Solution: Let $x-\cos x=t$ $\begin{array}{l} \Rightarrow d(x-\cos x)=d t \\ \Rightarrow(1+\sin x) d x=d t \end{array}$ $\therefore$ By substituting $\mathrm{t}$ and dt in given equation we obtain...
Solution: Let $\sin ^{-1} \mathrm{x}=\mathrm{t}$ $\begin{array}{l} \Rightarrow \mathrm{d}\left(\sin ^{-1} \mathrm{x}\right)=\mathrm{dt} \\ \Rightarrow...
Solution: Let $\cot x=t$ $\begin{array}{l} \Rightarrow \mathrm{d}(\cot x)=d t \\ \Rightarrow-\operatorname{cosec}^{2} x \cdot d x=d t \\ \Rightarrow d x=\frac{-d t}{\csc ^{2} x} \end{array}$...
with respect to 
, if, (i) 
(ii) 
(i) Let (ii) Let
Solution: Let $1+\mathrm{e}^{\mathrm{x}}=\mathrm{t}$ $\Rightarrow d\left(1+e^{x}\right)=d t$ $\Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$ $\therefore$ By substituting $t$ and $dt$...
![\int \sqrt[3]{\cos ^{2} x} \sin x d x](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3895073084535c9d91f30a3f8969a9ee_l3.png "Rendered by QuickLaTeX.com")
Solution: Let $\cos x=t$ $\begin{array}{l} \Rightarrow \mathrm{d}(\cos x)=d t \\ \Rightarrow-\sin x d x=d t \\ \Rightarrow d x=\frac{-d t}{\sin x} \end{array}$ $\therefore$ On substituting...
with respect to 
if, (iii) 
(iii) Let
Solution: Let $1+e^{x}=t$ $\begin{array}{l} \Rightarrow \mathrm{d}\left(1+\mathrm{e}^{\mathrm{x}}\right)=\mathrm{dt} \\ \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}=\mathrm{d}...
with respect to if, (i) 
(ii) 
(i) Let (ii)
Solution: Let $1+v x=t$ $\begin{array}{l} \Rightarrow d(1+v x)=d t \\ \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \\ \Rightarrow \frac{1}{\sqrt{x}} d x=2 d t \end{array}$ $\therefore$ On substituting...
with respect to 
, if (i) 
(ii) 
(i) Given sin-1 √ (1-x2) (ii) Given sin-1 √ (1-x2)
Solution: Let $\log \left(1+\frac{1}{\mathrm{x}}\right)=\mathrm{t}$ $\begin{array}{l} \Rightarrow \operatorname{d}\left(\log \left(1+\frac{1}{\mathrm{x}}\right)\right)=\mathrm{dt} \\ \Rightarrow...
Solution: Let $\log x=t$ $\begin{array}{l} \Rightarrow d(\log x)=d t \\ \Rightarrow \frac{1}{x} d x=d t \end{array}$ By substituting $\mathrm{t}$ and $dt$ in above equation we obtain...
Solution: It is better to eliminate the denominator, in order to solve these equations. $\Rightarrow \int \frac{\sin (x-a)}{\sin (x-b)} d x$ Now, add and subtract $b$ in $(x-a)$ $\begin{array}{l}...
Solution: Suppose $\mathrm{I}=\int \frac{\cos 2 x}{(\cos \mathrm{x}+\sin \mathrm{x})^{2}} d x$ On substituting the formula, we obtain $=\int \frac{\cos ^{2} x-\sin ^{2} x}{(\cos x+\sin x)^{2}} d x$...
and 
, prove that 
.
Solution: First of all we need to convert sec $x$ in terms of $\cos x$ It is known that $\Rightarrow \sec x=\frac{1}{\cos x}, \sec 2 x=\frac{1}{\cos 2 x}$ So, the above equation becomes,...
Given, $x=2 t /\left(1+t^{2}\right)$ On differentiating $x$ with respect to t using quotient rule, $$ \begin{array}{l} \frac{\mathrm{dx}}{\mathrm{dt}}=\left[\frac{\left(1+\mathrm{t}^{2}\right)...
Solution: Given that, $\int \frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}} d x$ It is known that $\begin{array}{l} 1-\operatorname{Cos} x=2 \sin ^{2} \frac{x}{2} \\ 1+\cos x=2 \cos ^{2} \frac{x}{2}...
