Quantitative Aptitude PYQs

Concept:

Minimum value of \(m{{\sec }^{2}}\theta +n~cose{{c}^{2}}\theta =~{{\left( \sqrt{m}+\sqrt{n} \right)}^{2}}~\)

Calculation:

\(\Rightarrow \frac{{{a}^{2}}}{{{\cos }^{2}}x}+\frac{{{b}^{2}}}{{{\sin }^{2}}x}={{a}^{2}}{{\sec }^{2}}x+{{b}^{2}}~cose{{c}^{2}}x~\)

As we know that, minimum value of \(m{{\sec }^{2}}\theta +n~cose{{c}^{2}}\theta =~{{\left( \sqrt{m}+\sqrt{n} \right)}^{2}}\)

Here, m = a² and n = b².

So, the minimum value of \(\frac{{{a}^{2}}}{{{\cos }^{2}}x}+\frac{{{b}^{2}}}{{{\sin }^{2}}x}={{a}^{2}}{{\sec }^{2}}x+{{b}^{2}}~cose{{c}^{2}}x={{\left( a+b \right)}^{2}}\)

Concept:

If A, B and C are angles and a, b and c are the sides of a ΔABC, then: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C~}\)

Calculation:

Let us consider Δ ABC with A, B and C as its angles and a, b and c as its sides.

Given: Angles are in AP i.e A, B and C are in AP.

⇒ 2 × B = A + C

As we know that, A + B + C =180°

⇒ 3 × B = 180°     ----(∵ 2 × B = A + C)

⇒ B = 60°

As we know that, if A, B and C are angles and a, b and c are the sides of a Δ ABC, then:

\(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C~}\)

\(\Rightarrow \frac{b}{c}=\frac{\sin B}{\sin C}\)

As it is given that b : c = √3 : √2 and sin C = sin 60° = √ 3 / 2.

\(\Rightarrow \frac{\sqrt{3}}{\sqrt{2}}=\frac{\frac{\sqrt{3}}{2}}{\sin C}~\Rightarrow \sin C=\frac{1}{\sqrt{2}}\)

⇒ C = 45°

∵ A + B + C = 180° ⇒ A + 60° + 45° = 180°

⇒ A = 75°.

Concept:

\(\cot \left( A-B \right)=\frac{\cot A\times \cot B+1}{\cot B-\cot A}\)

Calculation:

Given: tan A - tan B = x and cot B - cot A = y

∵ cot B - cot A = y

\(\Rightarrow \cot B-\cot A~=\frac{1}{\tan B}-\frac{1}{\tan A}=\frac{\tan A-\tan B}{\tan B\times \tan A}=\frac{x}{\tan B\times \tan A}=y\)

\(\Rightarrow \frac{x}{y}=\tan A\times \tan B\)

\(\Rightarrow \frac{y}{x}=\cot A\times \cot B\)

As we know that, \(\cot \left( A-B \right)=\frac{\cot A\times \cot B+1}{\cot B-\cot A}\)

\(\Rightarrow \cot \left( A-B \right)=\frac{\frac{y}{x}+1}{y}=\frac{1}{x}+\frac{1}{y}\)

Concept:

sin (α + β) = sin α × cos β + sin β × cos α

sin (α - β) = sin α × cos β - sin β × cos α

Calculation:

⇒ sin (α + β) - 2sin α cos β + sin (α - β)

= (sin α × cos β + sin β × cos α) - 2sin α cos β + (sin α × cos β - sin β × cos α)

= 0

Concept:

\(\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\times \tan B}\)

Calculation:

Given: 2 tan A = 3 tan B = 1

⇒ tan A = 1 / 2 and tan B = 1 / 3.

