Quantitative Aptitude PYQs

Calculation:

Given: \(\frac{{dy}}{{dx}} = \cos \left( {y - x} \right) + 1\)

Let y – x = t

Now, by differentiating both the sides w.r.t x, we get

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

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

So, the given differential equation can be written as

\(⇒ 1 + \frac{{dt}}{{dx}} = \cos t + 1\)

\(⇒ \frac{{dt}}{{dx}} = \cos t\)

\(⇒ \sec t;dt = dx\)

Now, by integrating both the sides we get

\(\smallint \sec t;dt = \smallint dx\)

As we know that, \(\smallint \sec x;dx = \log \left| {sec;x + \tan x} \right|; + C\)

\(⇒ \log \left| {\sec t + \tan t} \right| = x + C\)

Taking ant-log both the sides we get

\(⇒ \sec t + \tan t = {e^x} \times {e^C}\)

Now, substitute t = y – x in the above equation, we get

\(⇒ {e^x} = c \times \left[ {\sec \left( {y - x} \right); + \tan \left( {y - x} \right)} \right],;where;c = {e^{ - C}}\)

We know that sec² A - tan² A = 1

⇒ (sec A - tan A) (sec A + tan A) = 1

⇒  (sec A + tan A) = \(\frac{1}{\sec A - \tan A}\)

∴ sec (y - x) + tan (y - x) = \(\frac{1}{\sec (y - x) - \tan (y -x)}\)

Now, \({e^x} = c \times \left[ {\sec \left( {y - x} \right); + \tan \left( {y - x} \right)} \right]\)

⇒ \({e^x} = c \times \frac{1}{\sec (y - x) - \tan (y - x)}\)

⇒ ex [sec (y – x) – tan (y – x)] = c

Concept Used:

When 0 < x < π4, then sinx < cosx 

⇒ sinx - cosx < 0 

⇒ |sinx - cosx| = - (sinx - cosx)

When π/4 < x < π/2, then sinx > cosx 

⇒ sinx - cosx > 0 

⇒ |sinx - cosx| = (sinx - cosx)

Calculation:

According to the formula used

\(\mathop \smallint \limits_0^{\frac{π }{2}} |\sin x - \cos x|dx = ;\mathop \smallint \nolimits_0^{\frac{π }{4}} \left( {\cos x - \sin x} \right);dx + ;\mathop \smallint \nolimits_{\frac{π }{4}}^{\frac{π }{2}} \left( {\sin x - \cos x} \right);dx\)

\(⇒ \mathop \smallint \nolimits_0^{\frac{π }{4}} \left( {\cos x - \sin x} \right);dx = \left[ {\left( {\sin \frac{π }{4} - \sin 0} \right) + \left( {\cos \left( {\frac{π }{4}} \right) - \cos 0} \right)} \right] \)

\(\Rightarrow \frac{2}{{\sqrt 2 }} - 1 = \sqrt 2 - 1\)

\( ⇒ \mathop \smallint \nolimits_{\frac{π }{4}}^{\frac{π }{2}} \left( {\sin x - \cos x} \right);dx = ;\left[ { - ;\left( {\cos \frac{π }{2} - \cos \frac{π }{4}} \right) - \left( {\sin \frac{π }{2} - \sin \frac{π }{4}} \right)} \right] \)

\(\Rightarrow \frac{2}{{\sqrt 2 }} - 1 = ;\sqrt 2 - 1\)

\(\therefore \mathop \smallint \limits_0^{\frac{π }{2}} |\sin x - \cos x|dx = 2\left( {\sqrt 2 - 1} \right)\)

Calculation:

Given: y = a cos 2x + b sin 2x

Now, by differentiating both the sides w.r.t x, we get

\(\Rightarrow \frac{{dy}}{{dx}} = ; - 2a\sin 2x + 2b\cos 2x\)

Now, again by differentiating both the sides w.r.t x of the above equation, we get

\(\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = ; - ;4a\cos 2x - 4b\sin 2x\)

\(\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = ; - 4;\left( {a\cos 2x + b\sin 2x} \right) = ; - ;4y\)