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Solution: Given that, $\int \frac{\sqrt{1+\cos 2 x}}{\sqrt{1-\cos 2 x}} d x$ It is known that $\begin{array}{l} 1+\cos 2 x=2 \cos ^{2} x \\ 1-\cos 2 x=2 \sin ^{2} x \end{array}$ On substituting...
Solution: Given that $\int \frac{1}{\sqrt{1-\cos 2 x}} d x$ In the equation given $\cos 2 x=\cos ^{2} x-\sin ^{2} x$ Also it is known that $\cos ^{2} x+\sin ^{2} x=1$ On substituting the values in...
Let the $3-digit$ number be $ABC,$ where $C$ is at the unit’s place, $B$ at the tens place and $A$ at the hundreds place. As the number has to even, the digits possible at $C$ are $2$ or $4$ or $6.$...
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$\left| \begin{matrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix} \right|$ Let $\vartriangle =\left| \begin{matrix} 1 & a...
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$\left| \begin{matrix} a+b & 2a+b & 3a+b \\ 2a+b & 3a+b & 4a+b \\ 4a+b & 5a+b & 6a+b \\ \end{matrix} \right|$ Let $\vartriangle =\left| \begin{matrix} a+b & 2a+b &...
$\left| \begin{matrix} \frac{1}{a} & {{a}^{2}} & bc \\ \frac{1}{b} & {{b}^{2}} & ac \\ \frac{1}{c} & {{c}^{2}} & ab \\ \end{matrix} \right|$ Let $\vartriangle =\left|...
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$\left| \begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \\ \end{matrix} \right|$ Let $\vartriangle =\left| \begin{matrix} 2 & 3 & 7 \\ 13...
$\left| \begin{matrix} 6 & 3 & -2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \\ \end{matrix} \right|$ Let $\vartriangle =\left| \begin{matrix} 6 & 3 & -2 \\ 2 & -1 & 2 ...
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As per the given information it states, $\left| \begin{matrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \\ \end{matrix} \right|$ Let $\vartriangle =\left| \begin{matrix} 8...
$QR$ at $X$ and $PR$ at $Z.$ $OZ,$ $OX,$ $OY$ are perpendicular to the sides $PR,$ $QR,$ $PQ.$ Here $PQR$ is an isosceles triangle with sides $PQ = PR$ and also from the figure, \[\Rightarrow...
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$Δ ABC$ is an isosceles triangle such that $AB = AC.$ The vertical angle $BAC = 2θ$ Triangle is inscribed in the circle with center $O$ and radius $a.$ Draw $AM$ perpendicular to $BC.$ Since, $Δ...
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Let the radius and height of cone be r and h respectively \[Radius\text{ }of\text{ }sphere\text{ }=\text{ }R\] \[{{R}^{2}}~=\text{ }{{r}^{2}}~+\text{ }{{\left( h\text{ }-\text{ }R \right)}^{2}}\]...
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Let y = (log x)cos x Taking log both the sides, we get
Let the length and breadth of rectangle $ABCD$ be $2x$ and $y$ respectively \[Radius\text{ }of\text{ }semicircle\text{ }=\text{ }r\text{ }\left( given \right)\] In triangle $OBA,$ where is the...
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Let the dimensions of the rectangle be $x$ and $y.$ Therefore, the perimeter of window \[=\text{ }x\text{ }+\text{ }y\text{ }+\text{ }x\text{ }+\text{ }x\text{ }+\text{ }y\text{ }=\text{ }12\]...
Let the radius of semicircle, length and breadth of rectangle be $r,$ $x$ and $y$ respectively \[AE\text{ }=\text{ }r\] \[AB\text{ }=\text{ }x=2r\text{ }\left( semicircle\text{ }is\text{...
Let the length, breath and height of tank be $I, b$ and $h$ respectively. Also, assume volume of tank as $V$ \[h\text{ }=\text{ }2\text{ }m\text{ }\left( given \right)\] \[V\text{ }=\text{ }8\text{...
, prove that 
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.
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, prove that 
differentiating the equation on both sides with respect to x, we get,
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Given length of rectangle sheet \[=\text{ }45\text{ }cm\] Breath of rectangle sheet \[=\text{ }24\text{ }cm\] Let the side length of each small square be $a.$ If by cutting a square from each corner...
differentiating the equation on both sides with respect to x, we get,
Given side length of big square is $18 cm$ Let the side length of each small square be $a.$ If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with Length,...
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It is given that two sides of a triangle have lengths $a$ and $b$ and the angle between them is $θ.$ Let the area of triangle be $A$
differentiating the equation on both sides with respect to x, we get,
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Let us say the sum of perimeter of square and circumference of circle be $L$ Given sum of the perimeters of a square and a circle. Assuming, \[side\text{ }of\text{ }square\text{ }=\text{ }a\text{...
Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length $x$ and $y$ respectively. Then, \[x\text{ }+\text{ }y\text{ }=\text{ }20\text{ }\Rightarrow...
Suppose the given wire, which is to be made into a square and a circle, is cut into two pieces of length $x$ and $y$ $m$ respectively. Then, \[x\text{ }+\text{ }y\text{ }=\text{ }28\text{...
Solution: As per the given question,
Solution: Given set A = {x ∈ Z; 0 ≤ x ≤ 12} Also given that relation R = {(a, b): a = b} is defined on set A Now we have to check whether the given relation is equivalence or not. To prove...
Let $r$ $and$ $h$ be the radius and height of the cylinder, respectively. Then, Volume $(V)$ of the cylinder \[=\text{ }\pi {{r}^{2}}~h\] \[\to \text{ }100\text{ }=\text{ }\pi {{r}^{2}}~h\] \[\to...
Solution: First let R be a relation on A It is given that set A of ordered pair of integers defined by (x, y) R (u, v) if xv = y u Now we have to check whether the given relation is equivalence or...
Solution: Given that m is said to be related to n if m and n are integers and m − n is divisible by 13 Now we have to check whether the given relation is equivalence or not. To prove equivalence...
Solution: Given R = {(a, b): a, b ∈ Z and a + b is even} is a relation defined on R. Also given that Z be the set of integers To prove equivalence relation it is necessary that the given relation...
Let the given two numbers be $x$ and $y$. Then, \[x\text{ }+\text{ }y\text{ }=\text{ }15\text{ }\ldots ..\text{ }\left( 1 \right)\] \[y\text{ }=\text{ }\left( 15\text{ }-\text{ }x \right)\] Now we...
Solution: Given (a, b) ∈ R ⇔ a − b is divisible by n is a relation R defined on Z. To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and...
As per the given question, \[1\text{ }+\text{ }3{{\left( {\scriptscriptstyle 1\!/\!{ }_2}\text{ }-\text{ }a \right)}^{2}}~\left( -1 \right)\text{ }=\text{ }0\] \[1\text{ }-\text{ }3{{\left(...
Solution: Given relation R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5 To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive....
Solution: Given R = {(a, b): 2 divides a – b} is a relation defined on Z. To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive. Let us...
Let the two positive numbers be $a$ and $b$ Given \[a\text{ }+\text{ }b\text{ }=\text{ }64\text{ }\ldots \text{ }\left( 1 \right)\] We have, \[{{a}^{3}}~+\text{ }{{b}^{3}}\] is minima Assume,...
Solution: Given R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is a relation To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive. Let...
Which implies $S$ is minimum when \[a\text{ }=\text{ }15/2\text{ }and\text{ }b\text{ }=\text{ }15/2.\]
Solution: (i) The relation on A having properties of being reflexive, transitive, but not symmetric is R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)} Relation R satisfies reflexivity and transitivity....
Solution: Let A be a set. Then, Identity relation IA=IA is reflexive, since (a, a) ∈ A ∀a The converse of it need not be necessarily true. Consider the set A = {1, 2, 3} Here, Relation R = {(1, 1),...
Let $f(x)=2 x^{3}-24 x+107$ $ \therefore f^{\prime}(x)=6 x^{2}-24=6\left(x^{2}-4\right) $ Now, for local maxima and local minima we have $f^{\prime}(x)=0$ $ \begin{array}{l} \Rightarrow...
Solution: Given R = {(a, b): a ≤ b3} It is observed that (1/2, 1/2) in R as 1/2 > (1/2)3 = 1/8 ∴ R is not reflexive. Now, (1, 2) ∈ R (as 1 < 23 = 8) But, (2, 1) ∉ R (as 2 > 13 = 1) ∴ R is...
