Find the value
Question: Find the value $x^{3}-12 x(x-4)-64$ Solution: $=x^{3}-12 x^{2}+48 x-64$ $=x^{3}-3 \times x^{2} \times 4+3 \times 4^{2} \times x-4^{3}$ $=(x-4)^{3}$ $\left[\therefore a^{3}-b^{3}-3 a^{2} b+3 a b^{2}=(a-b)^{3}\right]$ $=(x-4)(x-4)(x-4)$ $\therefore x^{3}-12 x(x-4)-64=(x-4)(x-4)(x-4)$...
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Question: Find the value $8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}$ Solution: $=(2 a)^{3}-(3 b)^{3}-3 \times(2 a)^{2} \times 3 b+3 \times 2 a \times(3 b)^{2}$ $=(2 a-3 b)^{3}$ $\left[\therefore a^{3}-b^{3}-3 a^{2} b+3 a b^{2}=(a-b)^{3}\right]$ $=(2 a-3 b)(2 a-3 b)(2 a-3 b)$ $\therefore 8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}$ $=(2 a-3 b)(2 a-3 b)(2 a-3 b)$...
Read More →What is the difference between a collection and a set?
Question: What is the difference between a collection and a set? Give reasons to support your answer? Solution: Well-defined collections are sets. Example: The collection of good teachers in a school is not a set, It is a collection. Thus, we can say that every set is a collection, but every collection is not necessarily a set. The collection of vowels in English alphabets is a set....
Read More →Find the intervals in which the function
Question: Find the intervals in which the function $f$ given by $f(x)=x^{3}+\frac{1}{x^{3}}, x \neq 0$ is (i) increasing (ii) decreasing Solution: $f(x)=x^{3}+\frac{1}{x^{3}}$ $\therefore f^{\prime}(x)=3 x^{2}-\frac{3}{x^{4}}=\frac{3 x^{6}-3}{x^{4}}$ Then, $f^{\prime}(x)=0 \Rightarrow 3 x^{6}-3=0 \Rightarrow x^{6}=1 \Rightarrow x=\pm 1$ Now, the points $x=1$ and $x=-1$ divide the real line into three disjoint intervals i.e., $(-\infty,-1),(-1,1)$, and $(1, \infty)$. In intervals $(-\infty,-1)$ a...
Read More →If X
Question: If $X=\left\{8^{n}-7 n-1: n \in N\right\}$ and $Y=\{49(n-1): n \in N\}$, then prove that $X \subseteq Y$. Solution: Given: $X=\left\{8^{n}-7 n-1: n \in N\right\}$ and $Y=\{49(n-1): n \in N\}$ To prove: $X \subseteq Y$ Let: $x_{n}=8^{n}-7 n-1, n \in N$ $\Rightarrow x_{1}=8-7-1=0$ For any $n \geqslant 2$, we have : $x_{n}=8^{n}-7 n-1=(1+7)^{n}-7 n-1$ $\Rightarrow x_{n}={ }^{n} C_{0}+{ }^{n} C_{1} \cdot 7+{ }^{n} C_{2} \cdot 7^{2}+{ }^{n} C_{3} \cdot 7^{3}+\ldots+{ }^{n} C_{n} \cdot 7^{n}...
Read More →Solve the following systems of equations graphically:
Question: Solve the following systems of equations graphically: $x+y=4$ $2 x-3 y=3$ Solution: The given equations are $x+y=4 \quad \ldots \ldots .(i)$ $2 x-3 y=3$ ..(ii) Putting $x=0$ in equation $(i)$, we get: $\Rightarrow 0+y=4$ $\Rightarrow y=4$ $\therefore x=0, \quad y=4$ Putting $y=0$ in equation $(i,$, we get $\Rightarrow x+0=4$ $\Rightarrow x=4$ $\therefore x=4, \quad y=0$ Use the following table to draw the graph. Draw the graph by plotting the two points $A(4,0)$ and $B(4,0)$ from table...
