Quantitative Aptitude PYQs

Concept:

\(\displaystyle \int \ x^n\ dx = \frac{x^{n+1}}{n+1}+C, n\neq -1\)

Calculation:

\(\displaystyle \int^{\frac{π}{4}}_0 \frac{dx}{(\sin x + \cos x)^2}\)

We know that  \(\displaystyle tan\ x =\frac{sin\ x}{cos\ x}\) ⇒ \(\displaystyle sin\ x =\frac{tan\ x}{sec\ x}\) and \(\displaystyle cos\ x =\frac{1}{sec\ x}\)

⇒ \(\displaystyle I= \int^{\frac{π}{4}}_0 \frac{dx}{(\frac{tan\ x}{sec\ x} +\frac{1}{sec\ x})^2}\)

⇒ \(\displaystyle I= \int^{\frac{π}{4}}_0 \frac{{sec^2\ x}\ dx}{(\tan x + 1)^2}\)

Let tan x = t ⇒ sec²x dx = dt

When x = 0 ⇒ t = tan 0 ⇒ t = 0

When x = π/4 ⇒ t = tan π/4 ⇒ t = 1

Let us now substitute the 't' value and dt value in the given expression for integration.

⇒ \(\displaystyle I= \int^{1}_0 \frac{dt}{(1+t)^2}\)

Let (1 + t) = v ⇒ dv =  dt 

When t = 0 ⇒ v = 1 + 0 ⇒ v = 1

When t = 1 ⇒ v = 1 + 1 ⇒ v = 2

Let us now substitute the 'v' value and dv value in the given expression for integration.

⇒ \(\displaystyle I= \int^{2}_1 \frac{dv}{v^2} \)

Using \(\displaystyle \int \ x^n\ dx = \frac{x^{n+1}}{n+1}+C, n\neq -1\),

⇒ \(\displaystyle I=\left[ \frac{-1}{v}\right ]_1^2 \)

⇒ \(\displaystyle I=\left[ \frac{-1}{2}- \frac{(-1)}{1}\right ] \) = 1/2

∴ \(\displaystyle \int^{\frac{π}{4}}_0 \frac{dx}{(\sin x + \cos x)^2}=\frac{1}{2}\)

Concept:

\(\displaystyle \int \ x^n\ dx = \frac{x^{n+1}}{n+1}+C, n\neq -1\)

Calculation:

\(\displaystyle I=\int (\sin x)^{-1/2} (\cos x)^{-3/2}dx \)

⇒ \(\displaystyle I=\int \frac{1}{(\sqrt{sin \ x.cos^3\ x})}dx \)

⇒ \(\displaystyle I=\int \frac{1}{cos\ x\sqrt{sin \ x.cos\ x}}dx \)

Multiplying and dividing cosec²x in numerator and denominator, we get,

⇒ \(\displaystyle I=\int \frac{cosec^2x}{cosec^2x.cos\ x\sqrt{sin \ x.cos\ x}}dx \)

⇒ \(\displaystyle I=\int \frac{cosec^2x}{\frac{cos\ x}{sin\ x}\sqrt{\frac{sin \ x.cos\ x}{sin^2x}}}dx \)

⇒ \(\displaystyle I=\int \frac{cosec^2x}{cot\ x\sqrt{cot\ x}}dx \)

Let t = cot x ⇒ dt = - cosec²x

⇒ \(\displaystyle I=\int \frac{-dt}{t^\frac{3}{2}}\)

⇒ \(\displaystyle I=\int -t^{\frac{-3}{2}}dt=\frac{-t^{\frac{-1}{2}}}{\frac{-1}{2}}\)

⇒ \(\displaystyle I= \frac{2}{\sqrt{t}}+C\)

⇒ \(\displaystyle I= \frac{2}{\sqrt{cot\ x}}+C\)

⇒ \(I=2\sqrt{tan\ x}+C\)

∴ \(\displaystyle \int (\sin x)^{-1/2} (\cos x)^{-3/2}dx =2\sqrt{tan\ x}+C\)

