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Let $\left [ t \right ]$ denote the greatest integer $\leq t$ and $\lim_{x\rightarrow 0}x\left [ \frac{4}{x} \right ]=A.$ Then the function, $f\left ( x \right )=\left [ x^{2} \right ]\sin \left ( \pi x \right )$ is discontinuous, when x is equal to:
Option: 1 $\sqrt{A+1}$

Option: 2 $\sqrt{A}$

Option: 3 $\sqrt{A+5}$

Option: 4 $\sqrt{A+21}$

$\\\lim _{x \rightarrow 0} x\left(\frac{4}{x}-\left\{\frac{4}{x}\right\}\right)=A \quad \Rightarrow \lim _{x \rightarrow 0} 4-x\left\{\frac{4}{x}\right\}=A\\\Rightarrow 4-0=A$

$\\\text { (1) } x=\sqrt{A+1} \Rightarrow x=\sqrt{5} \Rightarrow \text { discontinuous }\\\text { (2) } x=\sqrt{A+21} \Rightarrow x=5 \Rightarrow \text { continuous }\\\text { (3) } x=\sqrt{A} \Rightarrow x=2 \Rightarrow \text { continuous }\\\text { (4) } x=\sqrt{A+5} \Rightarrow x=3 \Rightarrow \text { continuous }$

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Let $f$ and $g$ be differentiable functions on $R$ such that $fo g$ is the identity function. If for some $a,b\epsilon \textbf{R},g^{'}(a)=5\: \: and\: \: g(a)=b$ then $f^{'}(b)$ is equal to :
Option: 1 $\frac{2}{5}$
Option: 2 $5$
Option: 3 $1$
Option: 4 $\frac{1}{5}$

Rules of Differentiation (Chain Rule) -

Rules of Differentiation (Chain Rule)

Chain Rule or Derivation of Composite Function:

The chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

Let f and g be functions. For all x in the domain of g for which g is differentiable at x and f is differentiable at g(x), the derivative of the composite function

h(x) = (f?g)(x) = f (g(x)) Is given by

h′(x) = f’(g(x))?g’(x)

Composites of Three or More Functions

For all values of x for which the function is differentiable, if k(x) = h(f(g(x)))Then,

$k^{\prime}(x)=h'\left ( f(g(x)) \right )\cdot f'\left ( g(x) \right )\cdot g'(x)$

$\\f(g(x))=x \\\Rightarrow f^{\prime}(g(x)) \cdot g^{\prime}(x)=1 \\Put \;\;x=a \\\Rightarrow f^{\prime}(g(a)) g^{\prime}(a)=1 \\\Rightarrow f^{\prime}(b) \times 5=1 \Rightarrow f^{\prime}(b)=\frac{1}{5}$

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If $x=2\sin \theta -\sin 2\theta$ and $y=2\cos\theta -\cos 2\theta ,\: \theta \equiv \left [ 0,2\pi \right ]$ , then $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}\: at\: \theta =\pi$ is :
Option: 1 $\frac{3}{8}$
Option: 2 $\frac{3}{4}$
Option: 3 $\frac{3}{2}$
Option: 4 $-\frac{3}{4}$

Differentiation of Function in Parametric Form -

Differentiation of Function in Parametric Form

Sometimes, x and y are given as functions of a single variable, i.e., x = f(t) and y = g(t) are two functions and t is a variable. In such cases x and y are called parametric functions or parametric equations and t is called the parameter.

To find $\frac{dy}{dx}$ in such cases, first find the relationship between x and y by eliminating the parameter t and then differentiate with respect to t.

But sometimes it is not possible to eliminate t, then in that case use

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{f'(t)}{g'(t)}$

${\frac{d x}{d \theta}=2 \cos \theta-2 \cos 2 \theta} \\ {\frac{d y}{d \theta}=-2 \sin \theta+2 \sin 2 \theta} \\ {\therefore \frac{d y}{d x}=\frac{\sin 2 \theta-\sin \theta}{\cos \theta-\cos 2 \theta}} \\ {=\frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}}$

${\frac{d^{2} y}{d x^{2}}=\frac{-3}{2} \csc ^{2} \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}} \\ {\frac{d^{2} y}{d x^{2}}=\frac{-\frac{3}{2} \csc ^{2} \frac{3 \theta}{2}}{2(\cos \theta-\cos 2 \theta)}} \\ {\frac{d^{2} y}{d x^{2}}|_{\theta=\pi}=-\frac{3}{4(-1-1)}=\frac{3}{8}}$

