Quantitative Aptitude PYQs

Concept:

Order of a differential equation is the highest order of derivative that occurs in the differential equation.

Degree of a differential equation is the highest power of the highest order derivative that occurs in the equation, after all the derivatives are converted into rational and radical free form.

 

Calculation:

y2 + 2cy - cx + c2 = 0

⇒ (y + c)² = cx

⇒ 2(y + c)y' = c

⇒ c = \(\rm \frac{2yy'}{1-2y'}\)

Substituting this in the given equation, we get:

(y + \(\rm \frac{2yy'}{1-2y'}\))2 = \(\rm \frac{2yy'}{1-2y'}\)x

⇒ y² = 2xyy'(1 - 2y')

⇒ y² - 2xyy' + 4xy(y')² = 0

The highest derivative in it is y', therefore its order is 1.

And, the highest power of this derivative in the equation is 2, therefore its degree is 2.

Concept:

Order of a differential equation is the highest order of derivative that occurs in the differential equation.

Degree of a differential equation is the highest power of the highest order derivative that occurs in the equation, after all the derivatives are converted into rational and radical free form.

 

Calculation:

Getting rid of the radicals by raising both the sides to power 2 will give us:

\(\rm x^2=1+(\frac{d^2y}{dx^2})^1\)

The highest derivative in it is \(\rm \frac{d^2y}{dx^2}\), therefore its order is 2.

∴ The highest power of this derivative in the equation is 1, therefore its degree is 1.

Concept:

• Differential Equation: A differential equation is an equation that relates one or more functions and their derivatives.
• e.g. \(\rm \frac{dy}{dx}\) + x = 2y + 3, etc.
• \(\rm \dfrac{d}{dx}e^{f(x)}=\dfrac{d}{dx}f(x)e^{f(x)}\)

 

Calculation:

y = aex + be-x

⇒ \(\rm \frac{dy}{dx}=\frac{d}{dx}ae^x +\frac{d}{dx}be^{-x}\)

⇒ \(\rm \frac{dy}{dx}\) = aex - be-x

⇒ \(\rm \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(ae^x -be^{-x})\)

⇒ \(\rm \frac{d^2y}{dx^2}\) = aex + be-x = y

∴ The general solution of y = aex + be-x is \(\rm \frac{d^2y}{dx^2}\) - y = 0.

Formula Used:

If lnx = n, then x = eⁿ

**∫**e^x dx = e^x + c

Calculation:

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

⇒ \(\rm \ln\left(\frac{dy}{dx} \right)\) = x - y

⇒ \(\rm \frac{dy}{dx}\) = e^{x - y} = \(\rm \frac{e^x}{e^y}\), which is in variable separable form.

⇒ e^x dx = e^y dy

Integrating both sides, we get:

⇒ e^x = e^y + c

∴ The solution of the differential equation is ex - ey = c.

Concept:

• e^{ln x} = x.
• ∫ xn dx = \(\rm \frac{x^{n+1}}{n+1}+c\).

 

Calculation:

Let I = \(\rm \int e^{\left(2\ln x+\ln x^2\right)}dx\)

⇒ I = \(\rm \int e^{\left(\ln x^2+\ln x^2\right)}dx\)

⇒ I = \(\rm \int e^{2\ln x^2}dx\)

⇒ I = \(\rm \int e^{\ln x^4}dx\)

⇒ I = ∫ x⁴ dx

∴ The value of the integral \(\rm \int e^{\left(2\ln x+\ln x^2\right)}dx\) is \(\rm \frac{x^5}{5}+c\).

Concept:

Arithmetic Mean (A. M.): The Arithmetic Mean of a set of n numbers is given by:

A. M. = \(\rm \frac{\sum a_i}{n}\).

Geometric Mean (G. M.): The Geometric Mean of a set of n numbers is given by:

G. M. = \(\rm \sqrt[n]{\prod a_i}\).

Calculation:

Both the Arithmetic and Geometric Means use all the data for their calculations.

Concept:

In a pie chart, the central angle of each slice of the pie is proportional to its value as compared to the total value.

θ = \(\rm \frac{\text{Value of Slice.}}{\text{Total Value.}}\times360^\circ\)

 

Calculation:

Angle corresponding to the Science graduates = θ = \(\rm \frac{\text{Number of Science graduates.}}{\text{Total number of graduates.}}\times360^\circ\).