As we know that, \(\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\times \tan B}\)

\(\Rightarrow \tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\times \tan B}=\frac{\frac{1}{2}-\frac{1}{3}}{1+\frac{1}{2}\times \frac{1}{3}}=~\frac{1}{7}\)

Concept:

\(\cos A; + \cos B = 2\cos \left( {\frac{{A + B;}}{2}} \right) \times \cos \left( {\frac{{A - B}}{2}} \right)\)

Calculation:

As we know that, \(\cos A; + \cos B = 2\cos \left( {\frac{{A + B;}}{2}} \right) \times \cos \left( {\frac{{A - B}}{2}} \right)\)

\( \Rightarrow \cos 80^\circ + \cos 40^\circ = 2;\cos 60^\circ \times \cos 20^\circ = \cos 20^\circ \)

\( \Rightarrow ;\cos 80^\circ + \cos 40^\circ - \cos 20^\circ = \cos 20^\circ - \cos 20^\circ = 0\)

Concept:

tan (90° - θ) = cot θ

Calculation:

Consider the Δ ABC, right angled at C as shown in the figure below:

Let A = θ and it is given that C = 90°

∵ A + B + C = 180°

⇒ B = 90° - θ

⇒ tan A + tan B = tan θ + tan (90° - θ) = tan θ + cot θ \(=\frac{1}{\cos \theta \times \sin \theta }\)

From the diagram given above, we get

\(\Rightarrow \cos \theta =\frac{b}{c}~and\sin \theta =\frac{a}{c}\)

\(\Rightarrow \tan A+\tan B=~\frac{1}{\cos \theta \times \sin \theta }\)

\(=\frac{{{c}^{2}}}{ab}\)

Concept:

cos 2A = cos² A - sin² A

sin 2A = 2 sin A × cos A

Calculation:

\(\Rightarrow \cot \left( \frac{A}{2} \right)-\tan \left( \frac{A}{2} \right)=\frac{\cos \left( \frac{A}{2} \right)}{\sin \left( \frac{A}{2} \right)}-\frac{\sin \left( \frac{A}{2} \right)}{\cos \left( \frac{A}{2} \right)}=\frac{{{\cos }^{2}}\left( \frac{A}{2} \right)-{{\sin }^{2}}\left( \frac{A}{2} \right)}{\sin \left( \frac{A}{2} \right)\times \cos \left( \frac{A}{2} \right)}\)

As we know that, cos 2A = cos² A - sin² A

\(\Rightarrow \cot \left( \frac{A}{2} \right)-\tan \left( \frac{A}{2} \right)=\frac{{{\cos }^{2}}\left( \frac{A}{2} \right)-{{\sin }^{2}}\left( \frac{A}{2} \right)}{\sin \left( \frac{A}{2} \right)\times \cos \left( \frac{A}{2} \right)}=\frac{\cos A}{\frac{1}{2}\times 2\times \sin \left( \frac{A}{2} \right)\times \cos \left( \frac{A}{2} \right)~}\)

As we know that, sin 2A = 2 sin A × cos A

\(\Rightarrow \cot \left( \frac{A}{2} \right)-\tan \left( \frac{A}{2} \right)=\frac{\cos A}{\frac{1}{2}\times 2\times \sin \left( \frac{A}{2} \right)\times \cos \left( \frac{A}{2} \right)~}=~2\frac{\cos A}{\sin A}=2\cot A\)

Concept:

cos 2A = cos² A - sin² A = 2 cos² A - 1

sin 2A = 2 sin A × cos A

Calculation:

\(\Rightarrow \cot A+cosec~A=\frac{\cos A}{\sin A}+\frac{1}{\sin A}\)

\(\Rightarrow \cot A+cosec~A=\frac{1+\cos A}{\sin A}\)

As we know that,

cos 2A = cos² A - sin² A = 2 cos² A - 1

\(\Rightarrow \cot A+cosec~A=\frac{1+\cos A}{\sin A}=\frac{2\times {{\cos }^{2}}\left( \frac{A}{2} \right)}{\sin A}\)

