Quantitative Aptitude PYQs

Concept:

Step to find inverse of a function:

• Put y = f(x)
• Solve the equation for x.
• x = f^-1 (y)

Properties of logarithm used:

• ln M^p = p × ln M
• ln M = ln N, then M = N

 

Calculation:

Given: y = 5^{In x}

Let’s find the inverse function.

For the given function,

• Domain: x > 0. Since, x is in logarithmic function.
• Range: y > 0. Since, 5 to the power any arbitrary number will result a positive number.
• The inverse function domain will be the range of the original function and range will be the domain of the original function.

Hence, for inverse function, domain y > 0 and range x > 0.

y = 5^{In x}

Taking ln on both sides,

⇒ ln y = ln x × ln5

\(\Rightarrow \;\frac{1}{{\ln 5}} \times \ln y = \ln x\)

⇒ ln y^{(1/ln 5)} = ln x

∴ \(x = {\rm{\;}}{y^{\frac{1}{{In{\rm{\;}}5}}}}\)

Hence Option 1^st is correct.

Additional info:

• The domain of an inverse functions becomes the range of the original function and the range becomes domain of the original function.
• For an inverse function to exist, for every y there is only one x in the original function.

Concept:

• We say f(x) is continuous at x = c if

LHL = RHL = value of f(c) i.e., \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{c}} \right)\)

• \(\sqrt {{{\rm{x}}^2}} = \left| {\rm{x}} \right|\)
• \({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right|{\rm{\;}}\left\{ {\begin{array}{*{20}{c}} { - x,\;x < 0}\\ {x,\;x \ge 0} \end{array}} \right.\)

 

Calculation:

Given: \({\rm{f}}\left( {\rm{x}} \right):\left\{ {\begin{array}{*{20}{c}} { - \frac{{\rm{x}}}{{\sqrt {{{\rm{x}}^2}} }},{\rm{\;x}} \ne 0}\\ {0.{\rm{\;x}} = 0} \end{array}} \right.\) 

\({\rm{f}}\left( {\rm{x}} \right) = - \frac{{\rm{x}}}{{\sqrt {{{\rm{x}}^2}} }}\)

\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = - \frac{{\rm{x}}}{{\left| {\rm{x}} \right|}}\)

\({\rm{LHL}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = - \frac{{\rm{x}}}{{ - {\rm{x}}}} = 1\)                      (∵ |x| = -x, if x < 0)

\({\rm{RHL}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = - \frac{{\rm{x}}}{{\rm{x}}} = - 1\)                    (∵ |x| = x, if x > 0)

Here, LHL ≠ RHL, so f(x) is discontinuous.

Hence, option (3) is correct.

Concept:

Logarithmic formula used:

• ln M^p = p × ln M

Differentiation formulas used:

• \(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)
• \(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f'\left( x \right)g\left( x \right) + f\left( x \right)g'\left( x \right)\)
• Chain rule: \(\frac{d}{{dx}}(f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)\)

 

Calculation:

Given: \(y = {\left( {\cos x} \right)^{{{(\cos x)}^{{{(\cos x)}^.}{{^.}^.}^\infty }}}}\)

To find: \(\frac{{dy}}{{dx}}\)

\(y = {\left( {\cos x} \right)^{{{(\cos x)}^{{{(\cos x)}^.}{{^.}^.}^\infty }}}}\)

⇒ y = (cos x)^y

Taking ln of both sides,

⇒ ln y = ln (cos x)^y

⇒ ln y = y × ln (cos x) (Using ln M^p = p × ln M)

Differentiating on both sides w.r.t x

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

\(\Rightarrow \frac{{dy}}{{dx}}\left( {\frac{1}{y} - \ln \left( {\cos x} \right)} \right) = - y\tan x\)

\(\Rightarrow \frac{{dy}}{{dx}}\left( {\frac{{1 - y\ln \left( {\cos x} \right)}}{y}} \right) = - y\tan x\)

\( \Rightarrow \frac{{dy}}{{dx}} = - \frac{{{y^2}\tan x}}{{1 - y\ln \left( {\cos x} \right)}}\)

Concept:

• A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right) = f\left( a \right)\) exists or its graph is a single unbroken curve.

f (x) is Continuous at x = a ⇔ \(\mathop {\lim }\limits_{x \to {{\rm{a}}^ + }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ - }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right)\)

Calculation:

Checking the continuity of given functions:

1. For f(x) = x + x²

\(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 0\)

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

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

Hence, the function is continuous at x = 0. 2. For \(f\left( x \right) = x + \cos \frac{1}{x}\)

\(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 0 + \cos \left( {\frac{1}{0}} \right)\)

The value cannot be determined as (1/0) is infinite and cos (∞) cannot be determined.