= \(\rm \frac{30}{30+70+50}\times360^\circ\)

= \(\rm \frac{30}{150}\times360^\circ\)

= 72°.

Concept:

• When constructing a histogram with non-uniform (unequal) class widths, we must ensure that the areas of the rectangles are proportional to the class frequencies.
• Remember that the histogram differs from a bar chart in that it is the area of the bar that denotes the value, not the height. This means that we would need to consider the widths in order to determine the height of each rectangle.

 

Calculation:

From the definition of histogram, it is clear that the frequency of a class is proportional to the area of the rectangle, and NOT to its height.

Concept:

Co-efficient of Correlation (r):

In simple linear regression analysis, the co-efficient of correlation is a statistic which indicates an association between the independent variable and the dependent variable. The co-efficient of correlation is represented by "r" and its value lies between -1.00 and +1.00.

• When the co-efficient of correlation is positive, such as +0.80, it means the dependent variable is increasing/decreasing when the independent variable is increasing/decreasing. A negative value indicates an inverse association; the dependent variable is increasing/decreasing when the independent variable is decreasing/increasing.
• A co-efficient of correlation of +0.8 or -0.8 indicates a strong correlation between the independent variable and the dependent variable. An r of +0.20 or -0.20 indicates a weak correlation between the variables. When the co-efficient of correlation is 0.00, there is no correlation.
• r = \(\rm \frac{\sum\left(x_i-\bar x\right)\left(y_i-\bar y\right)}{\sqrt{\sum\left(x_i-\bar x\right)^2\sum\left(y_i-\bar y\right)^2}}\).

 

Calculation:

From the properties/nature of the co-efficient of correlation, we know that the correlation coefficient is independent of the choice of origin and scale.

Concept:

Median of 'n' observations is the value at the central position when the data is arranged in ascending or descending order.

• For, odd n, median is the value at the \(\rm \left(\frac{n+1}{2}\right)\) position.
• For even n, median is the mean of values at \(\rm \left(\frac{n}{2}\right)\) and \(\rm \left(\frac{n}{2}+1\right)\) position.

 

Calculation:

Writing the cumulative frequencies, we get

 

Since n = 198 is even, the median will be the mean of the values at \(\rm \left(\frac{198}{2}\right)\) = 99^th position and 99 + 1 = 100^th position.

From the above table, both the 99^th and 100^th values are 3.

Therefore, the median of the above data is 3.

Concept:

 

Indeterminate Forms: Any expression whose value cannot be defined, like \(\rm \frac00\), \(\rm \pm \frac{\infty}{\infty}\), 0⁰, ∞⁰ etc.

• L'Hospital's Rule: For the differentiable functions f(x) and g(x), the \(\rm\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}\), if f(x) and g(x) are both 0 or ±∞ (i.e. an Indeterminate Form) is equal to the \(\rm\mathop {\lim }\limits_{x \to c}\frac{f'(x)}{g'(x)}\) if it exists.
• For the indeterminate form ∞ - ∞, first rationalize by multiplying with the conjugate and then divide the terms by the highest power of the variable to get terms so that \(\rm \frac1{x}\) → 0 as x → ∞.
• For the indeterminate form \(\rm \frac00\), first try to rationalize by multiplying with the conjugate, or simplify by cancelling some terms in the numerator and denominator. Else, use the L'Hospital's rule.

 

Calculation:

\(\rm\mathop {\lim }\limits_{x \to a} \frac{a^x -x^a}{x^x -a^a}\)

= \(\rm \frac{a^a - a^a}{a^a - a^a}\)

= \(\frac00\), an indeterminate form.

Applying L'Hospital's rule, we get:

\(\rm\mathop {\lim }\limits_{x \to a} \frac{a^x -x^a}{x^x -a^a}\)

= \(\rm\mathop {\lim }\limits_{x \to a} \frac{a^x\log a -ax^{a-1}}{x^x(\log x +1)}\)

= \(\rm \frac{a^a\log a -a\cdot a^{a-1}}{a^a(\log a +1)}\)

= \(\rm \frac{\log a -1}{\log a +1}\)

According to the question, \(\rm \frac{\log a -1}{\log a +1}\) = -1.

⇒ log a - 1 = - log a - 1

⇒ 2 log a = 0

⇒ a = 1.