Solution: Given R = {(a, b): b = a + 1} Now for this relation we have to check whether it is reflexive, transitive and symmetric Reflexivity: Let a be an arbitrary element of R. Then,...
(ii) Solution: (i) Given function is $f(x)=3 x^{4}-8 x^{3}+12 x^{2}-48 x+25$ on $[0,3]$ On differentiating we get $ \begin{array}{l} f^{\prime}(x)=12 x^{3}-24 x^{2}+24 x-48 \\...
Solution: (i) Consider aRb if a – b > 0 Now for this relation we have to check whether it is reflexive, transitive and symmetric. Reflexivity: Let a be an arbitrary element of R. Then, a ∈ R...
Solution: Consider R1 Given R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)} Reflexivity: Here, (1, 1), (2, 2), (3, 3) ∈R So, R1 is reflexive. Symmetry: Here, (2, 1) ∈ R1, But (1, 2) ∉ R1...
Solution: (i) Given R1 on Q0 defined by (a, b) ∈ R1 ⇔ a = 1/b. Reflexivity: Let a be an arbitrary element of R1. Then, a ∈ R1 ⇒ a ≠1/a for all a ∈ Q0 So, R1 is not reflexive. Symmetry: Let...
(i) (ii) Given function is \[f\left( x \right)\text{ }=~{{\left( x\text{ }-\text{ }1 \right)}^{2}}~+\text{ }3\] On differentiation we get \[\Rightarrow \text{ }f'\left( x \right)~=\text{ }2\left(...
Solution: (i) Consider R1 Given R1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)} Now we have check R1 is reflexive, symmetric and transitive Reflexive: Given (a, a), (b, b) and...
Solution; (iii) Given R = {(x, y): x is wife of y} Now we have to check whether the relation R is reflexive, symmetric and transitive. First let us check whether the relation is reflexive:...
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Solution: (i) Given R = {(x, y): x and y work at the same place} Now we have to check whether the relation is reflexive: Let x be an arbitrary element of R. Then, x ∈R ⇒...
Given \[y\text{ }=\text{ }a\text{ }log\text{ }x\text{ }+\text{ }b{{x}^{2}}~+\text{ }x\] On differentiating we get Given that extreme values exist at \[x\text{ }=\text{ }1,\text{ }2\] \[a\text{...
Given $f(x)=-(x-1)^{3}(x+1)^{2}$ $ \begin{array}{l} f^{\prime}(x)=-3(x-1)^{2}(x+1)^{2}-2(x-1)^{3}(x+1) \\ =-(x-1)^{2}(x+1)(3 x+3+2 x-2) \\ =-(x-1)^{2}(x+1)(5 x+1) \\ f^{\prime...
Solution: (i) $ \begin{array}{l} \text { Given } f(x)=(x-1)(x-2)^{2} \\ f(x)=(x-2)^{2}+2(x-1)(x-2) \\ =(x-2)(x-2+2 x-2) \\ =(x-2)(3 x-4) \\ f^{\prime}(x)=(3 x-4)+3(x-2) \end{array} $ For maxima and...
Given $\mathrm{f}(\mathrm{x})=\mathrm{x} \mathrm{e}^{x}$ $ \begin{array}{l} f(x)=e^{x}+x e^{i}=e^{x}(x+1) \\ f^{\prime}(x)=e^{2}(x+1)+e^{t} \\ =e^{2}(x+2) \end{array} $ For maxima and minima, $...
(i) Given $f(x)=(x-1)(x+2)^{2}$ $ \begin{array}{l} \therefore \mathrm{f}(\mathrm{x})=(\mathrm{x}+2)^{2}+2(\mathrm{x}-1)(\mathrm{x}+2) \\ =(\mathrm{x}+2)(\mathrm{x}+2+2 \mathrm{x}-2) \\...
(i) Given $f(x)=x^{4}-62 x^{2+} 120 x+9$ $ \begin{array}{l} \therefore \mathrm{f}(\mathrm{x})=4 \mathrm{x}^{\mathrm{a}}-124 x+120=4\left(\mathrm{x}^{a}-31 \mathrm{x}+30\right) \\...
![\[\left| \begin{matrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \\ \end{matrix} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-feb3f1cab1c309d62bd5bbd07cc463c8_l3.png "Rendered by QuickLaTeX.com")
as per the question given it states $\left| \begin{matrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \\ \end{matrix} \right|$ Let $\vartriangle =\left| \begin{matrix} 102...