Hence, the limit cannot be determined. And therefore, function is not continuous at x = 0. 3. For \(f\left( x \right) = {x^2} + \cos \frac{1}{x}\)

Here, also at x = 0, cos (∞) is generated and that cannot be determined.

Hence, limit cannot be evaluated and therefore, the function is not continuous at x = 0.

Hence, statement 1 and 2 only are correct.

Concept:

• slope of any given curve at a point on it is the rate of change of one variable with respect to the other,

So, \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) at a point on the curve gives slope of the tangent at that point.

• \({\rm{acceleration\;}}\left( {\rm{a}} \right) = \frac{{{\rm{rate\;of\;change\;of\;velocity}}}}{{{\rm{time\;}}}} = \frac{{{\rm{dv}}}}{{{\rm{dt}}}}\), v = velocity
• \({\rm{velocity\;}}\left( {\rm{v}} \right) = \frac{{{\rm{dispacement}}}}{{{\rm{time}}}} = \frac{{{\rm{ds}}}}{{{\rm{dt}}}}\), s = displacement

 

Calculation:

1. \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) at a point on the curve gives slope of the tangent at that point.
2. \({\rm{a}} = \frac{{{\rm{dv}}}}{{{\rm{dt}}}}\)

 

\(\Rightarrow \smallint {\rm{dv}} = \smallint {\rm{adt}} + {\rm{c}}\)                        (integrating)

1. We know \({\rm{v}} = \frac{{{\rm{ds}}}}{{{\rm{dt}}}},{\rm{\;and\;a}} \ne \frac{{{\rm{ds}}}}{{{\rm{dt}}}}\)

So, statement (3) is not correct.

Hence, option (1) is correct.

Concept:

• \({\sec ^{ - 1}}{\rm{A}} = {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{1}{{\rm{A}}}} \right),\)
• \({\sin ^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2} - {\cos ^{ - 1}}{\rm{x}} \Rightarrow {\sin ^{ - 1}}{\rm{x}} + {\cos ^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2}\)

 

Calculation:

Given: \({\rm{y}} = {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\)                    ….

(1) We know, \({\cos ^{ - 1}}{\rm{x}} = {\sec ^1}\left( {\frac{1}{{\rm{x}}}} \right)\)

\(\therefore {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) = {\cos ^{ - 1}}\left( {\frac{1}{{\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}}}} \right)\)

\(= {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\)

\({\rm{y}} = {\sec ^{ - 1}}\left( {\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\)

\(\Rightarrow {\rm{y}} = {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\)

\(\Rightarrow {\rm{y}} = {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right) + {\sin ^{ - 1}}\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)\)               \(\left( {{{\sin }^{ - 1}}{\rm{x}} + {{\cos }^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2}} \right)\)

\(\Rightarrow {\rm{y}} = \frac{{\rm{\pi }}}{2}\)

\(\therefore \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = 0\)

Hence, option (1) is correct.

Concept:

Trigonometric formulas used:

• sin² A + cos² A = 1
• cos 2A = cos² A – sin² A = 2cos² A - 1
• \(\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\)

Integral formula used:

• \(\smallint {x^n}\;dx = \frac{{{x^n}^{ + 1}}}{{n + 1}}\)

 

Calculation:

I = ∫ tan^-1 (sec x + tan x) dx

\(\Rightarrow I = \smallint {\tan ^{ - 1}}\left( {\frac{{1 + \sin x}}{{\cos x}}} \right)dx\)

\( \Rightarrow I = \smallint {\tan ^{ - 1}}\left( {\frac{{{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos \frac{x}{2}}}{{{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}}} \right)dx\)

\( \Rightarrow I = \smallint {\tan ^{ - 1}}\left( {\frac{{{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}}}{{\left( {{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}} \right)}}} \right)dx\)

\( \Rightarrow I = \smallint {\tan ^{ - 1}}\left( {\frac{{{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}}}{{\left( {{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}} \right)}}} \right)dx\)

\( \Rightarrow I = \smallint {\tan ^{ - 1}}\left( {\frac{{{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}}}{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}}} \right)dx\)

\( \Rightarrow I = \smallint {\tan ^{ - 1}}\left( {\frac{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}}{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}}} \right)dx\)

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

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

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

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

\( \Rightarrow I = \smallint \left( {\frac{\pi }{4} + \frac{x}{2}} \right)dx\)

\(\Rightarrow I = \frac{{\pi x}}{4} + \frac{{{x^2}}}{4} + c\)

Concept:

• A function is differentiable at a point when there's a defined derivative at that point.

Calculation:

Given: \({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {1 - {{\rm{x}}^2}{\rm{\;for\;}}0 < {\rm{x}} \le 1}\\ {{\rm{In\;x\;for\;}}1 < {\rm{x}} \le 2}\\ {{\rm{In\;}}2 - 1 + 0.5{\rm{x\;for\;}}2 < {\rm{x}} < \infty } \end{array}} \right.\)

Referring to the graph,

• The given function is continuous through out the given domain but we will get the sharp corner at x = 1 when we will draw the graph of the function given function so at x = 1, the function is not differentiable.

Here, \({\rm{f}}\left( {\rm{x}} \right) = 1 - {{\rm{x}}^2}{\rm{\;for\;}}0 < {\rm{x}} < 1\)

\(\therefore {\rm{f'}}\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)\)

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

\(= - 2{\rm{x\;for\;}}0 < {\rm{x}} <1\)

• So option 3 is correct.

Concept:

Consider a function y = f(x) on a defined interval of x.

The function attains extreme values (the value can be maximum or minimum or both).

For maxima:

• Local maxima: A point is local maxima of a function if there is some other point where the maximum value is greater than the local maxima but that point doesn’t exist nearby local maxima.
• Global maxima: It is the point where there is no other point has in the domain for which function has more the value than global maxima.

For minima: Same as in the maxima but in place of maximum value minimum values are considered.

Stationary Points: Points where the derivative of the function is zero i.e., f’(x) = 0. The points can be:

• Inflection point
• Local maxima
• Local minima

Second derivative test: Let the function has a stationary point x = a

• If \({\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} < 0\) then x = a, is point of maxima.
• If \({\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} > 0\) then x = a, is point of minima.

 

Calculation:

Given: f(x) = x (x - 1) (x + 1)

Let’s find out the stationary point i.e., f’(x) = 0

f (x) = x (x - 1) (x + 1)

⇒ f (x) = x (x² – 1)

⇒ f (x) = x³ – x

Differentiating both sides w.r.t x, we get

⇒ f’(x) = 3x² – 1

Again differentiating w.r.t x, we get

⇒ f’’(x) = 6x

For maxima and minima f’(x) = 0

⇒ 3x² – 1 = 0

∴ x = ± 1/√3

 

At x = 1/√3 function gives minimum value ⇒ f (x = 1/√3) = -6/√3

At x = - 1/√3 function gives maximum value ⇒ f (x = -1/√3) = +6/√3

Hence, the local maximum value (6/√3) is greater than the local minimum value (-6/√3).

Calculation:

Let’s analyze each statement

1. Let, f(x) = |x|,

\({\rm{f}}\left( {\rm{x}} \right) = {\rm{\;}}\left| {\rm{x}} \right|{\rm{\;}} = \left\{ {\begin{array}{*{20}{c}} { - x,\;x < 0}\\ {x,\;x \ge 0} \end{array}} \right.\)

\({\rm{f'}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} { - 1,\;x < 0}\\ {1,\;x \ge 0} \end{array}} \right.\)

Left hand derivative ≠ Right hand derivative

At x = 0, f’(x) doesn’t exist, so statement 1 is true. 2. Let, f(x) = tan^-1(x)

Now, f’(x) = 1/ (1 + x²) = finite value (At some points)

So Derivative of f(x) may exist finitely at some point.