\(\therefore The \space \space required \space \space differential \space equation \space \space is \space \frac{{{d^2}y}}{{d{x^2}}} + 4y = 0\)

Calculation:

Let V be the volume of the half cylinder

\(\Rightarrow V = \frac{1}{2};\pi {r^2};h\), where r is the radius and h is the height of the cylinder

\(\Rightarrow h = \frac{{2V}}{{\pi ;{r^2}}}\)      ----(1)

Total surface area = area of rectangular base + area of 2 semi-circular ends + area of curved surface

\(\Rightarrow S = 2rh + 2 \times ;\frac{1}{2};\pi {r^2}; + \frac{1}{2} \times 2\pi rh\)

\(\Rightarrow S = 2rh + \pi {r^2} + \pi rh\)

By substituting (1) in the above equation, we get

\(S = 2r\left( {\frac{{2V}}{{\pi ;{r^2}}}} \right) + ;\pi {r^2} + \pi rh\)

Now, by differentiating the above equation w.r.t r, we get

\(\Rightarrow \frac{{dS}}{{dr}} = 2\pi r - \frac{1}{{{r^2}}};\left( {\frac{{4V}}{\pi } + 2V} \right)\)      ----(2)

By equating the above equation to zero , we get

\(\Rightarrow \frac{{dS}}{{dr}} = 2\pi r - \frac{1}{{{r^2}}};\left( {\frac{{4V}}{\pi } + 2V} \right) = 0\)

\(\Rightarrow {r^3} = \frac{{2V}}{{{\pi ^2}}}; + \frac{V}{\pi }\)

Again, by differentiating the equation (2) w.r.t x we get

\(\Rightarrow \frac{{{d^2}S}}{{d{r^2}}} = 2\pi + \frac{1}{{{r^3}}};\left( {\frac{{4V}}{\pi } + 2V} \right) > 0\)

So by substituting the value of V in the above equation we get,

\(\Rightarrow r = \frac{h}{{2\pi }};\left( {2 + \pi } \right)\)

\(\Rightarrow \frac{{2r}}{h} = \frac{\pi }{{\pi + 2}}\)

Concept Used:

\(\rm \int e^{f(x)}. f'(x) dx = e^{f(x)} +c\)

Calculation:

Let sin x = t

When x = 0 then t = 0     

And when x = π/2 then t = 1.

Now by differentiating both the sides w.r.t x, we get

\(⇒ \cos x;dx = dt\)

Now, We have to find the value of \(\mathop \smallint \limits_0^{\frac{\pi }{2}} {e^{\sin x}}\cos x;dx\)

\(⇒ ;\mathop \smallint \nolimits_0^1 {e^t};dt\)

\(⇒ \left[ {{e^1} - {e^0}} \right] \)

⇒ e - 1

∴ The value of the integral \(\mathop \smallint \limits_0^{\frac{\pi }{2}} {e^{\sin x}}\cos x;dx\) is  e - 1.

Given: \(f\left( x \right) = \frac{{x - 2}}{{x + 2}},;x \ne - 2\)

Let \(y = \frac{{x - 2}}{{x + 2}}\)

\(\begin{array}{l} \Rightarrow xy + 2y = x - 2\ \Rightarrow 2y + 2 = x;\left( {1 - y} \right) \end{array}\)

\(\Rightarrow x = \frac{{2y + 2}}{{1 - y}}\)

\(\Rightarrow x = \frac{{2;\left( {y + 1} \right)}}{{1 - y}}\)

Now, by interchanging x and y in above equation we get,

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

Concept:

\(\smallint u.v \space dx= u.\smallint v \space dx - \smallint \frac {du}{dx}.\smallint v \space dx\)

ILATE Rule: 

I-Inverse function, L-Logarithmic function, A-Algebraic function,

T-Trigonometric function & E-Exponential function

Calculation:

We have to find the value of ∫ In (x²) dx

Let x² = u

⇒ x = √u

⇒ dx = 1/(2√u) du

On putting the value of x and dx in the given integral 

∫ In (x2)