Formula Used:

\(\int \frac {1}{x} dx = lnx + c\)

At t = 0, Distance covered (x) = 0

Distance covered by the particle (x) = 24 m 

Calculation:

The given differential equation \(\rm \frac{dx}{dt}=x+1\) is in variable separable form.

On separating the variables, we get

⇒ \(\rm \left(\frac{1}{x+1}\right)dx = dt\)

On integrating both sides, we get:

⇒ ln (x + 1) = t + C    ---(i)

According to the question at t = 0, x = 0.

⇒ ln (0 + 1) = 0 + C

⇒ C = 0

Now, From (i), we get 

⇒ ln (x + 1) = t 

Again according to the question

⇒ t = ln (24 + 1)

⇒ t = ln 25

⇒ t = ln 5²

∴ The required time is 2 ln 5.

Concept:

Definite Integrals:

\(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).

If f(x) = f(2a - x), then \(\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx\).

A function f(x) is:

• Even, if f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
• Odd, if f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
• Periodic, if f(np ± x) = f(x), for some number p and n ∈ Z.

Calculation:

We know that \(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).

Let I = \(\rm \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx\)

⇒ \(\rm \int_0^a \frac{f[(a+0)-(a-x)]}{f[(a+0)-x]+f[(a+0)-(a-x)]}\ dx\)

⇒ \(\rm \int_0^a \frac{f(x)}{f(a-x)+f(x)}\ dx\)

⇒ 2I = \(\rm \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx\) + \(\rm \int_0^a \frac{f(x)}{f(a-x)+f(x)}\ dx\)

⇒ 2I = \(\rm \int_0^a 1\ dx\)

⇒ 2I = a.

∴ I = \(\rm\frac a2\).

Explanation

Given

\(\rm\mathop {\lim }\limits_{x \to 2} \frac{x^3 + x^2}{x^2 + 3x + 2}\)

as x → 2 [ Substituting x = 2 ]

= \(\rm \frac{2^3 + 2^2}{2^2 + 3\cdot2 + 2}\)

= \(\frac{12}{12}\)

= 1.

∴ Limit value is 1

Additional Information Continuity of a function:

• A function f(x) is said to be continuous at a point x = a in its domain, if \(\rm\mathop {\lim }\limits_{x \to a}f(x)\) exists or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ \(\rm\mathop {\lim }\limits_{x \to a^+}f(x)=\rm\mathop {\lim }\limits_{x \to a^- }f(x)=\rm\mathop {\lim }\limits_{x \to a}f(x)=f(a)\).

Concept:

Definite Integrals:

\(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).

If f(x) = f(2a - x), then

\(\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx\).

A function f(x) is:

• Even, if f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
• Odd, if f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
• Periodic, if f(np ± x) = f(x), for some number p and n ∈ Z.

Calculation:

\(\rm\int_0^a[f(x)+f(-x)].dx=\int_{-a}^ag(x).dx \)

\(\rm =\int^0_{-a}g(x).dx+\int_0^ag(x).dx \)

\(\rm =\int_0^ag(-x).dx+\int_0^ag(x).dx \)

\(\rm \int_0^a[g(x)+g(-x)].dx \)

∴ f(x) = g(x)

Concept:

• The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral \(\rm \left|\int_a^b f(x)\ dx\right|\), for curves which are entirely on the same side of the x-axis in the given range.
• If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.
• Definite integral: If ∫ f(x) dx = g(x) + C, then \(\rm \int_a^b f(x)\ dx = [ g(x)]_a^b\) = g(b) - g(a).
• \(\rm \int {\sqrt{a^2-x^2}}\ dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C\).

 

Calculation:

Let's first find the points where the curve meets the x-axis (y = 0).

⇒ \(\rm y=\sqrt{16-x^2}=0\)

⇒ x = ±4

Now, since the curve \(\rm y=\sqrt{16-x^2}\) is entirely on one side of the x-axis in the given range x = -4 to x = 4, we have:

The required area = \(\rm \int_{-4}^{\ \ 4} \sqrt{4^2-x^2}\ dx\).

= \(\rm \left[\frac{x}{2}\sqrt{4^2-x^2}+\frac{4^2}{2}\sin^{-1}\frac{x}{4}\right]_{-4}^{\ \ \ 4}\)

= \(\rm \left[\frac{4}{2}\sqrt{4^2-4^2}+\frac{4^2}{2}\sin^{-1}\frac{4}{4}\right] - \left[\frac{-4}{2}\sqrt{4^2-(-4)^2}+\frac{4^2}{2}\sin^{-1}\frac{-4}{4}\right]\)

= \(\rm 8\frac{\pi}{2}+8\frac{\pi}{2}\)

= 8π square units.

