Quantitative Aptitude PYQs

Concept: 

\(\rm \int x^{n}\space dx = \frac{x^{n + 1}}{n + 1} + C\)

\(\rm \int f(x) \space dx^2\) = \(\rm \int (1 + x^{2} + x^{4}) \space d(x^2)\)      ....(i)

Calculation:

Let, x² = u

From equation (i)

​\(\rm \int f(x) \space dx^2\) = \(\rm \int (1 + u + u^{2}) \space du\)

⇒ u + \(\rm \frac{u^{2}}{2}\) + \(\rm \frac{u^{3}}{3}\)+ C

Now putting the value of u,

​⇒ \(\rm \int f(x)dx^2\) = x² +​ \(\rm \frac{x^{4}}{2}\) + \(\rm \frac{x^{6}}{3}\) + C

∴ The required integral is x2 +​ \(\rm \frac{x^{4}}{2}\) + \(\rm \frac{x^{6}}{3}\) + C.

Concept:

Criteria of Maxima

The followings are the criteria of maxima of a function.

(i) The derivative  dy/dx is + ve just before x = a and – ve just after it.

(ii) The derivative dy/dx = 0, just at the point x = a, since it can change its sign from + ve to –ve by passing through zero only.

Criteria of Minima

The followings are the criteria of minima of a function.

(i) The derivative dy/dx is - ve just before x = a and + ve just after it.

(ii) The derivative, dy/dx is equal to zero just at the point x = 0, since it can change its sign from – ve to + ve by passing through zero only.

Calculation:

f(x) = x2 + 1

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

For maxima or minima

\(\rm \frac{d}{dx}f(x)\) = 0

⇒ 2x = 0

⇒ x = 0

The minimum value of this function is at x = 0 but we have to find the minimum and maximum value in the interval [1, 2]

f(1) = 12 + 1 = 1 + 1 = 2

f(2) = 22 + 1 = 4 + 1 = 5

So, at x = 1, minimum value = 2

and at x = 2, maximum value = 5

Hence, both statements are correct.

Given:

f(1) = f(4)

Concept Used:

(i) Integration or Anti Differentiation or Primitive: 

If f & F are function of x such that F' (x) = f(x) then the function F is called a Primitive or Antiderivative or Integral of f(x) w.r.t. x and is written symbolically as 

\(\int F(x) dx = f(x) + c\)  or \(\int f'(x) dx = f(x) + c\) , Where c = Constant of integration

(ii) \(\int^b_af'(x) dx\) = f(b) - f(a) 

Here, f(a) is the lower limit value of the integral and f(b) is the upper limit value of the integral.

Calculation:

According to the question we have to find the value of \(\rm \int^4_1f'(x) dx\)

⇒ \(\rm \int^4_1f'(x) dx\) = \(\rm[ f(x)] \frac{4}{1}\) (Integration is the 'inverse' of differentiation)

⇒ \(\rm \int^4_1f'(x) dx\) = f(4) - f(1)

⇒ f(4) - f(4) = 0

∴ \(\rm \int^4_1f'(x) dx\) equal to 0.

Concept Used:

• \(\rm \int cos x \space dx = sin x \)
• \(e^{ln(f(x))} = f(x)\)

Calculation:

Let I = \(\rm \int^\frac{\pi}{2}_0 e^{ln(cos x)} dx\)

According to the concept used

⇒ I = \(\rm \int^\frac{\pi}{2}_0 (cos x) dx\)

⇒  \(\rm [sinx]_{0}^{\frac{\pi }{2}}\)

⇒  \(\rm sin \frac{\pi }{2} - sin 0\)

⇒ 1 - 0 = 1

∴ The value of the integral \(\rm \int^\frac{\pi}{2}_0 e^{ln(cos x)} dx\)  is 1.

Concept:

sin²x + cos²x = 1

\(\rm \int sin x dx = -cosx \)

\(\rm \int cos x dx = sin x \)

Calculation:

We have \(\rm \int \sqrt{1 - sin 2x} \space dx\) = A sinx + B cosx + C

⇒ \(\rm \int (\sqrt{sin^{2}x + cos^{2}x - 2sinx cosx} )dx\) = A sinx + B cosx + C

⇒ \(\rm \int (\sqrt{(sinx - cosx)^{2}})dx\) = A sinx + B cosx + C

⇒ \(\rm \int (|(sinx - cosx)|)dx\) = A sinx + B cosx + C     ----(i)

If 0 < x < \(\frac{\pi}{4}\), then sinx < cosx

⇒ |sinx - cosx| = -sinx + cosx    -----(ii)

Now from (i) and (ii), we get 

⇒ \(\rm \int (-\space sinx + cosx) \space dx\) = A sinx + B cosx + C

⇒ cosx + sinx + C = A sinx + B cosx + C

On comparing A = 1, B = 1 and C = 0

Hence, A + B - 2 = 0 is correct.