\(⇒ ∫ln \space u \space. \space \frac {1} {(2√u)} du \)

\(⇒ \frac {1}{2} ∫ln \space u \space. \space \frac {1} {(√u)} du \)

\(⇒ [\frac {1}{2}] \space [ln \space u \space.∫\frac {1} {(√u)} du - ∫ \frac {1}{u} . ∫\frac {1} {(√u)} du] \)

\(⇒ [\frac {1}{2}] \space [lnu \space. 2(√u) - ∫ \frac {1}{u} . 2√u) du] \)

\(⇒ [\frac {1}{2}] \space [ 2(√u) \space lnu - 4√u) ] + c\)

Now, On putting u = x², we get 

\(⇒ [\frac {1}{2}] \space [ 2x \space lnx^2 - 4x) ] + c\)

⇒ x lnx² - 2x + c

⇒ 2x lnx - 2x + c

∴ The integral of ∫ In (x2) dx is 2x lnx - 2x + c.

Given:

y² = 8x and (4, 2)

Calculation:

Let y = a is a point of the parabola y2 = 8x

Now, On substituting y = a in the above equation, we get

⇒ x = a²/8

So, The point (a²/8, a) lies on the curve y² = 8x.

Let the distance between (a²/8, a) and (4, 2) be given by

\({D^2} = \frac{1}{{64}};{\left( {{a^2} - 32} \right)^2} + {\left ( {a - 2} \right)^2}\)     ---(1)

By differentiating the above equation w.r.t a, we get

\(\Rightarrow 2D\frac{{d;\left( D \right)}}{{da}} = \frac{1}{{64}}\times 4a\left( {{a^2} - 32} \right) + 2\left( {a - 2} \right)\)

\(\Rightarrow 2D\frac{{d;\left( D \right)}}{{da}} = \frac{1}{{16}} \times a\left( {{a^2} - 32} \right) + 2\left( {a - 2} \right)\)

Now, we know that in order to find the minimum distance

\(\frac{{d\left( D \right)}}{{da}} = 0\)

\(\Rightarrow 0 = \frac{{{a^3}}}{{16}} - 2a + 2a - 4\)

⇒ a³ = 64

⇒ a = 4

So, by substituting a = 4 in equation (1), we get

⇒ D² = 8 ⇒ D = 2√2

∴ The minimum distance between (4, 2) and the curve is 2√2.

Calculation:

As we know that a circles touching y – axis at origin has the centre on x – axis.

So, the equation of circle touching y – axis at origin is: (x - a)² + y² = a²

⇒ x² + y² – 2ax = 0       ---(1)

So, by differentiating the above equation w.r.t x, we get

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

\(\Rightarrow 2a = 2;\left( {x + y\frac{{dy}}{{dx}}} \right)\)

So, by substituting the value of 2a in equation (1), we get

⇒ x2 + y2 – \(2x;\left( {x + y\frac{{dy}}{{dx}}} \right)\)= 0

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

\(\therefore {x^2} - {y^2} + 2xy\frac{{dy}}{dx} = 0\)

Concept:

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

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:

\(\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} + 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 9y = x\)

⇒ In above differential equation highest derivative is two and its power is one

∴ Order = 2 and Degree  = 1

Both 1 and 2 are correct.

Concept:

Differential Equations by Variable Separable Method

If the coefficient of \(dx\) is only function of x and coefficient of \(dy\) is only a function of y in the given differential equation then we can separate both \(dx\) and \(dy\) terms and integrate both separately.

\(\Rightarrow \smallint f\left( x \right)dx = \smallint g\left( y \right)dy\)

Calculation:

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

\(\Rightarrow \frac{{dy}}{{dx}} = - \frac{x}{y}\)

\(\Rightarrow y;dy = - x;dx\)

Integrating both sides

\(\Rightarrow \smallint y;dy = - \smallint x;dx\)

\( \Rightarrow \frac{{{y^2}}}{2} = - \frac{{{x^2}}}{2} + C\)

\(\Rightarrow \frac{{{y^2}}}{2} + \frac{{{x^2}}}{2} = C\)

\(\Rightarrow \frac{{{y^2} + {x^2}}}{2} = C\)

\(\Rightarrow {y^2} + {x^2} = 2C\)

\(\therefore The \space \space required \space \space equation\space \space is \space \space {y^2} + {x^2} = c\).