Hence statement 2 is also true

3. We know that derivative of function is undefined at some points it means derivative of f(x) may be infinite (geometrically) at some point.

So, statement 3 is also true

Hence, option (4) is correct.

Concept:

• \(\ln {\rm{e}} = 1\)
• \({\ln _{\rm{a}}}{\rm{b}} = {\rm{c}} \Rightarrow {\rm{b}} = {{\rm{a}}^{\rm{c}}}\)

Following steps to finding maxima and minima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have find second derivative.

I. f``(x) is less than 0 then the given function is said to be maxima

II. If f``(x) Is greater than 0 then the function is said to be minima

Quotient rule in derivative:

• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {\frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}} \right] = \frac{{{\rm{f'}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{x}} \right){\rm{g'}}\left( {\rm{x}} \right)}}{{{{\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)}^2}}}\)

 

Calculation:

Given: Let f(x) = \(\frac{{{\rm{In\;x}}}}{{\rm{x}}}\) 

Differentiating both sides, we get

\(\begin{array}{l} {\rm{f'}}\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{\rm{In\;x}}}}{{\rm{x}}}} \right)\\ = \frac{{\frac{1}{{\rm{x}}}\left( {\rm{x}} \right) - 1\left( {{\rm{lnx}}} \right)}}{{{{\rm{x}}^2}}}\\ \Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = \frac{{1 - \ln {\rm{x}}}}{{{{\rm{x}}^2}}} \end{array}\)

\(\begin{array}{l} {\rm{f''}}\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f'}}\left( {\rm{x}} \right)} \right)\\ = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{1 - \ln {\rm{x}}}}{{{{\rm{x}}^2}}}} \right)\\ = \frac{{{{\rm{x}}^2}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {1 - \ln {\rm{x}}} \right) - \left( {1 - \ln {\rm{x}}} \right)\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right)}}{{{{\left( {{{\rm{x}}^2}} \right)}^2}}} \end{array}\)

\(\begin{array}{l} = \frac{{{{\rm{x}}^2}\left( { - \frac{1}{{\rm{x}}}} \right) - \left( {1 - \ln {\rm{x}}} \right)\left( {2{\rm{x}}} \right)}}{{{{\rm{x}}^4}}}\\ = \frac{{ - {\rm{x}} - 2{\rm{x}} + 2{\rm{x}}\left( {\ln {\rm{x}}} \right)}}{{{{\rm{x}}^4}}}\\ = \frac{{{\rm{x}}\left( { - 3 + 2\left( {\ln {\rm{x}}} \right)} \right)}}{{{{\rm{x}}^4}}}\\ \Rightarrow {\rm{f''}}\left( {\rm{x}} \right) = \frac{{ - \left( {3 - 2\ln {\rm{x}}} \right)}}{{{{\rm{x}}^3}}} \end{array}\)

 

To find the value of x,

\(\begin{array}{l} {\rm{f'}}\left( {\rm{x}} \right) = 0\\ \Rightarrow \frac{{1 - \ln {\rm{x}}}}{{{{\rm{x}}^2}}} = 0\\ \Rightarrow 1 - \ln {\rm{x}} = 0 \end{array}\)

\(\Rightarrow \ln {\rm{x}} = 1{\rm{\;}}\)

\(\Rightarrow {\rm{x}} = {{\rm{e}}^1} = {\rm{e}}\)                  \(\left( {{{\ln }_{\rm{a}}}{\rm{b}} = {\rm{c}} \Rightarrow {\rm{b}} = {{\rm{a}}^{\rm{c}}}} \right)\)

Now, at x = e,

\(\begin{array}{l} {\rm{f''}}\left( {\rm{e}} \right) = \frac{{ - \left( {3 - 2\ln {\rm{e}}} \right)}}{{{{\rm{e}}^3}}}\\ = \frac{{ - 3 + 2}}{{{{\rm{e}}^3}}}\\ = \frac{{ - 1}}{{{{\rm{e}}^3}}} < 0 \end{array}\)

At x = e, maximum value of f(x) obtain

\(\begin{array}{l} \therefore {\rm{f}}\left( {{\rm{x}} = {\rm{e}}} \right) = {\rm{\;}}\frac{{{\rm{In\;x}}}}{{\rm{x}}}\\ = \frac{{\ln {\rm{e}}}}{{\rm{e}}} \end{array}\)

\(= \frac{1}{{\rm{e}}}\)                         (∵ ln e = 1)

Hence, option (2) is correct

Concept:

• Odd function if f(-x) = -f(x),

Even function if f(-x) = f(x).