Formula used:

• \(\rm \frac{d}{dx} x^{n} = nx^{n - 1}\)
• The standard equation of an ellipse referred to its principal axes along the co-ordinate axes is \(\rm \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\)
• The order of a differential equation is the order of the highest differential coefficient occurring in it

Calculation:

We have the equation of ellipses \(\rm \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\)

On differentiating with respect to x, we get

\(⇒ \rm \frac{2x}{a^{2}} + \frac{2y y^{'}}{b^{2}} = 0\)  \(⇒ \rm \frac{x}{a^{2}} + \frac{yy^{'}}{b^{2}} = 0\)      ----(i)

\(⇒ \rm \frac{x}{a^{2}} = -\frac{yy^{'}}{b^{2}} \) \(⇒ \rm \frac{b^{2}}{a^{2}} = -\frac{yy^{'}}{x} \)   ----(ii)

Again on differentiating equation (i) with respect to x, we get 

\(⇒ \rm \frac{1}{a^{2}} + \frac{y.y^{''} + \space y'.y'}{b^{2}} = 0\) \(⇒ \rm \frac{b^{2}}{a^{2}} =- [\frac{y.y^{''} + \space y'.y'}{1}] \) ----(iii)

From (ii) and (iii), we get

\(⇒ -\frac{yy^{'}}{x} = - [\frac{y.y^{''} + \space y'.y'}{1}] \) ⇒ x.y.y'' + (y')2 - y y' = 0 

∴ According to the formula used order of the differential equation is 2.

Formula used:

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

Concept Used:

DEGREE OF DIFFERENTIAL EQUATION: The exponent of the highest order differential coefficient, when the differential equation is expressed as a polynomial in all the differential coefficient.

Calculation:

According to the question 

Center of circle = (a, a), Radius of circle = a

So, The equation of circle = (x − a)²  + (y − a)² = a²     -----(1)

On differentiating with respect to x, we get 

2(x - a) + 2(y - a) \(\rm \frac{dy}{dx}\) = 0 ⇒ x + y \(\rm \frac{dy}{dx}\) = a(1 + \(\rm \frac{dy}{dx}\))

⇒ \(\rm a = \frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}}\)      ----(2)

From (i), On expanding the equation of circle, we get

⇒ x² + a² - 2xa + y² + a² - 2ya = a² ⇒ x² + y²  + a2  = 2a(x + y)

On, Putting the value of a in eq(1), we get

x2 + y2  + \(\rm (\frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}})^2\) = 2\(\rm \frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}}\) (x + y)

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

∴ According to the concept used the degree of the differential equation is 2.

Formula used:

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

Calculation:

We have \(\rm y = A -\frac{B}{x}\)

On differentiating the above equation with respect to x, we get 

⇒ \(\rm y_{1} = \frac{B}{x^{2}}\)

⇒ x² \(\rm y_{1} = B\)

again differentiate with respect to x

⇒ \(\rm x^{2} y_{2} + 2x y_{1} = 0\)

⇒ x\(\rm (x y_{2}+ 2 y_{1}) = 0\)

⇒ \(\rm (x y_{2}+ 2 y_{1}) = 0\)

∴ The differential equation of \(\rm y = A -\frac{B}{x}\) is \(\rm (x y_{2}+ 2 y_{1}) = 0\).

Formula used:

\(\rm \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx\)

tan(π - θ) = - tan θ  

Calculation:

Let

I = \(\int^π _0 ln\left(tan\frac{x}{2}\right)dx\)        ---(1)

According to the formula used

I = \(\int^π _0 ln\left(tan (π -\frac{x}{2})\right)dx\)

⇒ I = - \(\int^π _0 ln\left(tan\frac{x}{2}\right)dx\)

From equation (1)

⇒ I = -I

⇒ 2I = 0

⇒ I = 0

∴ The value of the integral \(\int^π _0 ln\left(tan\frac{x}{2}\right)dx\) is 0.

Formula used:

\(\rm \frac{d}{dx} e^{x} = e^{x}\)

Calculation:

equation of curve is y = e^x

tangent to the curve is

\(\rm \frac{dy}{dx} = e^{x}\)

Slope of tangent at (0, 1) is

\(\rm \frac{dy}{dx}_{(0,1)} = e^{0}\)

\(\rm \frac{dy}{dx}_{(0,1)} = 1\)

Now, Equation of tangent at (0, 1) is 

(y - 1) = 1 (x - 0)

⇒ x - y + 1 = 0

On putting y = 0 we get x = -1 

∴ The tangent will meet x axis at (-1, 0).