Concept:

f(x) is Continuous at x = 0

\(\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right); = ;\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right); = f(0)\)

Calculation:

f(x) is Continuous at x = 0

\(\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right); = ;\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right); = f(0)\)

\( \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} \sin x; = ;\mathop {\lim }\limits_{x \to {0^ - }} \sin x; = k\)

\(\Rightarrow \mathop {\lim }\limits_{h \to 0} \sin \left( {0 + h} \right); = ;\mathop {\lim }\limits_{h \to 0} \sin \left( {0 - h} \right); =k\)

\(\Rightarrow k = 0\)

\(\therefore ;k = 0\)

Concept:

AM-GM inequality:

If x1.  . . Xn ≥ 0, then

\(\Rightarrow \frac{{x1 + x2 + \ldots + xn}}{n}; \ge ;\sqrt[n]{{x1.x2 \ldots xn}}\)

Calculation:

Let a²x and b²y are two numbers.

Then apply AM-GM inequality

\(AM \ge GM\)

\(\Rightarrow \frac{{{{\rm{a}}^2}{\rm{x;}} + {\rm{;}}{{\rm{b}}^2}{\rm{y}}}}{2}; \ge ;\sqrt[2]{{{{\rm{a}}^2}{\rm{x;}}{{\rm{b}}^2}{\rm{y}}}}\)

\(\Rightarrow {{\rm{a}}^2}{\rm{x;}} + {\rm{;}}{{\rm{b}}^2}{\rm{y;}} \ge 2\sqrt[2]{{{{\rm{a}}^2}{{\rm{b}}^2}xy}}\)

\(Given;xy = {c^2}\)

\(\Rightarrow {{\rm{a}}^2}{\rm{x;}} + {\rm{;}}{{\rm{b}}^2}{\rm{y}} \ge 2\sqrt[2]{{{{\rm{a}}^2}{{\rm{b}}^2}{c^2}}}\)

\( \Rightarrow {{\rm{a}}^2}{\rm{x;}} + {\rm{;}}{{\rm{b}}^2}{\rm{y;}} \ge {\rm{;}}2{\rm{abc}}\)

∴ minimum value of a²x + b²y is 2abc

Concept:

\( {e^{\ln f(x)}} = f(x)\)

\( \smallint {a^x}dx = \frac{{{a^x}}}{{\ln a}} + c\)

xln(a) = ln(a)^x

Calculation:

\(\smallint {e^{x\ln a}}dx = \smallint {e^{\ln {a^x}}}dx\)

\(\Rightarrow \smallint {e^{x\ln a}}dx = \smallint {a^x}dx\)

\(\Rightarrow \smallint {e^{x\ln a}}dx = \frac{{{a^x}}}{{\ln a}} + c\)

∴ The value of the integral is \(\frac{{{a^x}}}{{In;\left( a \right)}} + c\).

Concept:

Area under a Curve by Integration

Find the area under this curve is by summing horizontally

In this case, we find the area is the sum of the rectangles, heights y = f(x) and width dx.

\(\therefore ;{\bf{A}}; = ;\mathop \smallint \nolimits_{\bf{a}}^{\bf{b}} {\bf{ydx}}; = ;\mathop \smallint \nolimits_{\bf{a}}^{\bf{b}} {\bf{f}}\left( {\bf{x}} \right){\bf{dx}}\)

Calculation:

\({\rm{Area;under;curve;;}} = {\rm{;}}\mathop \smallint \nolimits_0^{\rm{\pi }} ({\rm{c}}\sin {\rm{x}}){\rm{dx}}\)

\(= c\mathop \smallint \nolimits_0^\pi \sin xdx\)

\(= c\left[ { - \cos x} \right]_0^{\rm{\pi }}\)

\(= - c\left[ {\cos \pi - \cos 0} \right]\)