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Let $f$ be any function continuous on $[a,b]$ and twice differentiable on $(a,b).$ If for all $x\epsilon \left ( a,b \right ),f^{'}(x)>0$ and $f^{''}(x)<0,$ then for any $c\epsilon (a,b),\frac{f(c)-f(a)}{f(b)-f(c)}$ is greater than :
Option: 1 $\frac{b-c}{c-a}$
Option: 2 $1$
Option: 3 $\frac{c-a}{b-c}$
Option: 4 $\frac{b+a}{b-a}$

Lagrange’s Mean Value Theorem -

Lagrange’s Mean Value Theorem

Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions f defined on a closed interval [a, b] with f(a) = f(b) . The Mean Value Theorem generalized Rolle’s theorem by considering functions that do not necessarily have equal value at the endpoints.

Statement

Let f (x) be a function defined on [a, b] such that

it is continuous on [a, b],
it is differentiable on (a, b).

Then there exists a real number c (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Cauchy’s mean value Theorem

Cauchy's mean value theorem, also known as the extended mean value theorem. It states that if both function f(x) and g(x) are continuous on the closed interval [a, b] and differentiable on the open interval (a, b) and g’(x) is not zero on that open interval, then there exists c in (a, b) such that

$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$

Use LMVT for x $\in$ [a,c]

$\frac{\mathrm{f}(\mathrm{c})-\mathrm{f}(\mathrm{a})}{\mathrm{c}-\mathrm{a}}=\mathrm{f}^{\prime}(\alpha), \alpha \in(\mathrm{a}, \mathrm{c})$

also use LMVT for x $\in$ [c,b]

$\frac{f(b)-f(c)}{b-c}=f^{\prime}(\beta), \beta \in(c, b)$

$\because$ f ''(x) < 0 $\Rightarrow$ f '(x) is decreasing

$\\ {f^{\prime}(\alpha)>f^{\prime}(\beta)} \\ {\frac{f(c)-f(a)}{c-a}>\frac{f(b)-f(c)}{b-c}} \\ {\frac{f(c)-f(a)}{f(b)-f(c)}>\frac{c-a}{b-c}(\because f(x)\text { is increasing) } }$

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A spherical iron ball of $10cm$ radius is coated with a layer of ice of uniform thickness that melts at a rate of $50cm^{3}/min$ . When the thickness of ice is $5cm$ , then the rate (in cm / min.) at which of the thickness of ice decreases, is :
Option: 1 $5/6\pi$
Option: 2 $1/54\pi$
Option: 3 $1/36\pi$
Option: 4 $1/18\pi$

Derivative as Rate Measure -

Derivative as Rate Measure

We have seen that for quantities that are changing over time, the rates at which these quantities change are given by derivatives. If two related quantities are changing over time, the rates at which the quantities change are related. For example, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing.

If a variable quantity y depends on and varies with a quantity x, then the rate of change of y with respect to x is $\frac{dy}{dx}$ .

A rate of change with respect to time is simply called the rate of change.

For example, the rate of change of displacement (s) of an object w.r.t. time is velocity (v). $\mathit{v=\frac{ds}{dt}}$

When limit Δt? 0 is applied, the rate of change becomes instantaneous and we get the rate of change of displacement (s) w.r.t. time at an instant.

$\text{i.e. }\lim _{\Delta t \rightarrow 0} \mathit{\frac{\Delta s}{\Delta t}=\frac{d s}{d t}}$

${V=\frac{4}{3} \pi(10+a)^{3}} \\ {\frac{\partial V}{\partial t}=4 \pi(10+a)^{2} \cdot \frac{\partial a}{\partial t}=50 \frac{c m^{3}}{\min }} \\ {4 \pi(10+5)^{2} \cdot \frac{\partial a}{\partial t}=50 \frac{c m^{3}}{\min }} \\ {\frac{\partial a}{\partial t}=\frac{1}{18 \pi} \frac{c m}{\min }}$