Formula used:

\(∫ x^ndx = \frac{x^{n+1}}{n+1}+C\)

∫ dx = x + C

Calculation:

\(I_1 =∫\frac{e^x dx}{e^x + e^{-x}}\)       

\(⇒ I_1 =∫\frac{e^x dx}{e^x +\frac{1}{ e^{x}}}\)

\(⇒ I_1 =∫\frac{e^{2x} dx}{e^{2x }+1}\)        ----(1)

 \(I_2 =∫\frac{dx}{e^{2x} + 1},\)          ----(2)

Add equation (1) & (2)

⇒ I = I₁ + I₂

⇒ I = \(∫\frac{e^{2x} dx}{e^{2x }+1}\) + \(∫\frac{dx}{e^{2x} + 1}\)

\(⇒ I = ∫[\frac{e^{2x} }{e^{2x }+1}+ \frac{1}{e^{2x}+1}]dx\)

\(⇒ I = ∫[\frac{e^{2x}+1 }{e^{2x }+1}]dx\)

⇒ I = ∫dx  

Using the formula given above 

∴ I1 + I2  = x + C

Concept:

f(x) = |x| will be equal to 

• x, if x > 0
• -x, if x < 0
• 0, if x = 0

∫ dx = x + C  (C is a constant)

∫ xⁿ dx = xⁿ⁺¹/n+1 + C

Calculation:

Let, \(I = \int^{-1}_{-2}\frac{x}{|x|}dx\)

Using the above concept, as x ∈ (-2, -1)

⇒ \(I=∫^{-1}_{-2}\frac{x}{-x}dx\)

⇒ \(I=-1∫^{-1}_{-2}(1)dx\)

⇒ \(I=-[x]^{-1}_{-2}\)  

⇒ I = -[-1 - (-2)] 

∴  \(\int^{-1}_{-2}\frac{x}{|x|}dx\) = -1

Concept:

First Order Derivative:

Let’s say we have a function f which is continuous at the critical point, defined in an open interval I and

f’(c) = 0 (slope is 0 at c). Then we check the value of f'(x) at the point left to the curve and right to the

curve and check the nature of f'(x), then we can say, that the given point will be:

Local maxima: If f’(x) changes sign from positive to negative as x increases via point c, then f(c) gives

the maximum value of the function in that range.

Local minima: If f’(x) changes sign from negative to positive as x increases via point c, then f(c) gives

the minimum value of the function in that range.

Maxima and minima are points where a function reaches a highest or lowest value, respectively. These

maxima and minima of the function together are called extremes.

Calculation;

Let f(x) = sin4x + 2x        \(0 < x < \frac{\pi}{2} \)

⇒ f′(x) = 4 cos4x + 2

⇒ f′(x) = 0                       

⇒ 4cos4x + 2 = 0

⇒ cos4x = \(\displaystyle\frac{-1}{2}\)​

⇒ 4x = \(\displaystyle \frac{2\pi}{3}\), ​\(\displaystyle \frac{4\pi}{3}\)​

⇒ x = ​\(\displaystyle \frac{\pi}{6}\), \(\displaystyle \frac{\pi}{3}\)

∴ sin4x + 2x have 2 extreme values where \(0 < x < \frac{\pi}{2} \)

Formula used:

• sin θ/cos θ = tan θ
• cos θ/sin θ = cot θ
• sin2θ + cos2θ = 1
• 2sin θ cos θ = sin 2θ

 

Calculation:

\(f(x) = \frac{1}{\tan x+\cot x},\)    \(0 < x < \frac{\pi}{2}?\)

⇒ \(f(x) = \frac{1}{\tan x+\cot x},\)

⇒ \(f(x) = \frac{1}{\frac{sin x}{cosx}+\frac{cosx}{sinx}}\)

⇒ \(f(x) =\frac{sinx\ cosx}{sin^2x+cos^2x}\)

⇒ f(x) = sin x cos x (\(\frac{2}{2}\))    [∵ sin²θ + cos²θ = 1]  

⇒ f(x) = \(\frac{1}{2}\)sin 2x        [∵ 2sin θ cosθ = sin 2θ]

We know that, -1 ≤ sin θ ≤ 1 

⇒  -1 ≤ sin 2x ≤ 1 

∴  f(x)_max = \(\frac{1}{2}\)(1) = 1/2

Formula used:

(a - b)(a + b) = a² - b²

Calculation:

\(4f(x) - f \left(\frac{1}{x}\right)=\left(2x+\frac{1}{x}\right)\left(2x-\frac{1}{x}\right)\)

By using the above formula

⇒ \(4f(x) - f \left(\frac{1}{x}\right)=4x^2-\frac{1}{x^2}\)       -----(1)

Multiply by 4 on both sides

\(16f(x) - 4f \left(\frac{1}{x}\right)=16x^2-\frac{4}{x^2}\)     ----(2)    

Replace x with 1/x in equation (1)

⇒ \(4f(\frac{1}{x}) - f \left(x\right)=4(\frac{1}{x})^2-x^2\)      -----(3)

Add equation (2) & (3)

⇒ 16f(x) - f(x) = (16x² - 4/x²) + (4/x² - x²)

⇒ 15f(x) =  15x² 

⇒ f(x) = x2 

Put x = 2 

∴  f(2) = 4

Concept:

For any two functions f and g, f o g is defined as f[g(x)].

Calculation:

Given that,

f(x) = 4x + 3   

Using the above concept

⇒ fof(x) = 4(4x + 3) + 3

⇒ fof(x) = 16x + 15 

Again using the same concept

⇒ fofof(x) = 16(4x + 3) + 15

⇒ fofof(x) = 64x + 63

Put x = -1 in the above function

⇒ fofof(-1) = 64(-1) + 63 = -1

∴  f o f o f(-1) equal to -1.

Formula used :

1. log mn = log m + log n
2. log mⁿ = n log m
3. Differentiation by part: \(\frac{d}{dx}(uv) = v\frac{du}{dx}+u\frac{dv}{dx}\)
4. \(\frac{d}{dx}x^n=nx^{n-1}\)
5. \(\frac{d}{dx}log\ x= \frac{1}{x}\)

 

Calculation :

Given that

xyyx = 1

Taking the log on both sides

log( xyyx) = log 1

Since, log mn = log m + log n 

⇒ log x^y +  log y^x = log 1

Since log mn = n log m

⇒ y log x + x log y = 0        (∵ log 1 = 0)

Diff. with resepct to x

⇒ \(\frac{d}{dx}\)(y log x) + \(\frac{d}{dx}\) (x log y) = 0 

Using differentiation by part

⇒ \(y\frac{d}{dx}log\ x+ log \ x \frac{dy}{dx}+ x\frac{d}{dx}log\ y+ log \ y \frac{dx}{dx}=0\)

By using the formula (4) & (5)

⇒ \(\frac{y}{x}+ log \ x \frac{dy}{dx}+ \frac{x}{y}\frac{dy}{dx}+ log \ y=0\)

⇒ \(\frac{y}{x}+log\ y + \frac{dy}{dx}(\frac{x}{y}+log\ x) =0\)

⇒ \( \frac{dy}{dx}=-\frac{\frac{y}{x}+log\ y}{\frac{x}{y}+log\ x}\)

Put x = y = 1 in above differential

⇒ \(\frac{dy}{dx}=-\frac{1 + log\ 1}{1 + log \ 1}= -1\)

 \(\frac{dy}{dx}\) at (1, 1) equal to -1.

Formula used :

1. log mn = n log m
2. Differentiation by part: \(\frac{d}{dx}(uv) = v\frac{du}{dx}+u\frac{dv}{dx}\)
3. \(\frac{d}{dx}x^n=nx^{n-1}\)
4. \(\frac{d}{dx}log\ x= \frac{1}{x}\)

 

Calculation :

Given that y = (xx)x,

Taking log on both sides

log y = log (xx)x

Using the formula (1)

⇒ log y = x log x^x

⇒ log y = x² log x

Diff. with respect to x 

\(\frac{d}{dx}\)(log y) = \(\frac{d}{dx}\)(x²log x)

Doing diff. by part as discussed in formula (2)

⇒ \(\frac{1}{y}\frac{dy}{dx} = x^2\frac{d}{dx}log\ x+ log \ x\frac{d}{dx}x^2\)

Since, \(\frac{d}{dx}x^n=nx^{n-1}\) & \(\frac{d}{dx}log\ x= \frac{1}{x}\)

⇒ \(\frac{1}{y}\frac{dy}{dx} = \frac{x^2}{x}+2x\ log \ x\)

⇒ \(\frac{1}{(x^x)^x}\frac{dy}{dx} = \frac{x^2}{x}+2x\ log \ x\)       (∵ y = (xx)x)

Put x = 1 in above equation

⇒ \(\frac{dy}{dx} \) = 1 + 2 × 1 log 1

⇒ \(\frac{dy}{dx} \) = 1          (∵ log 1 = 0)

∴ \(\frac{dy}{dx}\) at x = 1 is 1.

Concept:

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

The domain of [x] is R and the range is I.

Note:

Any function is differentiable only if it is continuous.

The floor function f(x) = ⌊x⌋ is differentiable in every open interval between integers, (n, n + 1) for any integer n.

Calculation:

Given that,

y = [x + 1], -4 < x < -3

We have to determine the derivative at y = [x + 1] at x = -3.5

We know that the floor function is differentiable at all points except integer points.

Hence, y = [x + 1] is differentiable at x = -3.5

⇒ y = [-3.5 + 1] = [-2.5] = -3

⇒ dy/dx = 0

∴ The derivative of y with respect to x at x = -3.5 is 0.

Formula used:

1. \(\int \frac{dy}{y} = ln \ y + C\)
2. log m + log n = log mn
3. log mⁿ = n log m

 

Calculation:

Given that, \(\frac{dy}{dx} = (\ln 5)y \)

⇒ \(\frac{dy}{y} = (\ln 5)dx\)

Doing integration on both sides

⇒ \(\int\frac{dy}{y} =\int (\ln 5)dx\)

Since, \(\int \frac{dy}{y} = ln \ y + C\) 

⇒ ln y = x ln 5 + ln C

Using the logarithmic property (2) & (3)

⇒ ln y = ln 5^x + ln C

⇒ ln y = ln C5^x

⇒ y = C ×  5x          ------(1)

According to the question, y(0) = ln 5

ln 5 = C × 5⁰ 

⇒ C = ln 5

Put this value in equation (1)

⇒ y = 5x ln 5 

Put x = 1

∴ y(1) = 5 ln 5

Concept:

The domain is the subset of R for which all operations in the function's formula make sense.

The derivative of a function at a given point is the slope of the tangent line at that point.

Formulae

\(\displaystyle \frac{dy}{dx}=a^x\ log \ a\)

Calculation:

Statement I: Its domain is (-∞, ∞).

The domain is the subset of R for which all operations in the function's formula make sense.

Since 10 is a positive real constant, it can be raised to any real power, so the domain is not limited. It is R.

So, Its domain is (-∞, ∞).

Statement II: It is a continuous function

We know that f(x) = 10^x 

Derivative of f(x) is f'(x) = 10^x log 10

And now we have to find the points at which the derivative of f(x) is zero.

⇒ f'(x) = 10x log 10 = 0 

As, we know that this can't happen.

So, 10x is a continuous function for all real numbers.

Statement II: It is differentiable at x = 0

The derivative of a function at a given point is the slope of the tangent line at that point.

Here, f(x) = 10x 

⇒ f'(x) = 10x log 10

At x = 0, f'(0) = 100 log 10 ⇒ log 10

The derivative of function f(x) = 10x at x = 0 is log 10.

So, It is differentiable at x = 0.

∴ Statement I, II, and III are correct.

Formula used:

\(\lim\limits_{x \to 0}\frac{sin \ x}{x} = 1\)

\(cosec\ x=\frac{1}{sin\ x}\)

Calculation:

Let the required value be y.