\(= - c\left[ { - 1 - 1} \right]\)

\(= 2c\)

Concept:

\(\Rightarrow \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\)

Calculation:

\( \Rightarrow ;\sin \theta + \cos \theta ; = ;\sqrt 2 ;\cos \theta\)

\( \Rightarrow ;\sin \theta = \sqrt 2 ;\cos \theta - \cos \theta \)

\(\Rightarrow ;\sin \theta = \left( {\sqrt 2 - 1} \right)\cos \theta\)

Multiple by \(\surd 2\) + 1 both sides;

\(\Rightarrow ;(\surd 2 + 1{\rm{;}})\sin \theta = \left( {\sqrt 2 + 1{\rm{;}}} \right)\left( {\sqrt 2 - 1} \right)\cos \theta \)

\(\Rightarrow ;\sqrt 2 \sin \theta + \sin \theta = \left( {2 - 1} \right)\cos \theta \)

\( \Rightarrow ;\sqrt 2 \sin \theta = \cos \theta - \sin \theta\)

\(\therefore ;\cos \theta - \sin \theta = \sqrt 2 \sin \theta\)

Concept:

Length of arc = \(\frac{\theta }{{360}} \times 2\pi r\)

Calculation:

Given Data;

Diameter = 44 cm

\({\rm{Radius}} = {\rm{}}\frac{{{\rm{Diameter}}}}{2}{\rm{}} = {\rm{}}\frac{{44}}{2}{\rm{}} = {\rm{;}}22{\rm{cm}}\)

Chord length = 22 cm

Figure

In triangle AOB;

OA = OB = Radius = 22 cm

AB = 22 cm

∵ All sides are equal so triangle is equilateral

∴ All angles are same, \(\theta = 60\)

\({\rm{Length;of;arc}} = {\rm{}}\frac{{\rm{\theta }}}{{360}} \times 2{\rm{\pi r}} = {\rm{}}\frac{{60}}{{360}} \times 2 \times \frac{{22}}{7} \times 22{\rm{}} = {\rm{}}\frac{1}{6} \times \frac{{968}}{7}{\rm{}} = {\rm{}}\frac{{484}}{{21}}{\rm{;cm}}\)

Concept:

1. In First quadrant all trigonometric identities are positive.

2. In second quadrant sin and cosec are positive.

3. In Third quadrant tan and cot are positive

4. In Fourth quadrant cos and sec are positive.

Explanation:

Given sinθ = -1/2 and tanθ = 1/√3

sinθ is negative this is possible only when 

θ lies in 3^rd quadrant or in 4^th quadrant 

and tanθ is positive, this is possible only when 

θ lies in 1^st quadrant  or in 3^rd quadrant 

From the above two statements, we can say that if θ lies in 3^rd quadrant then sin θ is negative and tan θ is positive.

Concept:

Arrangement of r things that can be done out of total n things. This is denoted by \({n_{{P_r};}}\)

\(\Rightarrow ;{n_{{P_r};}} = \frac{{n!}}{{\left( {n - r} \right)!}}\)

Calculation:

For even number unit digit can be filled in two ways (2 or 4 can be put on unit place)

Remaining two digits can be filled in \({\rm{}}{4_{{{\rm{P}}_2}{\rm{}}}}\)

Number of even numbers \({\rm{}} = {\rm{}}2 \times {\rm{}}{4_{{{\rm{P}}_2}{\rm{}}}}{\rm{}} = {\rm{}}2 \times 12{\rm{}} = {\rm{;;}}24\)

Calculation:

Let OP be the tower and let A and B be two points due south and east respectively of the tower such that ∠OAP = x and ∠OBP = y

Let OP = h

In △OAP, we have

⇒ tan x = h/OA

⇒ OA = h cot x      ....

(1) In △OBP, we have

⇒ tan y = h/OB

⇒ OB = h cot β       .... (2) Since OAB is a right angled triangle. Therefore,

AB² = OA² + OB²

⇒ z² = h²cot² x + h² cot²y

∴ z² = h² (cot² x + cot²y)