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if

$\small \lim_{n \to \infty }\frac{1^{a}+2^{a}+......+n^{a}}{\left ( n+1 \right )^{a-1}\left [ \left ( na+1 \right )+\left ( na+2 \right ) +.......+\left ( na+n \right )\right ]}=\frac{1}{60}$

for some positive real number a, then a is equal to :
Option: 1 7

Option: 2 8

Option: 3 $\frac{15}{2}$

Option: 4 $\frac{17}{2}$

$\large \lim _{n \rightarrow \infty} \frac{1^{a}+2^{a}+3^{a}+\cdots+n^{a}}{(n+1)^{a-1}[(n a+1)+(n a+2)+\cdots+(n a+n)]}=\frac{1}{60}$

$\large \Rightarrow \lim _{n \rightarrow \infty} \frac{1^{a}+2^{a}+\cdots+n^{a}}{(n+1)^{a-1}\left[n^{2} a+\frac{n(n+1)}{2}\right]}=\frac{1}{60}$

$\large \Rightarrow \lim _{n \rightarrow \infty} \frac{2\left[ \sum_{k=1}^{n} k^{a}\right]}{(n+1)^{a-1}\left[2 n^{2} a+n^{2}+n\right]}=\frac{1}{60}$

$\large \lim _{n \rightarrow \infty} \frac{2 \sum_{k=1}^{n}\left(\frac{k}{n}\right)^{a}}{\frac{(n+1)^{a-1}}{n^{a-1}}\left[\frac{2 n^{2} a+n^{2}+n}{n}\right]}=\frac{1}{60}$

$\large \lim _{n \rightarrow \infty} \frac{2 \sum_{k=1}^{n}\left(\frac{k}{n}\right)^{a}}{\left(1+\frac{1}{n}\right)^{a-1}[2 n a+n+1]}=\frac{1}{60}$

$\large \lim _{n \rightarrow \infty} \frac{\frac{2}{n} \sum_{k=1}^{n}\left(\frac{k}{n}\right)^{a}}{\left(1+\frac{1}{n}\right)^{a-1}\left[2 a+1+\frac{1}{n}\right]}=\frac{1}{60}$

$\large {\color{Blue} \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{1}^{n}\left(\frac{k}{n}\right)^{a}=\int_{0}^{1} x^{a} d x \quad[\text { Derivative as limit of a sum }]}$

$\large \therefore \frac{2 \int_{0}^{1} x^{a} d x}{2 a+1}=\frac{1}{60}$

$\large \\\Rightarrow \frac{\left.2\left(x^{a+1}\right)\right]_{0}^{1}}{(a+1)(2 a+1)}=\frac{1}{60} \\ \\\Rightarrow \frac{2}{(a+1)(2 a+1)}=\frac{1}{60} \\ \\\Rightarrow(a+1)(2 a+1)=120 \\ \\\Rightarrow 2 a^{2}+3 a+1=120 \\ \\\Rightarrow 2 a^{2}-3 a=119=0 \\ \\\Rightarrow a=\frac{-3 \pm \sqrt{9+952}}{4}$

$\large =\frac{-3 \pm \sqrt{961}}{4}=\frac{-3 \pm 31}{4} \quad(\text { since } a>0)$

$\large \\\text { Hence, the value of } a \text { is }\\ \\\frac{-3+31}{4}=\frac{28}{4}=7 \Rightarrow a=7$

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Let $f(x)=x\: \cos ^{-1}\left ( -\sin \left | x \right | \right ),x\: \epsilon \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ],$ then which of the following is true ?

Option: 1 $f'(0)=-\frac{\pi }{2}$ ,
Option: 2 $f'$ is decreasing in $\left ( -\frac{\pi }{2},0 \right )$ and increasing in $\left (0,\frac{\pi }{2} \right )$ ,
Option: 3 $f$ is not differentiable at $x=0$ ,
Option: 4 $f'$ is increasing in $\left ( -\frac{\pi }{2},0 \right )$ and decreasing in $\left (0,\frac{\pi }{2} \right )$ ,

Application of Monotonicity (Part 1 ) -

$\begin{array}{l}{f^{\prime}(x)=x\left(\pi-\cos ^{-1}(\sin |x|)\right)} \\ {=x\left(\pi-\left(\frac{\pi}{2}-\sin ^{-1}(\sin |x|)\right)\right)} \\ {=x\left(\frac{\pi}{2}+|x|\right)}\end{array}$

$\begin{array}{ll}{f(x)=} {\left\{\begin{array}{ll}{x\left(\frac{\pi}{2}+x\right)} & {x \geq 0} \\ {x\left(\frac{\pi}{2}-x\right)} & {x<0}\end{array}\right.} \\ {f^{\prime}(x)=\left\{\begin{array}{ll}{\frac{\pi}{2}+2 x} & {x \geq 0} \\ {\frac{\pi}{2}-2 x} & {x<0}\end{array}\right.}\end{array}$