⇒ y = \(\lim\limits_{x \to 0}x^3(\text{cosec} \ x)^2\)

⇒ y = \(\lim\limits_{x \to 0}\frac{x^3}{(sin \ x)^2}\)

⇒ y = \(\lim\limits_{x \to 0}(\frac{x}{sin \ x})^3sin\ x\)

We know that, \(\lim\limits_{x \to 0}\frac{sin \ x}{x} = 1\)

⇒ y = \(\lim\limits_{x \to 0}sin\ x\)

Taking the limit 

⇒ y = 0

∴ \(\lim\limits_{x \to 0}x^3(\text{cosec} \ x)^2\) = 0

Concept:

L’Hospital’s Rule:- 

L’Hospital’s Rule tells us that if we have an indeterminate form \(\rm \frac{0}{0} \) or \(\frac{\infty}{\infty}\)all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

\(\rm \displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \displaystyle \lim_{x \to a}\frac{f'(x)}{g'(x)}\)

Formula used:

\(\frac{d}{dx}x^n=nx^{n-1}\)

\(\frac{d}{dx}\sqrt x = \frac{1}{2\sqrt x} \)

Calculation:

Let the required value be y.

⇒ y = \(\lim\limits_{x \to 1}\frac{x^3 - 1}{\sqrt x - 1}\)

We can see, at x = 1, there is a 0/0 form.

Hence, applying the L'Hospital Rule

⇒ y = \(\lim\limits_{x \to 1}\frac{3x^2}{\frac{1}{2\sqrt x}} = \lim\limits_{x \to 1}{6x^{\frac{5}{2}}}\)

Taking the limit

⇒ y = 6

∴ \(\lim\limits_{x \to 1}\frac{x^3 - 1}{\sqrt x - 1}\) = 6

Concept:

Increasing and Decreasing Functions Definition

• Increasing Function: A function f(x) is said to be increasing on an interval I if for any two numbers x and y in I such that x < y, we have f(x) ≤ f(y).
• Decreasing Function: A function f(x) is said to be decreasing on an interval I if for any two numbers x and y in I such that x < y, we have f(x) ≥ f(y).
• Strictly Increasing Function: A function f(x) is said to be strictly increasing on an interval I if for any two numbers x and y in I such that x < y, we have f(x) < f(y).
• Strictly Decreasing Function: A function f(x) is said to be strictly decreasing on an interval I if for any two numbers x and y in I such that x < y, we have f(x) > f(y).

Calculation:

\(\displaystyle f(x) = \frac{x^3}{3}-\frac{7x^2}{2} + 6x + 5\)

⇒ \(\displaystyle f'(x)=x^2-7x+6\)

For the function to decreasing, f'(x) < 0

⇒ \(\displaystyle f'(x)=x^2-7x+6\)

⇒ f'(x) = x² - 6x - x + 6

⇒ f'(x) = x(x - 6) - 1(x - 6)

⇒ f'(x) = (x - 1) (x - 6)

From the graph, f'(x) is negative in the interval (1,6).

∴ The function is decreasing in (1, 6).

Calculation:

⇒ \(\displaystyle f(x) =\frac{m}{x} +2nx + 1\)

⇒ \(\displaystyle f'(x) =\frac{-m}{x^2} +2n\)

Since the derivative vanishes at x = 2, f'(2) = 0

⇒ \(\displaystyle f'(2) =\frac{-m}{2^2} +2n=0\)

⇒ \(\displaystyle f'(2) =\frac{{-m} +8n}{2^2}=0\)

⇒ m = 8n

∴ Value of m + 8n cannot be determined due to insufficient data.

Concept:

If f(x) and g(x) are continuous on [a, b] and g(x) < f(x) for all x in [a, b], then we have the following formula.

Area = \(\displaystyle \int_a^b \left[ f(x)−g(x) \right] dx\)

Calculation:

Given:

The line y = x lies above the curve y = x³ in the first quadrant and y = x³ lies above the line y = x in the third quadrant.