• \({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right|{\rm{\;}}\left\{ {\begin{array}{*{20}{c}} { - x,\;x < 0}\\ {x,\;x \ge 0} \end{array}} \right.\)

 

Calculation:

Given: f(x) = |x| - x³

\({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right| - {{\rm{x}}^3} = \left\{ {\begin{array}{*{20}{c}} { - x - {{\rm{x}}^3},\;x < 0}\\ {x - {{\rm{x}}^3},\;x \ge 0} \end{array}} \right.\)           (∵ |x| = -x, if x < 0 and |x| = x, if x ≥ 0)

\({\rm{f}}\left( { - {\rm{x}}} \right) = \left| {\rm{x}} \right| - {{\rm{x}}^3}={\rm{\;}}\left\{ {\begin{array}{*{20}{c}} {x + {{\rm{x}}^3},\;x < 0}\\ {x + {{\rm{x}}^3},\;x \ge 0} \end{array}} \right.\)

Here, f (-x) ≠ - f (x) so f(x) not odd function

Also, f(-x) ≠ f(x), so f(x) is not even function

Hence, option (4) is correct.

Concept:

• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)\) is defined as \({\rm{f'}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left( {\rm{x}} \right)}}{{\rm{h}}}\)

 

Calculation:

Here, \({{\rm{l}}_1} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{\sin {\rm{x}}}}} \right)\)       ….

(1) \({{\rm{l}}_2} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{{\rm{e}}^{\sin \left( {{\rm{x}} + {\rm{h}}} \right)}} - {{\rm{e}}^{\sin {\rm{x}}}}}}{{\rm{h}}}\)

\(= \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{\sin {\rm{x\;}}}}} \right)\)     ---(2)      \(\left( \because {\mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left( {\rm{x}} \right)}}{{\rm{h}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)} \right)\)

\({{\rm{l}}_3} = \smallint {{\rm{e}}^{\sin {\rm{x}}}}\cos {\rm{x\;dx}}\)

Put, sin x = t,

So, cos (x)dx = dt                              (taking derivative)

\(\therefore {{\rm{l}}_3} = \smallint {{\rm{e}}^{\sin {\rm{x}}}}\cos {\rm{x\;dx}}\)

\(\begin{array}{l} = \smallint {{\rm{e}}^{\rm{t}}}{\rm{dt}}\\ = {{\rm{e}}^{\rm{t}}} + {\rm{c}} \end{array}\)

\(= {{\rm{e}}^{{\rm{sinx}}}} + {\rm{c}}\)                         (∵ t = sin x)

Now take derivative, we get

\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{l}}_3}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{sinx}}}}} \right)\)                          (3)

\(\therefore {\rm{from\;}}\left( 1 \right),{\rm{\;}}\left( 2 \right),{\rm{\;}}\left( 3 \right){\rm{\;}}{{\rm{l}}_1} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{l}}_3}} \right) = {{\rm{l}}_2}\)

Hence, option (2) is correct.

Concept:

• Equation of conic section: ax² + 2hxy + by² + 2gx + 2fy + c = 0
• For Circle: Coefficient of x² = Coefficient of y² and Coefficient of xy = 0 (a = b & h = 0)

 

Calculation:

Given: \(\frac{{dy}}{{dx}} = \frac{{ax + h}}{{by + k}}\) 

⇒ (ax + h) dx = (by + k) dy

∫ (ax + h) dx = ∫ (by + k) dy     ...(integrating both sides)

⇒ ax²/2 + hx = by²/2 + ky

⇒ ax² + 2hx = by² + 2ky

⇒ ax² - by² + 2hx - 2ky = 0

For Circle: Coefficient of x² = Coefficient of y²

∴ a = -b ≠ 0

Hence, option 2 is correct.