$f'(x) \text { is increasing in }\left(0, \frac{\pi}{2}\right) \text { and decreasing in }\left(\frac{-\pi}{2}, 0\right)$

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Kuldeep Maurya

If c is a point at which Roll's theorem holds for the function, $f(x)=\log _{e}\left ( \frac{x^{2}+\alpha }{7x} \right )$ in the interval $\left [ 3,4 \right ],$ where $\alpha \; \epsilon \; R$ then $f''(c)$ is equal to :

Option: 1 $-\frac{1}{24}$
Option: 2 $-\frac{1}{12}$
Option: 3 $\frac{\sqrt{3}}{7}$
Option: 4 $\frac{1}{12}$

Rolle’s Theorem -

Rolle’s Theorem

Informally, Rolle’s theorem states that if the outputs of a differentiable function f are equal at the endpoints of an interval, then there must be an interior point c where f ′(c) = 0

Rolle’s Theorem Statement:

Let f be a real-valued function defined on the closed interval [a, b] such that

it is continuous on the closed interval [a, b],
it is differentiable on the open interval (a, b), and
f (a) = f (b).

There then exists at least one c ∈ (a, b) such that f ′(c) = 0.

${f(3)=f(4)} \\ {\frac{9+\alpha}{3}=\frac{16+\alpha}{4}}$

$\alpha=12$

$\\f'(x)=\frac{x^2-12}{x\left(x^2+12\right)}\\f'(c)=0\\\begin{array}{ll}{\therefore} & {c=\sqrt{12}} \\ {\therefore \quad} & {f^{\prime \prime}(c)=\frac{1}{12}}\end{array}$

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$\lim_{x\rightarrow 0}\left ( \frac{3x^{2}+2}{7x^{2}+2} \right )^{1/x^{2}}$ is equal to :

Option: 1 $e$
Option: 2 $\frac{1}{e^{2}}$
Option: 3 $\frac{1}{e}$
Option: 4 $e^{2}$

Limits of the form (1 power infinity) -

Limits of the form 1^∞ (1 power infinity)

To find the limit of the exponential of form 1^∞ , we will use the following results

If $\lim_{x\rightarrow a}\;f(x)=\lim_{x\rightarrow a}\;g(x)=0$

Then,

$\lim_{x\rightarrow a}\;\left [1+f(x) \right ]^{\frac{1}{g(x)}}=e^{\lim_{\;x\rightarrow a}\frac{f(x)}{g(x)}} .$

When, $\lim_{x\rightarrow a}\;f(x)=1 and \lim_{x\rightarrow a}\;g(x)=\infty$

Then,

$\\\lim_{x\rightarrow a}\;\left [f(x) \right ]^{{g(x)}}=\lim_{x\rightarrow a}\;\left [1+f(x)-1 \right ]^{{g(x)}}\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=e^{\lim_{\;x\rightarrow a}{\left (f(x)-1 \right )}{g(x)}}$

Some particular cases

$\\\text{(a)}\;\;\;\lim_{x\rightarrow 0}\;(1+x)^{\frac{1}{x}}=e\\\\\text{(b)}\;\;\;\lim_{x\rightarrow 0}\;\left (1+\frac{1}{x} \right )^{x}=e\\\\\text{(c)}\;\;\;\lim_{x\rightarrow 0}\;(1+cx)^{\frac{1}{x}}=e^c\\\\\text{(d)}\;\;\;\lim_{x\rightarrow 0}\;\left (1+\frac{c}{x} \right )^{x}=e^c$

Limits of the form (1 power infinity) -

Limits of the form 1^∞ (1 power infinity)

To find the limit of the exponential of form 1^∞ , we will use the following results

If $\lim_{x\rightarrow a}\;f(x)=\lim_{x\rightarrow a}\;g(x)=0$

Then,

$\lim_{x\rightarrow a}\;\left [1+f(x) \right ]^{\frac{1}{g(x)}}=e^{\lim_{\;x\rightarrow a}\frac{f(x)}{g(x)}} .$

When, $\lim_{x\rightarrow a}\;f(x)=1 and \lim_{x\rightarrow a}\;g(x)=\infty$

Then,

Some particular cases

$\\\lim _{x\to 0}\left(\frac{3x^2+2}{\:7x^2+2}\right)^{\frac{1}{x^2}}=\lim _{x\to 0}\left(1+\frac{3x^2+2}{\:7x^2+2}-1\right)^{\frac{1}{x^2}}\\\lim_{x\rightarrow 0}\left ( 1+\frac{-4x^2}{7x^2+2} \right )^{1/x^2}=e^{\lim _{x\to 0}\left(-\frac{4x^2}{7x^2+2}\right)\left(\frac{1}{x^2}\right)}\\=\frac{1}{e^2}$

Correct Option (2)

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Let $f(x)=\left\{\begin{matrix} max \left\{\left|x\right|,\:x^2\right\}, &|x|\leq 2 \\ 8-2|x| & 2<|x|\leq 4 \end{matrix}\right.$ Let S be the set of points in the interval (-4,4) at which f is not differentiable. Then S:

Option: 1 is an empty set
Option: 2 equals {-2,-1,0,1,2}
Option: 3 equals{-2,-1,1,2}
Option: 4 equals{-2,2}

Condition for differentiability -

A function f(x) is said to be differentiable at $x=x_{\circ }$ if $Rf'(x_{\circ })\:\:and\:\:Lf'(x_{\circ })$ both exist and are equal otherwise non differentiable

Geometrical interpretation of Derivative -

Let P be any point (x, y) on the curve y = f(x) and Q is a point in the neighbourhood of P on either side of P. such that the co-ordinate of the point Q are

$(x+\delta x, y+\delta y)$ satisfying the curve y = f(x)

$\therefore \:M_{T}=slope\:of\:tangent$

$=\lim_{\delta x\rightarrow \circ }\:\:\frac{(y+\delta y)-y}{(x+\delta x)-x}=\lim_{\delta x\rightarrow \circ }\:\:\frac{\delta y}{\delta x}$

$\therefore \:\:M_{T}=(\frac{dy}{dx})\:\:at \;\:(x,\:y)$

- wherein

Where (x, y) on the curve and M_T is tangent at (x, y).

$f(x) = \left\{\begin{matrix} 8+2x, & -4\leq x<-2 \\ x^{2} ,&-2\leq x\leq -1 \\ \left | x \right |, &-1<x<1 \\ x^{2}, & 1\leq x\leq 2\\ 8-2x, & 2< x\leq 4 \end{matrix}\right.$

Plot graph

f(x) is not differentiable at

x = -2,-1,0,1,2.

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Get Answers to all your Questions

All Questions

Let denote the greatest integer and Then the function, is discontinuous, when x is equal to: Option: 1 Option: 2 Option: 3 Option: 4

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Let and be differentiable functions on such that is the identity function. If for some then is equal to : Option: 1 Option: 2 Option: 3 Option: 4

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If and , then is : Option: 1 Option: 2 Option: 3 Option: 4

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Let be any function continuous on and twice differentiable on If for all and then for any is greater than : Option: 1 Option: 2 Option: 3 Option: 4

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JEE Main high-scoring chapters and topics

A spherical iron ball of radius is coated with a layer of ice of uniform thickness that melts at a rate of . When the thickness of ice is , then the rate (in cm / min.) at which of the thickness of ice decreases, is : Option: 1 Option: 2 Option: 3 Option: 4

Let $f$ and $g$ be differentiable functions on $R$ such that $fo g$ is the identity function. If for some $a,b\epsilon \textbf{R},g^{'}(a)=5\: \: and\: \: g(a)=b$ then $f^{'}(b)$ is equal to :
Option: 1 $\frac{2}{5}$
Option: 2 $5$
Option: 3 $1$
Option: 4 $\frac{1}{5}$

If $x=2\sin \theta -\sin 2\theta$ and $y=2\cos\theta -\cos 2\theta ,\: \theta \equiv \left [ 0,2\pi \right ]$ , then $\frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}\: at\: \theta =\pi$ is :
Option: 1 $\frac{3}{8}$
Option: 2 $\frac{3}{4}$
Option: 3 $\frac{3}{2}$
Option: 4 $-\frac{3}{4}$

$\lim_{x\rightarrow 0}\left ( \frac{3x^{2}+2}{7x^{2}+2} \right )^{1/x^{2}}$ is equal to :

Option: 1 $e$
Option: 2 $\frac{1}{e^{2}}$
Option: 3 $\frac{1}{e}$
Option: 4 $e^{2}$