Solve the curves to obtain the point of intersection y = x³, y = x ⇒ x³ = x.

Equating the values of y, we get,

⇒ x = x³

∴ x3 - x = 0 

⇒ x(x2 - 1) = 0 

⇒ x (x - 1) (x + 1) = 0 

⇒ x = {0, ± 1}

Required area in the first quadrant, A = \(\displaystyle \int^x_0(x -x^3)dx\)

⇒ A = \(\displaystyle \int^1_0 x \ dx -\int^1_0 x^3 \ dx\)

⇒ A = \(\displaystyle \left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1\)

⇒ A = \(\displaystyle \left(\frac{1}{2}-\frac{1}{4}\right)- \left(0-0\right)\)

⇒ A = \(\displaystyle\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)

∴ The area included in the first quadrant between the curves y = x and y = x³ is \(\displaystyle\frac{1}{4}\) square units.

Concept:

First Order Derivative:

Let’s say we have a function f which is continuous at the critical point, defined in an open interval I and f’(c) = 0 (slope is 0 at c). Then we check the value of f'(x) at the point left to the curve and right to the curve and check the nature of f'(x), then we can say, that the given point will be:

Local maxima: If f’(x) changes sign from positive to negative as x increases via point c, then f(c) gives the maximum value of the function in that range.

Local minima: If f’(x) changes sign from negative to positive as x increases via point c, then f(c) gives the minimum value of the function in that range.

Point of inflection: If the sign of f’(x) doesn’t change as x increases via c, and the point c is neither the maxima nor minima of the function, then the point c is called the point of inflection.

Second-Order Derivative:

If the second derivative of the function exists within the given range, then the given point will be:

Local maxima: If f''(c) < 0

Local minima: If f''(c) > 0

Test fails: If f''(c) = 0

Calculation:

xy = 4225 ⇒ y = \(\displaystyle \frac{4225}{x}\)      ------(i)​

Let S = x + y = x + \(\displaystyle \frac{4225}{x}\)​.

Then  \(\displaystyle \frac{ds}{dx}=\left(1-\frac{4225}{x^2}\right )\)

⇒ \(\displaystyle \frac{ds}{dx}=\left(\frac{x^2-4225}{x^2}\right) \)

⇒ \(\displaystyle \frac{ds}{dx}=0\) ​⇒ \(\displaystyle \left(\frac{x^2-4225}{x^2}\right) =0\)

⇒ x² = 4225 ⇒ x = 65

Also, from \(\displaystyle \frac{ds}{dx}=\left(1-\frac{4225}{x^2}\right) \), we get

 \(\displaystyle \frac{d^2s}{dx^2}=4225\left(\frac{2}{x^3}\right) =\frac{8450}{x^3}\)

\(\displaystyle \frac{d^2s}{dx^2}\) is positive at x = 65, it has a minimum value at x = 65.

Putting the value of x from equation (i), we get, 

⇒ y = \(\displaystyle \frac{4225}{x}=65\)      ------(ii)​

∴ S is minimum at x = 65 and minimum value of S = (65 + 65) = 130.

Concept:

A variable separable differential equation is any differential equation in which variables can be separated. 

i.e. The equation which can be written in the form : \(\displaystyle N(y)\frac{dx}{dy}​=M(x)\)

Formulae 

\(\displaystyle \int\frac{dx}{x}=ln\ x+C\)

Calculation:

\(\displaystyle x \frac{dy}{dx}-2y= 0\)

⇒ \(\displaystyle \frac{dy}{dx}-\frac{2y}{x}= 0\)

⇒ \(\displaystyle \frac{dy}{dx}=\frac{2y}{x}\)

⇒ \(\displaystyle \frac{dy}{y}=\frac{2dx}{x}\)

By integrating both sides, we get,

⇒ \(\displaystyle \int\frac{dy}{y}=\int\frac{2dx}{x}\)

⇒ lny = 2 lnx + lnc

⇒ y = x²c

∴ ​y = x2c represents a family of parabolas.