As we know that,

sin 2A = 2 sin A × cos A

\(\Rightarrow \cot A+cosec~A=\frac{1+\cos A}{\sin A}\)

\(=\frac{2\times {{\cos }^{2}}\left( \frac{A}{2} \right)}{\sin A}=\frac{2\times {{\cos }^{2}}\left( \frac{A}{2} \right)}{2\times \cos \left( \frac{A}{2} \right)\times \sin \left( \frac{A}{2} \right)}=\cot \left( \frac{A}{2} \right)\)

∴ The value of cot A + cosec A is \(\cot \left( \frac{A}{2} \right)\)

Concept:

\(\tan \left( A+B \right)=\frac{\tan A+\tan B}{1-\tan A\times \tan B}\)

tan (90° - θ) = cot θ

Calculation:

⇒ tan 25° tan 15° + tan 15° tan 50° + tan 25° tan 50° = tan 15° × (tan 25° + tan 50°) + tan 25° tan 50°

As we know that, \(\tan \left( A+B \right)=\frac{\tan A+\tan B}{1-\tan A\times \tan B}\)

⇒ (tan 25° + tan 50°) = tan 75° (1 - tan 25° tan 50°)

⇒ tan 25° tan 15° + tan 15° tan 50° + tan 25° tan 50° = tan 15° × tan 75° (1 - tan 25° tan 50°) + tan 25° tan 50°

As we know that, tan (90° - θ) = cot θ

⇒ tan 75° = tan (90° - 15°) = cot 15°

⇒ tan 25° tan 15° + tan 15° tan 50° + tan 25° tan 50° = tan 15°× cot 15°(1 - tan 25° tan 50°) + tan 25° tan 50°

⇒ tan 25° tan 15° + tan 15° tan 50° + tan 25° tan 50° = 1

Concept:

|x| < a ⇒ - a < x < a

Equation of x-axis is given by: y = 0

If y = a ≠ 0 and a > 0 , then y = a represents the equation of line parallel to x-axis and lying above the x-axis.

If y = a ≠ 0 and a < 0, then y = a represents the equation of line parallel to x-axis and lying below the x-axis.

Calculation:

Given: |x| < 5, y = 0 and y = 8

By plotting the above equations and in equation on the co-ordinate plane as shown in the figure below:

The shaded portion represents the region bounded by the given equations and in equation.

As we can see that the bounded region is a rectangle with length l = 10 units and breadth b = 8 units.

⇒ The area of the bounded region = l × b = 10 × 8 = 80 sq. units

Concept:

For a function say f, \(\underset{x\to a}{\mathop{\lim }},f(x)\) exists

\(\Rightarrow \underset{x\to {{a}^{-}}}{\mathop{\lim }},f\left( x \right)=~\underset{x\to {{a}^{+}}}{\mathop{\lim }},f\left( x \right)=l=~\underset{x\to a}{\mathop{\lim }},f(x)\), where l is a finite value.

Any function say f is said to be continuous at point say a if and only if \(\underset{x\to a}{\mathop{\lim }},f(x)=l=f\left( a \right)\), where l is a finite value.

Calculation:

Given: \(f\left( x \right)=~\left{ \begin{matrix} \sin \left( \frac{1}{x} \right),~x\ne 0 \ 0~,~x=0 \ \end{matrix} \right.\)

Statement 1:  \(\underset{x\to 0}{\mathop{\lim }},f(x)\) exists

As we know that, for a function say f, \(\underset{x\to a}{\mathop{\lim }},f(x)\) exists

\(\Rightarrow \underset{x\to {{a}^{-}}}{\mathop{\lim }},f\left( x \right)=~\underset{x\to {{a}^{+}}}{\mathop{\lim }},f\left( x \right)=l=~\underset{x\to a}{\mathop{\lim }},f(x)\) , where l is a finite value

\(\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }},f\left( x \right)=~\underset{h\to 0}{\mathop{\lim }},f\left( 0-h \right)=~\underset{h\to 0}{\mathop{\lim }},\sin \left( -\frac{1}{h} \right)=~-~\underset{h\to 0}{\mathop{\lim }},\sin \left( \frac{1}{h} \right)\)

∵ \(\underset{h\to 0}{\mathop{\lim }},\sin \left( -\frac{1}{h} \right)\) is an oscillating function between - 1 and 1.