Read More →Find the intervals in which the function f given by
Question: Find the intervals in which the functionfgiven by $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ is (i) increasing (ii) decreasing Solution: $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ $\therefore f^{\prime}(x)=\frac{(2+\cos x)(4 \cos x-2-\cos x+x \sin x)-(4 \sin x-2 x-x \cos x)(-\sin x)}{(2+\cos x)^{2}}$ $=\frac{(2+\cos x)(3 \cos x-2+x \sin x)+\sin x(4 \sin x-2 x-x \cos x)}{(2+\cos x)^{2}}$ $=\frac{6 \cos x-4+2 x \sin x+3 \cos ^{2} x-2 \cos x+x \sin x \cos x+4 \sin ^{2} x-2 x \sin x-...
Read More →What universal set (s) would you propose for each of the following:
Question: What universal set (s) would you propose for each of the following: (i) The set of right triangles. (ii) The set of isosceles triangles. Solution: (i) The set of all triangles in a plane (ii) The set of all triangles in a plane...
Read More →How many elements has P (A),
Question: How many elements has $\bar{P}(A)$, if $A=\phi$ ? Solution: Given : $A=\phi$ This means $P(A)=\{\phi\}$. Hence, $P(A)$ would have one element....
Read More →Find the value
Question: Find the value $8 a^{3}+27 b^{3}+36 a^{2} b+54 a b^{2}$ Solution: $=(2 a)^{3}+(3 b)^{3}+3 \times(2 a)^{2} \times 3 b+3 \times 2 a \times(3 b)^{2}$ $=(2 a+3 b)^{3}$ $\left[\therefore a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$ $=(2 a+3 b)(2 a+3 b)(2 a+3 b)$ $\therefore 8 a^{3}+27 b^{3}+36 a^{2} b+54 a b^{2}$ $=(2 a+3 b)(2 a+3 b)(2 a+3 b)$...
Read More →Prove that:
Question: Prove that: $A \subseteq B, B \subseteq C$ and $C \subseteq A \Rightarrow A=C$ Solution: Let $x \in A$ $\Rightarrow x \in B \quad(\because A \subseteq B)$ $\Rightarrow x \in C \quad(\because B \subseteq C)$ $\therefore x \in A \Rightarrow x \in C$ $\Rightarrow A \subseteq C$ ...(1) It is given that, $C \subseteq A$ ...(2) From $(1)$ and $(2)$, we have $A=C$...
Read More →Find the value
Question: Find the value $8 x^{3}+y^{3}+12 x^{2} y+6 x y^{2}$ Solution: $=(2 x)^{3}+(y)^{3}+3 \times(2 x)^{2} \times y+3(2 x) \times y^{2}$ $=(2 x+y)^{3}$ $\left[\therefore a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$ $=(2 x+y)(2 x+y)(2 x+y)$ $\therefore 8 x^{3}+y^{3}+12 x^{2} y+6 x y^{2}=(2 x+y)(2 x+y)(2 x+y)$...
Read More →Show that the normal at any point θ to the curve
Question: Show that the normal at any pointto the curve $x=a \cos \theta+a \theta \sin \theta, y=a \sin \theta-a \theta \cos \theta$ is at a constant distance from the origin. Solution: We havex=acos+asin. $\therefore \frac{d x}{d \theta}=-a \sin \theta+a \sin \theta+a \theta \cos \theta=a \theta \cos \theta$ $y=a \sin \theta-a \theta \cos \theta$ $\therefore \frac{d y}{d \theta}=a \cos \theta-a \cos \theta+a \theta \sin \theta=a \theta \sin \theta$ $\therefore \frac{d y}{d x}=\frac{d y}{d \thet...
Read More →Find the value
Question: Find the value $x^{3}+8 y^{3}+6 x^{2} y+12 x y^{2}$ Solution: $=(x)^{3}+(2 y)^{3}+3 \times x^{2} \times 2 y+3 \times x \times(2 y)^{2}$ $=(x+2 y)^{3}$ $\left[\therefore x^{3}+y^{3}+3 x^{2} y+3 x y^{2}=(x+y)^{3}\right]$ $=(x+2 y)(x+2 y)(x+2 y)$ $\therefore x^{3}+8 y^{3}+6 x^{2} y+12 x y^{2}=(x+2 y)(x+2 y)(x+2 y)$...