Concept:

Steps to solve the limit:

• Check the form by substituting the limit value in the function
• After substitution if indeterminate form (0/0, ∞/∞ etc.) obtain then solve it further

 

Calculation:

Given: \(\mathop {\lim }\limits_{{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \frac{{\sin {\rm{x}}}}{{\rm{x}}} = {\rm{l}}\) and \(\mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{\cos {\rm{x}}}}{{\rm{x}}} = {\rm{m}}\)

\({\rm{l}} = \mathop {\lim }\limits_{{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \frac{{\sin {\rm{x}}}}{{\rm{x}}}\)

\(= \frac{{\sin \left( {\frac{{\rm{\pi }}}{2}} \right)}}{{\frac{{\rm{\pi }}}{2}}}\)          (put x = π/2)

\(= \frac{1}{{\frac{{\rm{\pi }}}{2}}}\)                  (∵ sin(π/2) = 1)

\(\Rightarrow {\rm{l}} = \frac{2}{{\rm{\pi }}}\)

And \({\rm{m}} = \mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{\cos {\rm{x}}}}{{\rm{x}}}\)

\(= \frac{{\cos \left( \infty \right)}}{\infty }\)   (cos x lies between -1 to 1)

⇒ m = 0          (1/∞ = 0)

Hence, option (3) is correct.

Concept:

• \({\sin ^2}{\rm{x}} + {\cos ^2}{\rm{x}} = 1\)
• \(\sin 2{\rm{x}} = 2\sin {\rm{x}}\cos {\rm{x}}\)

 

Calculation:

Let, \({\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {1 + \sin \frac{{\rm{x}}}{2}} {\rm{dx}}\)

\(= \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {{{\sin }^2}\left( {\frac{{\rm{x}}}{4}} \right) + {{\cos }^2}\left( {\frac{{\rm{x}}}{4}} \right) + \sin \frac{{\rm{x}}}{2}} {\rm{dx}}\)                  (1 = \({\sin ^2}\left( {\frac{{\rm{x}}}{4}} \right) + {\cos ^2}\left( {\frac{{\rm{x}}}{4}} \right)\))

\(= \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {{{\sin }^2}\left( {\frac{{\rm{x}}}{4}} \right) + {{\cos }^2}\left( {\frac{{\rm{x}}}{4}} \right) + 2\sin \left( {\frac{{\rm{x}}}{4}} \right)\cos \left( {\frac{{\rm{x}}}{4}} \right)} {\rm{dx}}\)             \(\left( {\sin \frac{{\rm{x}}}{2} = 2\sin \left( {\frac{{\rm{x}}}{4}} \right)\cos \left( {\frac{{\rm{x}}}{4}} \right)} \right)\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {{{\left( {\sin \left( {\frac{{\rm{x}}}{4}} \right) + \cos \left( {\frac{{\rm{x}}}{4}} \right)} \right)}^2}} {\rm{dx}}\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \left| {\sin \left( {\frac{{\rm{x}}}{4}} \right) + \cos \left( {\frac{{\rm{x}}}{4}} \right)} \right|{\rm{dx}}\)

Now, put \(\frac{{\rm{x}}}{4} = {\rm{t}}\)

\(\Rightarrow \frac{{{\rm{dx}}}}{4} = {\rm{dt}}\)                           (taking derivative)

\(\Rightarrow {\rm{dx}} = 4{\rm{dt}}\)

When, x = 2π, t = x/4 = 2π/4 = π/2

When, x = 0, t = 0

\(\therefore {\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \left| {\sin \left( {\frac{{\rm{x}}}{4}} \right) + \cos \left( {\frac{{\rm{x}}}{4}} \right)} \right|{\rm{dx}}\)

\(\Rightarrow {\rm{I}} = 4\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \left| {\sin \left( {\rm{t}} \right) + \cos \left( {\rm{t}} \right)} \right|{\rm{dt}}\)

\(\Rightarrow {\rm{I}} = 4\left[ { - \cos {\rm{t}} + \sin {\rm{t}}} \right]_0^{\frac{{\rm{\pi }}}{2}}\)

\(= 4\left[ {\left( { - \cos \left( {\frac{{\rm{\pi }}}{2}} \right) + \sin \left( {\frac{{\rm{\pi }}}{2}} \right)} \right) - \left( { - \cos 0 + \sin 0} \right)} \right]\)

\(\Rightarrow {\rm{I}} = 4\left( {1 - \left( { - 1} \right)} \right)\)

\(= 4\left( 2 \right)\)

= 8

Hence, option (1) is correct.