\(\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }},f\left( x \right)~does~not~exist\)

Hence, Statement 1 is false.

Statement 2: f(x) is continuous at x = 0

As we know that, Any function say f is said to be continuous at point say a if and only if \(\underset{x\to a}{\mathop{\lim }},f(x)=l=f\left( a \right)\), where l is a finite value.

But as we have seen in the statement 1 that \(\underset{x\to {{0}^{-}}}{\mathop{\lim }},f\left( x \right)~does~not~exist\) ⇒ f(x) is not continuous at x = 0.

Hence, Statement 2 is false.

Concept:

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

Calculation:

Given: \(\underset{x\to 0}{\mathop{\lim }},\frac{\sin x{}^\circ }{\tan 3x{}^\circ }\)

We know that 180° = π Radian

∴ 1° = π /360 ⇒ x° = πx/180 

\(\\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin {\rm{x}}^\circ }}{{\tan 3{\rm{x}}^\circ }}= \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \frac{{{\rm{\pi x}}}}{{180}}}}{{\tan \frac{{3{\rm{\pi x}}}}{{180}}}}\ \)

= \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\frac{{\sin \frac{{{\rm{\pi x}}}}{{180}}}}{{\frac{{{\rm{\pi x}}}}{{180}}}}; \times \frac{{{\rm{\pi x}}}}{{180}}}}{{\frac{{\tan \frac{{3{\rm{\pi x}}}}{{180}}}}{{\frac{{3{\rm{\pi x}}}}{{180}}}}; \times \frac{{3{\rm{\pi x}}}}{{180}}}} = ;\)(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{\rm{\pi x}}}}{{180}}}}{{\frac{{3{\rm{\pi x}}}}{{180}}}} = \frac{1}{3}\)

∴ Option 2 is correct.

Concept:

The degree of a differential equation is represented by the power of the highest order derivative in the given differential equation.

Calculation:

Given: \(\frac{{{d}^{3}}y}{d{{x}^{3}}}+{{\left( \frac{dy}{dx} \right)}^{2}}-{{x}^{2}}\left( \frac{{{d}^{4}}y}{d{{x}^{4}}} \right)=0\)

For the given differential equation the highest order derivative is 4.

Now, the power of the highest order derivative is 1.

∴ The degree of the given differential equation is 1.

Concept:

General form of a second degree polynomial function is: P (x) = a₀ + a₁ × x + a₂ × x², where a₀, a₁ and a₂ are real coefficients and a₂ ≠ 0.

Calculation

As we know that, general form of a second degree polynomial function is: P (x) = a₀ + a₁ × x + a₂ × x², where a₀, a₁ and a₂ are real coefficients and a₂ ≠ 0.

Let f(x) = a × x² + b × x + c where a, b and c are real coefficients and a ≠ 0.

Given: f(0) = 5, f(-1) = 10 and f(1) = 6.

⇒ f(0) = c = 5

⇒ f(- 1) = a - b + 5 = 10          ----(1)

⇒ f(1) = a + b + 5 = 6          ---(2)

By adding (1) and (2), we get

⇒ 2a + 10 = 16 ⇒ a = 3.

Now by substituting a = 3 in equation (1) we get,

⇒ 3 - b + 5 = 10 ⇒ b = - 2.

Hence, f(x) = a × x² + b × x + c = 3x² - 2x + 5.

Concept:

If y = f(x) then slope of the curve at point a is given by y’ = f’(x = a)

Calculation:

Given: y = me^mx, intersects the y –axis at point P.

