Quantitative Aptitude PYQs

Concept:

\(\smallint \frac{1}{{1 + {x^2}}};dx = {\tan ^{ - 1}}x + C\)

Calculation:

By using completing the square method, we can write 2x² – 2x + 1 as (4x² – 4x + 1) + 1

\( \Rightarrow \smallint \frac{{dx}}{{2{x^2} - 2x + 1}} = ;2\smallint \frac{{dx}}{{\left( {4{x^2} - 4x + 1} \right) + 1}} = ;\smallint \frac{{2;dx}}{{{{\left( {2x - 1} \right)}^2} + 1}}\)

Let, 2x – 1 = t.

Now by differentiating both the sides with respect to x we get:

\(\frac{{d\left( {2x - 1} \right)}}{{dx}} = \frac{{dt}}{{dx}} \Rightarrow ;2dx = dt\)      ----(1)

By using equation (1), we get

\( \Rightarrow \smallint \frac{{dx}}{{2{x^2} - 2x + 1}} = ;\smallint \frac{{;dt}}{{1 + {t^2}}}\)

As we know that, \(\smallint \frac{1}{{1 + {x^2}}};dx = {\tan ^{ - 1}}x + C\)

\( \Rightarrow \smallint \frac{{dx}}{{2{x^2} - 2x + 1}} = ;\smallint \frac{{;dt}}{{1 + {t^2}}} = {\tan ^{ - 1}}\left( t \right) + C\)

By substituting, 2x – 1 = t in the above equation we get,

\( \therefore \smallint \frac{{dx}}{{2{x^2} - 2x + 1}} = {\tan ^{ - 1}}\left( {2x - 1} \right) + C\)

Concept:

\(\smallint {x^a};dx = \frac{{{x^{a; + ;1}}}}{{a + 1}} + C\)

Calculation:

\(\smallint \frac{{dx}}{{x{{\left( {1 + In;x} \right)}^n}}}\)

Let us consider 1 + ln (x) = t.

Now by differentiating both the sides with respect to x, we get

\( \Rightarrow \frac{{d\left( {1 + \ln x} \right)}}{{dx}} = \frac{{dt}}{{dx}} \Rightarrow \frac{{dx}}{x} = dt\)    -----(1)

Now by using the equation (1), we get

\( \Rightarrow \smallint \frac{{dx}}{{x{{\left( {1 + In;x} \right)}^n}}}; = ;\smallint \frac{{dt}}{{{t^n}}} = ;\smallint {t^{ - n}};dt\)

As we know that, \(\smallint {x^a};dx = \frac{{{x^{a; + ;1}}}}{{a + 1}} + C\)

\( \Rightarrow \smallint \frac{{dx}}{{x{{\left( {1 + In;x} \right)}^n}}}; = \smallint {t^{ - n}};dt = \frac{{{t^{1 - n}}}}{{1 - n}} + C\)

Now by substituting 1 + ln (x) = t in the above equation we get,

\( \Rightarrow \smallint \frac{{dx}}{{x{{\left( {1 + In;x} \right)}^n}}}; = \frac{{{t^{1 - n}}}}{{1 - n}} + C = ; - \frac{1}{{\left( {n - 1} \right) \times {{\left( {1 + \ln x} \right)}^{n - 1}}}} + C\)

Concept:

\(\frac{{d\left( {u;v} \right)}}{{dx}} = v \times \frac{{du}}{{dx}} + u \times \frac{{dv}}{{dx}}\)

Calculation:

Given: \(y = \frac{1}{{2{x^2} - c}}\)

\( \Rightarrow y = \frac{1}{{2{x^2} - c}} \Rightarrow y \times \left( {2{x^2} - c} \right) = 1\)

\( \Rightarrow 2{x^2}y - yc = 1\)      ------(1)

Now by differentiating equation (1) with respect to x, we get

\( \Rightarrow 4xy + 2{x^2}\frac{{dy}}{{dx}} - c\frac{{dy}}{{dx}} = 0\)

\( \Rightarrow \frac{{dy}}{{dx}} \times \left( {2{x^2} - c} \right) + 4xy = 0\)