Concept:

• \(\left| {\rm{x}} \right|{\rm{\;}} = \left\{ {\begin{array}{*{20}{c}} { - x,\;x < 0}\\ {x,\;x \ge 0} \end{array}} \right.\)
• Area of triangle = ½ × base × height

Calculation:

Given: |x| + |y| = 1

 

Now, we will get x + y = 1, x - y = 1, -x + y = 0, -x - y = 1

Area of ABCD = 4 (Area of OAD)

= 4 (½ × base × height)

= 4(½ × 1 × 1)

= 2 square unit

Hence, option 3 is correct.

Concept:

Let, y = f(x), then interval of y is the minimum and maximum value range of y.

 

Calculation:

Let, \({\rm{y}} = \frac{{{{\rm{x}}^2}}}{{1 + {{\rm{x}}^4}}}\),

For x = 0, y = 0, for x = -1, y = + ½ etc.

So here y ≥ 0

Now, \({\rm{y}} = \frac{{{{\rm{x}}^2}}}{{1 + {{\rm{x}}^4}}} = \frac{1}{{\frac{{1 + {{\rm{x}}^4}}}{{{{\rm{x}}^2}}}}} = \frac{1}{{\frac{1}{{{{\rm{x}}^2}}} + {{\rm{x}}^2}}}\)

For x = 1, y = ½

For x = 1/2, y = 4/17 < ½

For x = 2, y = 4/17 etc.

So here, y ≤ ½

∴ y = \((\frac{{{{\rm{x}}^2}}}{{1 + {{\rm{x}}^4}}})\) ϵ [0, ½]

Hence, option (2) is correct

Concept:

• \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{c}} \right)}}{{{\rm{x}} - {\rm{c}}}}\) OR \(\mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {{\rm{c}} - {\rm{h}}} \right) - {\rm{f}}\left( {\rm{c}} \right)}}{{ - {\rm{h}}}}\) is left hand derivative at x = c
• \(\sin \left( {{\rm{n\pi }}} \right) = 0\)
• The value of ⌊x⌋ is the largest integer that is less than or equal to x.

Example: [4.2] = 4, [-3.2] = -4

• Suppose [3 - 0.5] = [2.5] = (3 - 1) = 2,
• [k - h] = (k - 1), k = integer
• sin (nπ - θ) = (-1)^(n-1) sin(θ)
• \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin {\rm{x}}}}{{\rm{x}}} = 1\)

 

Calculation:

Given: f(x) = [x] sin (πx) at x = k

LHD

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\left[ {{\rm{k}} - {\rm{h}}} \right]\sin \left( {{\rm{\pi }}\left( {{\rm{k}} - {\rm{h}}} \right)} \right) - \left[ {\rm{k}} \right]\sin \left( {{\rm{k\pi }}} \right)}}{{ - {\rm{h}}}}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\left[ {{\rm{k}} - {\rm{h}}} \right]\sin \left( {{\rm{\pi }}\left( {{\rm{k}} - {\rm{h}}} \right)} \right)}}{{ - {\rm{h}}}}\)                  (∵ sin (kπ) = 0)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\left( {{\rm{k}} - 1} \right)\sin \left( {{\rm{\pi k}} - {\rm{\pi h}}} \right)}}{{ - {\rm{h}}}}\)                  (∵ [k - h] = (k - 1))

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\left( {{\rm{k}} - 1} \right){{\left( { - 1} \right)}^{{\rm{k}} - 1}}\sin \left( {{\rm{\pi h}}} \right)}}{{ - {\rm{h}}}}\)                               (∵ sin (nπ - θ) = (- 1)^(n-1) sin(θ))