Concept:

Slope of a curve y = f(x) at a point is the value of its first derivative at that point.

i.e. m = f'(x).

Maxima/Minima of a function y = f(x):

• Relative (Local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
• Relative (Local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
• At the points of relative (local) maxima or minima, f'(x) = 0.
• At the points of relative (local) maxima, f''(x) < 0.
• At the points of relative (local) minima, f''(x) > 0.

 

Calculation:

The slope of the curve y = -x3 + 3x2 + 2x - 27 will be given by:

m = y' = \(\rm\frac{d}{dx}\left(-x^3 + 3x^2 + 2x - 27\right)\) = -3x² + 6x + 2.

In order to maximize the slope, we must have m' = 0 and m'' < 0.

Now, m' = 0.

⇒ \(\rm\frac{d}{dx}\left(-3x^2 + 6x+2\right)\) = -6x + 6 = 0.

⇒ x = 1.

And m'' = -6 < 0.

Therefore, the slope is maximum at x = 1.

Concept:

The area of a triangle with a fixed perimeter is maximum when it is equilateral.

 

Calculation:

Using the property mentioned above, the area is maximum when the triangle is equilateral.

Height of the triangle = \(\rm \frac{\sqrt3}{2}a=\frac{\sqrt3}{2}8=4\sqrt3\) cm.

Additional Information

Area of a triangle is \(\rm\sqrt{s(s-a)(s-b)(s-c)}\) (Heron's formula), where s = (a + b + c)/2.

Since we are trying to maximize the Area, we can say that we are trying to maximize s(s - a)(s - b)(s - c).

Since s is a constant (because the perimeter is a fixed quantity), this is equivalent to maximizing (s - a)(s - b)(s - c).

Note that (s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 3s - 2s = s, which is a constant.

So, by AM-GM inequality, a product xyz is maximum, if sum x + y + z is a constant, when x = y = z.

∴ (s - a) = (s - b) = (s - c)

⇒ a = b = c, or the triangle is equilateral.

Concept:

• Differentiability of a Function: A function f(x) is differentiable at x = a in its domain if its derivative is continuous at a.

This means that f'(a) must exist, or equivalently: \(\rm \lim_{x\to a^+}f'(x)=\lim_{x\to a^-}f'(x)=\lim_{x\to a}f'(x)=f'(a)\).

• The Modulus Function '| |' is defined as: \( \rm |x|=\left{\begin{matrix}\rm \ \ \ x, &\rm x \ge 0\ \rm -x, &\rm x<0\end{matrix}\right.\).

 

Calculation:

Using the definition of Modulus Function, we have:

f(x) = ex, x ≥ 0.

And, f(x) = e-x, x < 0.

Using the first principle of derivatives, we find that:

\(\rm \lim_{x\to 0^+}f'(x)=\lim_{x\to 0}e^x\) = 1

And, \(\rm \lim_{x\to 0^-}f'(x)=\lim_{x\to 0}-e^{-x}\) = -1.

Since \(\rm \lim_{x\to a^+}f'(x)\neq\lim_{x\to a^-}f'(x)\), the given function is not differentiable at x = 0, or, f'(0) does not exist.

Concept:

• 1 + tan² x = sec2 x.
• \(\rm \int \tan x\ dx\) = ln |sec x| + C.
• \(\rm \int \sec x\ dx\) = ln |sec x + tan x| + C.

 

Calculation:

Let I = \(\rm\int \frac{dx}{\sec x+\tan x}\).

By rationalizing the denominator of the integrand we get:

⇒ I = \(\rm \int \frac{(\sec x - \tan x)}{(\sec x+\tan x) \cdot (\sec x -\tan x)}\ dx\)

⇒ I = \(\rm \int \frac{(\sec x - \tan x)}{(\sec^2 x-\tan^2 x)}\ dx\)

As we know that, sec2 x - tan2 x = 1, therefore:

⇒ I = \(\rm \int (\sec x -\tan x) \ dx\)

⇒ I = \(\rm \int \sec x \ dx - \int \tan x \ dx\)

∴ I = ln |sec x + tan x| - ln |sec x| + C.