∵ \(y = \frac{1}{{2{x^2} - c}}\)

\(\frac{{dy}}{{dx}} \times \frac{1}{y} + 4xy = 0 \Rightarrow \frac{{dy}}{{dx}} = ; - 4x{y^2}\)

Concept:

\({\log _a}{x^b} = b \times ;{\log _a}x;and;{\log _a}a = 1\)

\(\frac{{d\left( {u;v} \right)}}{{dx}} = v \times \frac{{du}}{{dx}} + u \times \frac{{dv}}{{dx}}\)

\(\frac{{d;\left( {\ln x} \right)}}{{dx}} = \frac{1}{x}\)

Calculation:

Given: x^y = e^x-y

By applying ln both the sides we get

⇒ y × ln x = ln (e^x-y) = x – y

⇒ y × ln x = x – y

⇒ x = y × (1 + ln x)      ------(1)

Now by differentiating both the sides with respect to x we get,

\( \Rightarrow 1 = \left( {1 + \ln x} \right) \times \frac{{dy}}{{dx}} + \frac{y}{x}\)

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{1 - \frac{y}{x}}}{{1 + \ln x}}\)

Now by using equation (1), we get

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{\ln x}}{{{{\left( {1 + \ln x} \right)}^2}}}\)

So, \(\frac{{dy}}{{dx}};at;x = 1;is;0\)

Concept:

\(\frac{d}{{dx}}\left( {\frac{u}{v}} \right) = \frac{{v \times \frac{{du}}{{dx}} - u \times \frac{{dv}}{{dx}}}}{{{v^2}}}\)

Calculation:

From previous question we got, \(\frac{{dy}}{{dx}} = \frac{{\ln x}}{{{{\left( {1 + \ln x} \right)}^2}}}\)

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{\ln x}}{{{{\left( {1 + \ln x} \right)}^2}}}} \right) = \frac{{\left( {1 + \ln x} \right) \times \frac{1}{x} - 2 \times \ln x \times \left( {1 + \ln x} \right) \times \frac{1}{x}}}{{{{\left( {1 + \ln x} \right)}^4}}}\)

So, \(\frac{{{d^2}y}}{{d{x^2}}};at;x = 1;is;1\)

Concept:

Let f(x) be a function and p ∈ R. If f(x + p) = f(x) ∀ x belonging to the domain of f. Then p is called as the period of the function f.

Calculation:

Given: \(;g\left( x \right) = \sin \left( {\frac{x}{4}} \right)\)

As we know that, period of sin x is 2π

So, period of \(g\left( x \right) = \sin \left( {\frac{x}{4}} \right);; \)

⇒ x/4 = 2π 

∴ x = 8π

Concept:

Let f(x) be a function and p ∈ R. If f(x + p) = f(x) ∀ x belonging to the domain of f. Then p is called as the period of the function f.

Calculation:

Given: \(h\left( x \right) = \cos \left( {\frac{{4x}}{5}} \right)\)

As we know that, period of cos x is 2π.

So, period of \(h\left( x \right) = \cos \left( {\frac{{4x}}{5}} \right);is\frac{5}{4} \times 2\pi = \frac{{5\pi }}{2}\)

Concept:

Let f(x) be a function and p ∈ R. If f(x + p) = f(x) ∀ x belonging to the domain of f. Then p is called as the period of the function f.

If f and g are two functions with period p and q respectively, then the period of the function f + g is lcm (p, q).

Calculation:

Given: \(g\left( x \right) = \sin \left( {\frac{x}{4}} \right)\) and \(h\left( x \right) = \cos \left( {\frac{{4x}}{5}} \right)\) and f(x) = g(x) + h(x)

We know that, period of g(x) is 8π and the period of h(x) is 5π / 2.

As we know that, If f and g are two functions with period p and q respectively, then the period of the function f + g is lcm (p, q).

So, period of f(x) is lcm (8π, 5π / 2) = LCM of (8π, 5π)/ HCF of (1, 2) = 40π

Calculation:

Given: F(x) = 3x⁴ – 20x³ – 12x² + 288x + 1

⇒ F’(x) = 12x³ – 60x² – 24x + 288

⇒ F’(x) = 12 × (x³ – 5x² – 2x + 24)

⇒ F’(x) = 12 × (x + 2) × (x - 3) × (x + 4)

To be increasing F’(x) > 0 i.e 12 × (x + 2) × (x - 3) × (x - 4) > 0

So, the function is increasing in the interval (-2, 3) ∪ (4, ∞)

Calculation:

Given: F(x) = 3x⁴ – 20x³ – 12x² + 288x + 1

⇒ F’(x) = 12x³ – 60x² – 24x + 288

⇒ F’(x) = 12 × (x³ – 5x² – 2x + 24)

⇒ F’(x) = 12 × (x + 2) × (x - 3) × (x + 4)

To be decreasing F’(x) < 0 i.e 12 × (x + 2) × (x - 3) × (x - 4) < 0

So, the function is decreasing in the interval: (-∞, - 2) ∪ (3, 4)

Concept:

If α and β are the roots of the equation f(x) = ax² + bx + c = 0. Then f(α) = 0 = f(β).

Calculation:

Given: f(x) = x² + 2x – 5 and g(x) = 5x + 30

⇒ g[f(x)] = g(x² + 2x – 5) = 5 (x² + 2x – 5) + 30 = 5x² + 10x + 5 = 0.

⇒ 5x² + 10x + 5 = 0

⇒ x² + 2x + 1 = (x + 1)² = 0

⇒ x = - 1, -1

Concept:

If p(x) = a₀ + a₁ x + …… + a_nxⁿ , where coefficients of x are real and if a_n ≠ 0 .Then p(x) is a polynomial of degree n.

Calculation:

Given: f(x) = x² + 2x – 5 and g(x) = 5x + 30

⇒ f[g(x)] = f(5x + 30) = (5x + 30)² + 2(5x + 30) – 5 = 25x² + 310x + 955

⇒ f[g(x)] is a polynomial of degree 2.

Hence statement 1 is false.

⇒ g[g(x)] = g(5x + 30) = 5(5x + 30) + 30 = 25x + 180

⇒ g[g(x)] is a polynomial of degree 1.

Hence statement 2 is wrong.

Calculation:

Given: h(x) = 5f(x) – xg(x), where f(x) = x² + 2x – 5 and g(x) = 5x + 30

⇒ h(x) = 5 × (x² + 2x – 5) – x × (5x + 30)

⇒ 5x2 + 10x – 25 – 5x2 - 30x

⇒ h(x) = -20 x – 25

⇒ h’(x) = - 20.

Concept:

\(\mathop \smallint \nolimits_a^b f\left( x \right);dx = ;\mathop \smallint \nolimits_a^b f\left( {a + b - x} \right)dx\)

\(\smallint {\sec ^2}x;dx = \tan x + C;and;\smallint (\sec x \times \tan x);dx = \sec x + C\)

Calculation:

Given: \({I_1} = \mathop \smallint \limits_0^π \frac{{xdx}}{{1 + \sin x}};;and;;{I_2} = \mathop \smallint \limits_0^π \frac{{\left( {π - x} \right)dx}}{{1 - \sin \left( {π + x} \right)} }\)

As we know that, \(\mathop \smallint \nolimits_a^b f\left( x \right);dx = ;\mathop \smallint \nolimits_a^b f\left( {a + b - x} \right) dx\)

\( ⇒ {I_1} = ;\mathop \smallint \nolimits_0^π \frac{{x;dx}}{{1 + \sin x}} \)      ----(1)

\( ⇒ {I_1} = ;\mathop \smallint \nolimits_0^π \frac{{(π - x)}dx}{{1 + \sin \left( x \right)}} ;\)      -----(2)

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

\( ⇒ 2{I_1} = ;π ;\mathop \smallint \nolimits_0^π \frac{{(1 - \sin x) dx}}{{{{\cos }^2 }x}}\)

\( ⇒ 2{I_1} = π ;\mathop \smallint \nolimits_0^π ({\sec ^2}x - \sec x \times \tan x)\ dx\)

\( ⇒ 2{I_1} = π ;\mathop \smallint \nolimits_0^π {\sec ^2}x;dx - π ;\mathop \smallint \nolimits_0^π (\sec x \times \tan x);dx\)

⇒ 2I₁ = π [tanx]₀^π  - π [secx]0π

\( ⇒ 2{I_1} = π ;\left( {0 - 0} \right) - π \left( { - ;1 - 1} \right) \)

⇒ 2I1 = 2π 

∴ I1 = π.

Concept:

\(\mathop \smallint \nolimits_a^b f\left( x \right);dx = ;\mathop \smallint \nolimits_a^b f\left( {a + b - x} \right)dx\)

Calculation:

Given: \({I_1} = \mathop \smallint \limits_0^π \frac{{xdx}}{{1 + \sin x}};;and;;{I_2} = \mathop \smallint \limits_0^π \frac{{\left( {π - x} \right)dx}}{{1 - \sin \left( {π + x} \right)} }\)

As we know that, \(\mathop \smallint \nolimits_a^b f\left( x \right);dx = ;\mathop \smallint \nolimits_a^b f\left( {a + b - x} \right)dx\)

\( \Rightarrow {I_1} = ;\mathop \smallint \nolimits_0^π \frac{{x;dx}}{{1 + \sin x}} = ;\mathop \smallint \nolimits_0^π \frac{{π - x}}{{1 + \sin \left( x \right)}} = {I_2}\)

\( ⇒ {I_1} = ;\mathop \smallint \nolimits_0^π \frac{{x;dx}}{{1 + \sin x}} \)    ----(1)

\( ⇒ {I_1} = ;\mathop \smallint \nolimits_0^π \frac{{(π - x)}dx}{{1 + \sin \left( x \right)}} ;\)       -----(2)

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

\( ⇒ 2{I_1} = ;\mathop \smallint \nolimits_0^π \frac{{π;dx}}{{1 + \sin x}} = π \mathop \smallint \nolimits_0^π \frac{{dx}}{{1 + \sin x}}\)

 

\( ⇒ 2{I_1} = π ;\mathop \smallint \nolimits_0^π ({\sec ^2}x - \sec x \times \tan x)\ dx\)

\( ⇒ 2{I_1} = π ;\mathop \smallint \nolimits_0^π {\sec ^2}x;dx - π ;\mathop \smallint \nolimits_0^π (\sec x \times \tan x);dx\)

⇒ 2I1 = π [tanx]0π  - π [secx]0π

\( ⇒ 2{I_1} = π ;\left( {0 - 0} \right) - π \left( { - ;1 - 1} \right) \)

⇒ 2I1 = 2π 

⇒ I1 = π.

And we know that I₁ = I₂ = π

∴, I1 + I2 = 2π.

Concept:

\(\frac{{d\left( {tanx} \right)}}{{dx}} = {\sec ^2}x\)

Calculation:

Given: tan y = c(1 - e^x)

Differentiating both the sides with respect to x, we get

\( \Rightarrow {\sec ^2}y \times \frac{{dy}}{{dx}} = - c{e^x}\)     ------(1)

∵ tan y = c(1 - e^x)

\( \Rightarrow c = \frac{{\tan y}}{{\left( {1 - {e^x}} \right)}}\)

By substituting the value of c in equation (1), we get

\( \Rightarrow {\sec ^2}y \times \frac{{dy}}{{dx}} = ; - \frac{{\tan y \times {e^x}}}{{\left( {1 - {e^x}} \right)}}\)

\(\Rightarrow \left( {1 - {e^x}} \right) \times {\sec ^2}y;dy + \tan y \times {e^x};dx = 0\)

Concept:

\(\frac{d}{{dx}};\left( {{a^x}} \right) = {a^x} \times \ln a\)

Calculation:

Given: \({2^{{{\left( {\sin x} \right)}^2}}}\)

As we know that, \(\frac{d}{{dx}};\left( {{a^x}} \right) = {a^x} \times \ln a\)

\( \Rightarrow \frac{d}{{d\left( {\sin x} \right)}};\left( {{2^{{{\left( {\sin x} \right)}^2}}}} \right) = ;{2^{{{\left( {\sin x} \right)}^2}}} \times \ln 2 \times \frac{{d\left( {{{\left( {\sin x} \right)}^2}} \right)}}{{d\left( {\sin x} \right)}} \)

\( \Rightarrow \frac{d}{{d\left( {\sin x} \right)}};\left( {{2^{{{\left( {\sin x} \right)}^2}}}} \right) = ;\sin x;{2^{{{\left( {\sin x} \right)}^2};}}\ ln;4\)

Concept:

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

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

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

Calculation:

Given: \(f\left( x \right) = ;\left{ {\begin{array}{*{20}{c}} {2x + \frac{1}{4},;;;;;x < 0}\ {k,;;;;;;;x = 0}\ {{{\left( {x + \frac{1}{2}} \right)}^2},;;;;;;;x > 0} \end{array}} \right.\)

Left hand limit:

\( \Rightarrow ;\mathop {\lim }\limits_{x \to 0 - } f\left( x \right); = ;\mathop {\lim }\limits_{h \to 0} \left( {2\left( { - ;h} \right) + \frac{1}{4}} \right) = \frac{1}{4}\)      ------(1)

Right hand limit:

\( \Rightarrow ;\mathop {\lim }\limits_{x \to 0 + } f\left( x \right) = ;\mathop {\lim }\limits_{h \to 0} \left( {{{\left( {h + \frac{1}{2}} \right)}^2}} \right) = ;\mathop {\lim }\limits_{h \to 0} \left( {{h^2} + h + \frac{1}{4}} \right) = \frac{1}{4}\)   ------(2)

As we know that, if for any function f if \(\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = ;\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = l \Rightarrow ;\mathop {\lim }\limits_{x \to a} f(x)\)

From equation (1) and (2), we get

\( \Rightarrow ;\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \frac{1}{4};;\)

As we know that any function f is said to be continuous at x = a if and only if \(\mathop {\lim }\limits_{x \to a} f(x) = f\left( a \right)\)

It is given that f(0) = k and f is continuous at x = 0.

\(\mathop { \Rightarrow \lim }\limits_{x \to 0} f(x) = f\left( 0 \right) \Rightarrow k = \frac{1}{4}\)

Calculation:

Given: y² = 2x and y = x

⇒ x² – 2x = 0    ------(∵ y = x)

⇒ x = 0 or 2

When x = 0 then y = 0 and when x = 2 then y = 2    ------(∵ y = x)

∴ The curve y² = 2x intersects the line y = x at: (0, 0) and (2, 2)

Let area bounded between the curve y² = 2x and the straight line y = x is given by A.

\( \Rightarrow A = ;\mathop \smallint \nolimits_0^2 \left( {\sqrt {2x} - x} \right);dx = \sqrt 2 \times ;\mathop \smallint \nolimits_0^2 {x^{\frac{1}{2}}};dx - ;\mathop \smallint \nolimits_0^2 x;dx\)

\( \Rightarrow A = \frac{{2\sqrt 2 }}{3} \times \left[ {{2^{\frac{3}{2}}} - 0} \right] - \frac{1}{2};\left[ {4 - 0} \right] = \frac{8}{3} - 2 = \frac{2}{3};sq.;units\)

Calculation:

Given: \(f\left( x \right) = \frac{{{x^3}}}{3} - \frac{{5{x^2}}}{2} + 6x + 7\) and f increases in the interval T and decreases in the interval S.

\( \Rightarrow f'\left( x \right) = {x^2} - 5x + 6 = \left( {x - 2} \right);\left( {x - 3} \right)\)

To be increasing, f’(x) > 0 ⇒ (x - 2) × (x - 3) > 0.

⇒ f’(x) increases when x < 2 or x > 3

⇒ T = (-∞, 2) ∪ (3, ∞)

Similarly, to be decreasing, f’(x) < 0 ⇒ (x - 2) × (x - 3) < 0.

⇒ f’(x) decreases when 2 < x < 3.

⇒ S = (2, 3)