Formula used:

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

Concept Used:

If f''(a) > 0 then x = a is a point of local minima

If f''(a) < 0 then x = a is a point of local maxima

Calculation:

Let f(x) = x + \(\frac{1}{x}\)​

\(\rm \frac{d f(x)}{dx} = 1 - \frac{1}{x^{2}}\)

\(\rm \frac{d f(x)}{dx} = 0\)

x = ±1

\(\rm \frac{d^{2} f(x)}{dx^{2}} = \frac{2}{x^{3}}\)

At x = 1, \(\rm \frac{d^{2} f(x)}{dx^{2}} = 2\) 

\(\rm \frac{d^{2} f(x)}{dx^{2}} > 0\)

At x = -1, \(\rm \frac{d^{2} f(x)}{dx^{2}} = -2\)

\(\rm \frac{d^{2} f(x)}{dx^{2}} < 0\)

Hence local maximum value of f(x) is at x = −1 and the local maximum value = −2

Local minimum value of f(x) is at x = 1 and local minimum value = 2.

Therefore, the local maximum value (−2) is less than the local minimum value 2.

∴ Statement 1 is only correct.

Shortcut TrickLet the length of the rectangle be L and its breadth be B

According to the question,

The diagonal of the rectangle = Diameter of the circle.

√(L2 + B2) = 4   

⇒ L2 + B2 = 42    ----(1)

For maximum area of rectangle 

Length should be equal to it breadth [ L = B]

So, from (1) 

L = B = √8

∴ Area of rectangle = L × B = √8 × √8 = 8

 

Detailed Mehtod

Formula used:

Area of rectangle = length × breadth

Diameter of circle = 2 × Radius of circle 

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

Calculation:

Let the length of the rectangle be L and its breadth be B

Since the rectangle is inscribed inside the circle, its diagonal has to be the diameter of the circle.

√(L² + B²) = 4   

⇒ L2 + B2 = 42    ----(1)

The area of the rectangle is given by Lb

Differentiating the condition w.r.t L,

2L + 2b \(\rm \frac{dB}{dL}\) = 0

⇒ \(\rm \frac{dB}{dL} = \frac{-L}{B}\)       ----(2)

Now, differentiating the area w.r.t l since we need to maximize it,

B + L \(\rm \frac{dB}{dL}\) = 0       ----(3)

Substituting equation (2) into equation (3), we have

⇒ B + L \(\rm \frac{-L}{B}\) = 0   

⇒ B² = L²

⇒ B = L

Substituting this relation in equation (1), we get

⇒ 2L² = 4² or L² = 8

⇒ area = LB = L² = 8 square units

∴ The maximum area of rectangle inscribed in the circle is 8 sq.units.

Formula used:

\(\rm \int \frac{1}{x} dx = log : x + C\)

log_ax - log_ay = \(\rm log_{a}\frac{x}{y}\)

Calculation:

\(\int \frac{dx}{x(x^2 + 1)}\)

⇒ \(\rm \int \frac{dx}{x} - \frac{xdx}{x^{2} + 1}\)

⇒ \(\rm \int \frac{dx}{x} - \frac{1}{2}\int \frac{2xdx}{x^{2} + 1}\)

⇒ ln x - \(\rm \frac{1}{2} ln |1 + x^{2}|\) + c

⇒ \(\frac{1}{2} \ln x^{2} -\)(\rm \frac{1}{2} ln |1 + x^{2}|\) + c

⇒ \(\rm \frac{1}{2}ln\left(\frac{x^2}{x^2 + 1}\right) + C\)

∴ \(\int \frac{dx}{x(x^2 + 1)}\) is equal to 

\(\rm \frac{1}{2}ln\left(\frac{x^2}{x^2 + 1}\right) + C\).

Formula used:

\(\rm \frac{d}{dx} e^{x} = e^{x}\)

Differentiation of y with respect to x is given by dy/dx.

Calculation:

Let, f(x) = \(\rm e^{e^x}\)

d[f(x)] = d[\(\rm e^{e^x}\)]

⇒ d[f(x)] = \(\rm e^{e^x} e^x\)     ----(1)

Again, let, g(x) = \(\rm e^x\)

⇒  d[g(x)] = d[\(\rm e^x\)]      -----(2)

Hence, the required derivative

 \(\rm \frac{d[f(x)]}{d[g(x)]}\)= \(\rm \frac{\rm e^{e^x} e^x}{e^x}\)

⇒ \(\rm \frac{d[f(x)]}{d[g(x)]}\) = \(\rm e^{e^x}\)

∴ The derivative of \(e^{e^x}\) with respect to ex is \(\rm e^{e^x}\)

Formula used:

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

Calculation:

f(x) = x3 + x2 + kx

f^'(x) = 3x² + 2x + k

For no extremum, f should have zero turning points.