As we that, If y = f(x) then slope of the curve at point a is given by y’ = f’(x = a)

\( \Rightarrow \frac{{d\left( y \right)}}{{dx}} = \frac{{d\left( {m \times {e^{m \times x}}} \right)}}{{dx}} = m \times {e^{mx}} \times \frac{{d\left( {m \times x} \right)}}{{dx}} = {m^2} \times ;{e^{mx}}\)

The x – coordinate of point P is 0 as it lies on the y axis.

\( \Rightarrow \frac{{d\left( {y = f\left( 0 \right)} \right)}}{{dx}} = ;{m^2} \times ;{e^0} = {m^2}\)

Hence the slope of given curve at point P is m².

Concept:

If y = f(x), then the slope of the curve at point a is given by y’ = f’(x = a) = tan θ, where θ is the angle that the tangent makes with the x – axis.

The angle that the tangent makes with the y axis is given by: (90° – θ)

tan^-1 x + cot^-1 x = π / 2

\({\sin ^{ - 1}}x = {\cot ^{ - 1}}\left( {\frac{{\sqrt {1 - {x^2}} }}{x}} \right)\)

Calculation:

As in the previous question, we have found the slope of the given curve at point P is m².

⇒ tan θ = m²

⇒ θ = tan^-1 (m²)

As we know that, the angle that the tangent makes with the y axis is given by: (90° – θ)

Let α = (90° – θ)

\(\Rightarrow \alpha = \frac{\pi }{2} - {\tan ^{ - 1}}\left( {{m^2}} \right)\)

As we know that, tan^-1 x + cot^-1 x = π / 2

\( \Rightarrow \alpha = \frac{\pi }{2} - {\tan ^{ - 1}}\left( {{m^2}} \right) = {\cot ^{ - 1}}\left( {{m^2}} \right)\)

As we know that, \({\sin ^{ - 1}}x = {\cot ^{ - 1}}\left( {\frac{{\sqrt {1 - {x^2}} }}{x}} \right)\)

\( \Rightarrow \alpha = {\sin ^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {m^4}} }}} \right)\)

Concept:

If y = f(x), then the slope of the curve at point a is given by y’ = f’(x = a) = tan θ = m.

The equation of tangent with slope m and passing through (x₀, y₀) is given by: y – y₀ = m × (x – x₀)

Calculation:

Given: curve y = me^mx where m > 0 intersects y-axis at a point P.

∵ the x coordinate of P is 0.

So by substituting x = 0 in the given equation of curve, we get:

⇒ y = m.

⇒ P = (0, m)

The slope of the given curve is m².

As we know that, the equation of tangent with slope m and passing through (x₀, y₀) is given by:

y – y₀ = slope × (x – x₀)

The equation of the tangent to the given curve is : y – m = m² (x - 0) ⇒ y = m² x + m

Concept:

If f :A → B and g : C → D. Then (fog) (x) will exist if and only if co-domain of g = domain of f i.e D = A and (gof) (x) will exist if and only if co-domain of f = domain of g i.e B = C.

Calculation:

Given: f(x) = x², g(x) = tan x and h(x) = In x

⇒ ho(gof) = [h(g(x²))] = h(tan (x²)) = ln (tan (x²))

\( \Rightarrow \left[ {ho\left( {gof} \right)} \right];\left( {\frac{{\sqrt {\pi ;} }}{2}} \right) = \ln \left( {\tan \left( {\frac{\pi }{4}} \right)} \right) = 0\)

Concept:

If f :A → B and g : C → D. Then (fog) (x) will exist if and only if co-domain of g = domain of f i.e D = A and (gof) (x) will exist if and only if co-domain of f = domain of g i.e B = C.

Calculation:

Given: f(x) = x², g(x) = tan x and h(x) = In x

⇒ fo(fof) = (fof) (2² =4) = f(4² = 16) = 16² = 256.