![\[\left| \begin{matrix} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \\ \end{matrix} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-93b9d7767d25ea45254643ae995d3a0f_l3.png "Rendered by QuickLaTeX.com")
It is given in the question that, $\left| \begin{matrix} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \\ \end{matrix}...
![\[\left| \begin{matrix} 6 & 3 & -2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \\ \end{matrix} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6d27a4ddf783cacc0add9d8ae8dbfbfc_l3.png "Rendered by QuickLaTeX.com")
As per the question it is given that, $\vartriangle =\left| \begin{matrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \\ \end{matrix} \right|$ Applying row operation...
![\[\left| \begin{matrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \\ \end{matrix} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-252ae45dec5628cdff43048e03e61e8a_l3.png "Rendered by QuickLaTeX.com")
As per the question it states, $\vartriangle =\left| \begin{matrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \\ \end{matrix} \right|$ By applying column operation...
![\[\left| \begin{matrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \\ \end{matrix} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-586d218344d18810311a1f3b56a6e729_l3.png "Rendered by QuickLaTeX.com")
As per the question it is given that, $\left| \begin{matrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \\ \end{matrix} \right|$ Let $\vartriangle =\left| \begin{matrix} 1...
![\[\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5beb48821e3c0a9c8a697d805298a907_l3.png "Rendered by QuickLaTeX.com")
According to the question it states $\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|$ Let $\vartriangle =\left| \begin{matrix} a & h...
![\[\left| \begin{matrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \\ \end{matrix} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8914dacb0395f7ab351fbfae01d9ef8e_l3.png "Rendered by QuickLaTeX.com")
As per the question it is given that, \[\left| \begin{matrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \\ \end{matrix} \right|\] Let $\vartriangle =\left|...
![\[\left| \begin{matrix} 1 & 5 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \\ \end{matrix} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7d240d5e7668cb48f2980c956673a9d9_l3.png "Rendered by QuickLaTeX.com")
it is given in the question that, \[\left| \begin{matrix} 1 & 5 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \\ \end{matrix} \right|\] Let $\vartriangle =\left| \begin{matrix} 1...
Given \[f\text{ }\left( x \right)\text{ }=\text{ }sin\text{ }2x\] Differentiate w.r.t x, we get \[f'\left( x \right)\text{ }=\text{ }2\text{ }cos\text{ }2x,\text{ }0\text{ }<\text{ }x\text{...
Given, $f(x)=x^{3}-6 x^{2}+9 x+15$ Differentiate with respect to $x$, we get, $f^{\prime}(x)=3 x^{2}-12 x+9=3\left(x^{2}-4 x+3\right)$ $=3(x-3)(x-1)$ For all maxima and minima, $ \begin{array}{l}...
As per the given question, Therefore \[x\text{ }=\text{ }0,\] now for the values close to \[x\text{ }=\text{ }0,\] and to the left of \[0,\text{ }f'\left( x \right)\text{ }>\text{ }0\] Also for...
Differentiate with respect to $x$, we get, $ \begin{array}{l} f(x)=(x+2)^{2}+2(x-1)(x+2) \\ =(x+2)(x+2+2 x-2) \\ =(x+2)(3 x) \end{array} $ For all maxima and minima, $ \begin{array}{l} f(x)=0 \\...
Given, $f(x)=x^{2}(x-1)^{2}$ Differentiate with respect to $x$, we get, $ \begin{array}{l} {\left x=3 x^{2}(x-1)^{2}+2 x^{2}(x-1)\right.} \\ =[x-1]\left(3 x^{2}(x-1)+2 x^{2}\right) \\ =[x-1]\left(3...
Given, $f(x)=x^{2}-3 x$ Differentiate with respect to $x$ then we get, $ f(x)=3 x^{2}-3 $ $\mathrm{Now}, \mathrm{f}(x)=0$ $ 3 x^{2}=3 \Rightarrow x=\pm 1 $ Again differentiate $f(x)=3 x^{2}-3$ $...
Given $f(x)=(x-5)^{4}$ Differentiate with respect to $x$ $ f(x)=4(x-5)^{2} $ For local maxima and minima $ \begin{array}{l} f(x)=0 \\ =4(x-5)^{2}=0 \\ =x-5=0 \\ x=5 \end{array} $ $f(x)$ changes from...