Read More →Find the value
Question: Find the value $a^{3}-3 a^{2} b+3 a b^{2}-b^{3}+8$ Solution: $=(a-b)^{3}+23$ $\left[\therefore a^{3}-b^{3}-3 a^{2} b+3 a b^{2}=(a-b)^{3}\right]$ $=(a-b+2)\left((a-b)^{2}-(a-b)^{2}+2^{2}\right)$ $\therefore\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$ $=(a-b+2)\left(a^{2}+b^{2}-2 a b-2(a-b)+4\right)$ $=(a-b+2)\left(a^{2}+b^{2}-2 a b-2 a+2 b+4\right)$ $\therefore a^{3}-3 a^{2} b+3 a b^{2}-b^{3}+8$ $=(a-b+2)\left(a^{2}+b^{2}-2 a b-2 a+2 b+4\right)$...
Read More →Find the equation of the normal to curve
Question: Find the equation of the normal to curve $y^{2}=4 x$ at the point $(1,2)$. Solution: The equation of the given curve is $y^{2}=4 x$. Differentiating with respect tox, we have: $2 y \frac{d y}{d x}=4$ $\Rightarrow \frac{d y}{d x}=\frac{4}{2 y}=\frac{2}{y}$ $\left.\therefore \frac{d y}{d x}\right]_{(1,2)}=\frac{2}{2}=1$ Now, the slope of the normal at point $(1,2)$ is $\left.\frac{\frac{-1}{d y}}{d x}\right]_{(1,2)}=\frac{-1}{1}=-1$. Equation of the normal at (1, 2) isy 2 = 1(x 1). $\Rig...
Read More →Find the equation of the normal to curve
Question: Find the equation of the normal to curve $y^{2}=4 x$ at the point $(1,2)$. Solution: The equation of the given curve is $y^{2}=4 x$. Differentiating with respect tox, we have: $2 y \frac{d y}{d x}=4$ $\Rightarrow \frac{d y}{d x}=\frac{4}{2 y}=\frac{2}{y}$ $\left.\therefore \frac{d y}{d x}\right]_{(1,2)}=\frac{2}{2}=1$ Now, the slope of the normal at point $(1,2)$ is $\left.\frac{\frac{-1}{d y}}{d x}\right]_{(1,2)}=\frac{-1}{1}=-1$. Equation of the normal at (1, 2) isy 2 = 1(x 1). $\Rig...
Read More →Find the value
Question: Find the value. $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$ Solution: $=(2 x)^{3}+(3 y)^{3}+3 \times(2 x)^{2} \times 3 y+3 \times(2 x)(3 y)^{2}$ $=(2 x+3 y)^{3}$ $\left[\therefore a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$ $=(2 x+3 y)(2 x+3 y)(2 x+3 y)$ $\therefore 8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$ $=(2 x+3 y)(2 x+3 y)(2 x+3 y)$...
Read More →Find the value
Question: Find the value $\frac{8}{27} x^{3}+1+\frac{4}{3} x^{2}+2 x$ Solution: $=\left(\frac{2}{3} x^{3}\right)^{3}+1^{3}+3 \times\left(\frac{2}{3} x\right)^{2} \times 1+3(1)^{2} \times\left(\frac{2}{3} x\right)$ $=\left(\frac{2}{3} x+1\right)^{3}$ $\left[\therefore x^{3}+b^{3}+3 x^{2} b+3 x b^{2}=(x+b)^{3}\right]$ $=\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)$ $\therefore \frac{8}{27} x^{3}+1+\frac{4}{3} x^{2}+2 x$ $=\left(\frac{2}{3} x+1\right)\left(\f...
Read More →Solve the following systems of equations graphically:
Question: Solve the following systems of equations graphically: $x-2 y=6$ $3 x-6 y=0$ Solution: The given equations are: $x-2 y=6 \quad \ldots \ldots .(i)$ $3 x-6 y=0 \quad \ldots \ldots . .(i i)$ Putting $x=0$ in equation $(i)$, we get: $\Rightarrow 0-2 y=6$ $\Rightarrow y=-3$ $x=0, \quad y=-3$ Putting $y=0$ in equation $(i)$ we get: $\Rightarrow x-2 \times 0=6$ $\Rightarrow y=6$ $x=6, \quad y=0$ Use the following table to draw the graph. