x ;dx + \int \sin x \cos x;dx\)

⇒ I = I₁ + I₂               ....

(1) Now, I₁ = \(\rm \int x\cos x ;dx\)

Applying by parts, we get

\(⇒ {\rm{x;}}\smallint {\rm{\cos x;dx}} - {\rm{;}}\smallint \left( {\frac{{{\rm{dx}}}}{{{\rm{dx}}}}{\rm{;}}\smallint {\rm{\cos x;dx}}} \right){\rm{dx}} \)

⇒ x sin x - \(\rm \int \sin x ;dx \) + c

⇒ x sin x + cos x + c

Now, I₂ = \(\rm \int \sin x \cos x;dx\)

Let sin x = t

Differentiating with respect to x, we get

⇒ cos x dx = dt

⇒ I₂ = \(\rm \int tdt = \frac{t^2}{2} + c\)

⇒ \(\rm \frac{\sin^2 x}{2} + c\)

Put the value of I₁ and I₂ in equation (1), we get

∴ The required integral (I) is \(\rm x \sin x + \cos x + \dfrac{\sin^2 x}{2}+c\).

Concept:

Domin of cos^-1 x is [-1, 1]

Adding or subtracting the same quantity from both sides of an inequality leaves the inequality symbol unchanged.

 

Calculation:

Given:  f(x) = cos^-1 (x - 2)

As we know, Domin of cos^-1 x is [-1, 1]

Therefore, -1 ≤ (x - 2) ≤ 1

Adding 2 in above inequality, 

⇒ -1 + 2 ≤ x - 2 + 2 ≤ 1 + 2

⇒ 1 ≤ x ≤ 3

∴ Domin of cos^-1 (x - 2) is [1, 3]

Concept:

The area under the curve y = f(x) between x = a and x = b,is given by,  Area = \(\rm \int_{x=a}^{x =b}f(x) ;dx\)

Calculation:

To find:  The area of the region enclosed between the curve y² = 2x and the straight line y = x, draw the diagram.

First find the point of intersection of the curve y² = 2x and the straight line y = x

Put y = x in y² = 2x , we get

⇒ x² = 2x

⇒ x² - 2x = 0

⇒ x (x - 2) = 0

⇒ x = 0 or x = 2

⇒ y = 0 or y = 2

Hence, (0, 0) and (2, 2) are the point of intersection of the curve  y² = 2x and the straight line y = x.

 

 

The area under the curve y = f(x) between x = a and x = b,is given by,  Area = \(\rm \int_{x=a}^{x =b}f(x) ;dx\)

Here f(x) = \((\sqrt {2x}-x)\)

Hence,  Area = \(\rm \int_{x=0}^{x =2} (\sqrt {2x}-x) ;dx\)

Area = \(\rm [\sqrt 2\cdot \dfrac {x^\frac 3 2}{\frac 3 2} - \dfrac {x^2}{2}]^2_0\)

Area = \(\rm [\dfrac {{2^\frac 1 2}\cdot {2^\frac 3 2}}{\frac 3 2} - \dfrac {2^2}{2}] - 0\)

Area = \(\dfrac 2 3\)

Hence, the area of the region enclosed between the curve y² = 2x and the straight line y = x is \(\dfrac 2 3\)

Calculation:

Given: f(x) = 2x - x²

To Find:  f(x + 2) + f(x - 2) when x = 0

Now,

f(x + 2) = 2(x + 2) - (x + 2)²

f(x - 2) = 2(x - 2) - (x - 2)²

Adding above equation,

f(x + 2) + f(x - 2) =  2(x + 2) - (x + 2)² + 2(x - 2) - (x - 2)²

Put x = 0

f(2) + f(-2) = 2(0 + 2) - (0 + 2)² + 2(0 - 2) - (0 - 2)²

= 4 - 4 - 4 - 4

= -8

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{;g}}\left( {\rm{x}} \right)} \right] = {\rm{;f'}}\left( {\rm{x}} \right){\rm{;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{;g'}}\left( {\rm{x}} \right)\)

 

Calculation:

Given: \(\rm x^m y^n =a^{m+n}\)

Differentiating with respect to x, we get

\(\Rightarrow \rm x^m\dfrac{dy^n}{dx}+ y^n\dfrac{dx^m}{dx} = \dfrac{da^{m+n}}{dx}\Rightarrow \rm x^m \times ny^{n-1} \dfrac{dy}{dx}+ y^n \times mx^{m-1} = 0\Rightarrow \rm x^m \times ny^{n-1} \dfrac{dy}{dx}=- y^n \times mx^{m-1} \ \Rightarrow\dfrac{dy}{dx}=\frac{- y^n \times mx^{m-1}}{\rm x^m \times ny^{n-1}}\\therefore \dfrac{dy}{dx}= \frac{-my}{nx} \)

Concept:

log m - log n = log \(\rm\dfrac {m}{n}\)

 

 

Calculations:

Given, I = \(\rm \displaystyle\int \dfrac{dx}{x(x^n+1)}\)

Multiply and divide the given integration by x^{n - 1} , we get

 I = \(\rm \displaystyle\int \dfrac{x^{n - 1}dx}{x^n(x^n+1)}\)              ....(1)

Put xⁿ = t

⇒ nx^{n - 1} dx = dt

⇒ \(\rm x^{n-1}dx = \dfrac {dt}{n}\)

Equation (1) becomes,

I = \(\rm \displaystyle \dfrac {1}{n}\int \dfrac{dt}{t(t+1)}\)            .....(2)

Consider, \(\rm\dfrac {1}{t(t+1)} = \dfrac{A}{t} + \dfrac{B}{t+1}\)         ....(3)

⇒\(\rm\dfrac {1}{t(t+1)} = \dfrac{A(t+1)+B(t)}{t(t+1)} \)

⇒\(\rm 1 = A(t+1)+B(t)\)

Put t = 0 to find A

⇒ A = 1

Put t = - 1 to find B

⇒ B = - 1

Put the values of A and B in equation (3)

\(\rm\dfrac {1}{t(t+1)} = \dfrac{1}{t} - \dfrac{1}{t+1}\)

Equation (2) becomes

⇒ I = \(\rm \displaystyle \dfrac {1}{n}\int (\dfrac{1}{t} - \dfrac{1}{t+1})dt\)

⇒ I = \(\rm\dfrac{1}{n} [log; t - log;(t+1)]\) + c

⇒ I = \(\rm\dfrac{1}{n} [log; \dfrac {t}{t+1}]\) + c

Substitute the value of t 

⇒ I = \(\rm\dfrac{1}{n} [log; \dfrac{ x^n}{x^n+1}]\) + c

Concept:

The minimum value of modulus functions is zero.

Calculations:

Given:

f(x) = |x - 1|

Since the minimum value of modulus functions is zero. 

Hence, the minimum value of |x - 1|, where x ∈ R is 0

Concept:

A rational function is one such that \(\rm f(x) = \dfrac{P(x)}{Q(x)}\) 

Where P(x) and Q(x) are polynomials and Q(x) ≠ 0

Calculations:

Let f(x) = (3x² + 8 - 4k)/x

We can solve this question by the help of option

Option 1) If we take k = 0 

f(x) = (3x2 + 8)/x

On doing integration f(x) with respect to x we get 

\(\rm\int f(x) dx = \int\dfrac{3x^2+8}{x}dx \)

⇒ \(\rm\int f(x) dx = \int (3x+ \dfrac 4 x)dx \)

⇒ \(\rm\int f(x) dx = \dfrac {3x^2}{2} +4logx \)

⇒ \(\rm\int f(x) dx = \dfrac {3x^2+8logx}{2} \)

Here Numerator is not a polynomial 

So, The function is not a rational function at k = 0.

Option 2) If we take k = 1 

f(x) = (3x2 + 4)/x

After integration we will get same function as in option 1)

So, The function is not a rational function at k = 1

Option 3) If we take k = 2

f(x) = \(\rm \dfrac{3x^2+8-4(2)}{x}\) = 3x

Integrating, we get 

\(\rm\int f(x) dx = \int 3xdx \)

⇒ \(\rm\int f(x) dx = \dfrac{3x^2}{2} \) = \(\rm \dfrac {P(x)}{Q(x)}\)

Here, P(x) and Q(x) both are polynomials with Q(x) ≠ 0

So, The function is a rational function at k = 3.

Option 4) If we take k = -2

f(x) = \(\rm \dfrac{3x^2+8-4(-2)}{x}\)

f(x) = \(\rm \dfrac{3x^2+16}{x}\)

After integration we will get same function as in option 1)

∴ The function is not a rational function at k = -2.

Hence the correct option is option 3.

Concept:

Difference between the number of students:

We take the modulus of the subtraction of the numbers and it is the difference between the two numbers.

 

Calculation:

We will take the difference between each class.

The difference for the class 10-20 is 10 - 8 = 2.

The difference for the class 20-30 is 21 - 11 = 10.

The difference for the class 30-40 is 38 - 30 = 8.

The difference for the class 40-50 is 26 - 15 = 11.

The difference for the class 50-60 is 15 - 10 = 5.

The difference for the class 60-70 is 10 - 6 = 4.

From the above observations, we see that the difference is maximum for the class 40-50.

Concept:

Modal value:

Modal value of the grouped frequency data is given as follows:

\(\begin{align*} \mbox{ mode } &= \rm l + \left(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)\times h\end{align*}\)

Where l = lower limit of the modal class

f₁ = frequency of modal class

f₀ = frequency of the preceding class

f2 = frequency of the succeeding class

 

Median:

The median of the grouped frequency data is given by the following:

\(\begin{align*} \mbox{ median } &= \rm l + \left(\dfrac{\frac{n}{2} - cf}{f}\right)\times h\end{align*}\)

where l = lower limit of the median class

n = number of frequency

cf = cumulative frequency of the preceding class

f = frequency of the median class

h = height of the median class

 

Calculation:

First of all, we will calculate the median for physics.

We will form the required table as follows:

 

Since the observations are 100 the median will be \(\dfrac{50+51}{2} = 50.5\) .

The median class will be 40 - 50.

Now we will calculate the median as follows:

\(\begin{align*} \mbox{ median } &= l + \left(\dfrac{\frac{n}{2} - cf}{f}\right)\times h\ &= 40 + \left(\dfrac{50-49}{26}\right)\times 10\ &= 40.38 \end{align*}\)

Therefore, the median of the physics is 40.38 .

Similarly, we will calculate the median of mathematics.

We will form the required table as follows:

 

Since the observations are 100 the median will be \(\dfrac{50+51}{2} = 50.5\) .

The median class will be 30-40.

Now we will calculate the median as follows:

\(\begin{align*} \mbox{ median } &= l + \left(\dfrac{\frac{n}{2} - cf}{f}\right)\times h\ &= 30 + \left(\dfrac{50-31}{38}\right)\times 10\ &= 35 \end{align*}\)

Therefore, the median of Maths is 35.

Thus, the median of physics is more than mathematics.

Now we will calculate the modal value for physics.

Therefore, the modal class is 30-40.

Now we will calculate the mode as follows:

\(\begin{align*} \mbox{ mode } &= l + \left(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)\times h\ &= 30 + \left(\dfrac{30-11}{2(30) - 11 - 26}\right)\times h\ &= 30 + \left(\dfrac{19}{23}\right)\times 10\ &= 38.26 \end{align*}\)

Therefore, modal value for physics is 38.26 which is greater than the class frequency of the interval 30-40.

Therefore, Both statement is incorrecct.

Concept:

Mean of observations:

If we have a grouped data with different class with frequencies f_i and if the classmark is x then the mean is calculated as follows:

\(\rm \mu = \dfrac{\sum fx}{\sum f}\)

 

Calculation:

We will make the table as follows:

 

Therefore, the mean is:

\(\begin{align*} \mu &= \rm \dfrac{\sum fx}{\sum f}\ &= \dfrac{4090}{100}\ &= 40.9 \end{align*}\)

Therefore, the mean is 40.9.

Concept:

Standard deviation:

The standard deviation of the observation set \(\rm {x_i,i=1,2,3,\cdots}\) is given as follows:

\(\rm \sigma=\sqrt{\dfrac{\sum\left(x_i-\mu\right)^2}{N}}\)

Where \(\rm N=\mbox{size of the observation set}\) and \(\rm \mu=\mbox{mean of the observations}\) .

 

Calculations:

First, we will calculate the mean of the given observations.

\(\begin{align*} \mu &= \dfrac{-\sqrt6-\sqrt5-\sqrt4-1+1+\sqrt4+\sqrt5+\sqrt6}{8}= 0 \end{align*}\)

Therefore, the numerator inside the square root term of the standard deviation formula will simply be equal to \(\rm (x_i-\mu)^2=x_i^2\) .

Now we observe that \(\rm N=8\) .

Therefore, the standard deviation is given as follows:

\(\begin{align*} \sigma &= \sqrt{\dfrac{\left(-\sqrt6\right)^2+\left(-\sqrt5\right)^2+\left(-\sqrt4\right)^2+\left(-1\right)^2+\left(1\right)^2+\left(\sqrt4\right)^2+\left(\sqrt5\right)^2+\left(\sqrt6\right)^2}{8}}\ &= \sqrt{\dfrac{32}{8}}\ &= \sqrt4\ &= 2 \end{align*}\)

Therefore, the standard deviation of the given observations is 2.

Concept:

Standard deviation:

Standard deviation is calculated as follows:

\(\rm \sigma=\sqrt{\left(\dfrac{\sum x_i^2}{n}\right)-\left(\dfrac{\sum x_i}{n}\right)^2}\)

 

Coefficient of variation:

Coefficient variation (CV) is calculated as follows:

\(\rm \mbox{CV}=\dfrac{\sigma}{\mu}\times 100\)

Where \(\mu\) is the mean.

 

Calculation:

First we will calculate the standard deviation as follows:

\(\begin{align*} \rm \sigma &=\rm \sqrt{\left(\dfrac{\sum x_i^2}{n}\right)-\left(\dfrac{\sum x_i}{n}\right)^2}\ &= \sqrt{\left(\dfrac{200}{10}\right)-\left(\dfrac{20}{10}\right)^2}\ &= \sqrt{20-4}\ &= \sqrt{16}\ &= 4 \end{align*}\)

Now the average of the data is given as follows:

\(\begin{align*} \mu &= \rm \dfrac{\sum x_i}{n}\ &= \dfrac{20}{10}\ &= 2 \end{align*}\)

Therefore, the coefficient of variation is given as follows:

\(\begin{align*} \mbox{CV} &= \dfrac{\sigma}{\mu}\times100\ &= \dfrac{4}{2}\times 100\ &= 200 \end{align*}\)

Therefore, the coefficient of variation is 200.