There will be no slope existing either it is +ve or -ve

For that above quadratic have no real roots.

⇒ f'(x) ≠ 0 which is only possible when,

2² - 4 × 3 × k < 0

⇒ 4 - 12k < 0

⇒ 4 < 12k

⇒ 1 < 3k

⇒ 3k > 1

∴ The condition that f(x) = x3 + x2 + kx has no local extremum is 3k > 1

Given:

f(x) = 2x

Formula used:

\(\rm \frac{d}{dx} a^{x} = (In: a).a^{x}\)

\(\rm \int 1.dx = x + C\)

\(\rm \int_{a}^{b} f(x) dx = F(b) - F(a)\), Where \(\rm \int f(x).dx = F(x) + C\)

Calculation:

We have f(x) = 2x     ----(i)

⇒ f^'(x) = 2x In(2)    ----(ii)

Now, We have to find the value of \(\rm \int^{10}_2\frac{f'(x)}{f(x)}dx\)

Form equation (i) and (ii), we get 

⇒ \(\rm \int^{10}_2 \frac{2^{x} In(2)}{2^{x}} dx\)

⇒ \(\rm \int^{10}_2 { In(2)} \space dx\)

⇒ \()In(2) [10 - 2]

⇒ 8 In(2)

**∴​ The value of \(\rm \int^{10}_2\frac{f'(x)}{f(x)}dx\) is equal to 8 In(2).**

Given:

\(\rm \int^0_{-2} f(x)dx=k\)

Formula used:

Properties of definite integrals:

\(\rm \int_{a}^{b} f(x) dx = F(b) - F(a)\) Where, \(\rm \int f(x).dx = F(x) + C\)

\(\rm \int_{a}^{b} k.f(x) = k \int_{a}^{b} f(x) dx\), (k = constant)

|x| = x, if x ≥ 0

|x| < 0, if x < 0 

Calculation:

We have to find the value of \(\rm \int^0_{-2}|f(x)|dx\)

\(\int^0_{-2}|f(x)|dx\) 

⇒ \(\rm \int^ {0}_{-2}|f(x)|dx \)     ----(i)

Since, The lower and upper limit is -2 and 0 which is less than zero,

So, |f(x)| = -f(x)

Now from (i), we get 

⇒ \(\int^0_{-2}|f(x)|dx = -\int^0_{-2}f(x)dx\) 

⇒ - k ≤ k [∵ We don't know the exact value of k (it can be zero, -ve, +ve)]

∴ \(\rm \int^0_{-2}|f(x)|dx \le k\).

Formula used:

Differentiation Formula:

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

Calculation:

f(x) = x2 - kx

⇒ f^'(x) = 2x - k

Since, f is monotonic increasing.

f'(x) > 0

⇒ 2x - k > 0

⇒ k < 2x     ----(1)

Since, we have 1 < x < ∞ 

⇒ 2 < 2x < ∞    ----(2)

From (1) and (2), we get

k < 2

∴ The correct relation is k < 2.

TFormula used:

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

\(\rm \int_{a}^{b} f(x) dx = F(b) - F(a) \space, Where \rm \int f(x).dx = F(x) + C\)

Calculation:

Required Area = \(| \rm \int_{-1.8}^{-1.5} [x] dx |\) , [∵ Area can never be negative]

Since, [x] represents G.I.F and  -1.8 < x < -1.5

So, The value of [x] for the given range will be - 2

⇒ Required Area = \(| \rm \int_{-1.8}^{-1.5} [x] dx | = | -2 \rm \int_{-1.8}^{-1.5} dx |\) 

⇒ Required Area = 2 [-1.5 + 1.8]

⇒ Required Area = 0.6

∴ The required area is 0.6 square root.

Formula used:

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

Calculation:

x2 = y 

So, differentiate with respect to x

\(\rm \frac{dy}{dx} = 2x\) 

⇒ \(\rm \frac{dy}{dx}_{(1,1)} = 2\)

⇒ tan θ = 2

We know that tan π/3 = √3 = 1.732 and tan π/2 = ∞

1.732 < 2 < ∞

⇒ tanπ/3 < tan θ <  tan π/2

∴ \(\rm \frac{\pi}{3} < \theta < \frac{\pi}{2}\) is the correct relation.