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\left( { - 1} \right)\left( {{\rm{k}} - 1} \right){{\left( { - 1} \right)}^{{\rm{k}} - 1}}\sin \left( {{\rm{\pi h}}} \right)}}{{\rm{h}}}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\left( {{\rm{k}} - 1} \right){{\left( { - 1} \right)}^{\rm{k}}}\sin \left( {{\rm{\pi h}}} \right)}}{{\rm{h}}}\)                  ((-1)(-1)^(k-1) = (-1)^(k-1+1) = (-1)^k)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\left( {{\rm{k}} - 1} \right){{\left( { - 1} \right)}^{\rm{k}}}\sin \left( {{\rm{\pi h}}} \right)}}{{{\rm{\pi h}}}} \times {\rm{\pi }}\)                          (multiply and divide by π)

\(= {\rm{\pi }}\left( {\left( {{\rm{k}} - 1} \right){{\left( { - 1} \right)}^{\rm{k}}}} \right)\mathop {\lim }\limits_{{\rm{h}} \to 0} \left( {\frac{{\sin \left( {{\rm{\pi h}}} \right)}}{{{\rm{\pi h}}}}} \right)\)

\(= {\rm{\pi }}\left( {\left( {{\rm{k}} - 1} \right){{\left( { - 1} \right)}^{\rm{k}}}} \right)\)                  \(\left( {\mathop {\lim }\limits_{{\rm{h}} \to 0} \left( {\frac{{\sin \left( {{\rm{\pi h}}} \right)}}{{{\rm{\pi h}}}}} \right) = 1} \right)\)

\(= {\left( { - 1} \right)^{\rm{k}}}\left( {{\rm{k}} - 1} \right){\rm{\pi }}\)

Hence, option (1) is correct

Concept:

1. Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is nearest and smaller integer to x. It is also known as floor of x

• In general, If \(n \le x\; \le n + 1\) Then [x] = n (n ∈ Integer)
• Means if x lies in [n, n + 1) then the Greatest Integer Function of x will be n.

 

 

2. A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right) = f\left( a \right)\) exists or its graph is a single unbroken curve.

• f (x) is Continuous at x = a ⇔ \(\mathop {\lim }\limits_{x \to {{\rm{a}}^ + }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ - }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right)\)

 

Calculation:

Given: f (x) = x/2 - 1 for [0, π]

For \(0\; \le x\; \le \;\pi\) 

\(\Rightarrow \frac{0}{2} \le \frac{x}{2} \le \frac{\pi }{2}\)

\(\Rightarrow 0 - 1 \le \;\frac{x}{2} - 1 \le \;\frac{\pi }{2} - 1\)

\(\Rightarrow - 1\; \le f\left( x \right) \le \;\frac{\pi }{2} - 1\)

Now, GIF of f (x)

\(\left[ {f\left( x \right)} \right] = \left\{ {\begin{array}{*{20}{c}} { - 1,\;0 \le x < 2}\\ {0,\;2\; \le \;x < \pi } \end{array}} \right.\)

\(\Rightarrow \tan \left[ {f\left( x \right)} \right] = \left\{ {\begin{array}{*{20}{c}} {\tan \left( { - 1} \right),\;0 \le x < 2}\\ {\tan 0,\;2 \le \;x < \pi } \end{array}} \right.\)

Checking continuity of tan [f(x)] in [0, π]

RHL = \(\mathop {\lim }\limits_{x \to {2^ + }} \tan \left[ {f\left( x \right)} \right] = \mathop {{\rm{lim}}}\limits_{x \to {2^ + }} \tan \left( { - 1} \right)\) = - tan 1

LHL = \(\mathop {\lim }\limits_{x \to {2^ - }} \tan \left[ {f\left( x \right)} \right] = \mathop {{\rm{lim}}}\limits_{x \to {2^ + }} \tan 0\) = 0

RHL ≠ LHL so, tan [f(x)] is not continuous at x = 2

Now, f (x) = x/2 -1 = (x – 2) /2

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

Clearly 1/f(x) is not continuous at x = 2.

So, tan [f(x)] and 1/f(x) both are discontinuous at x = 2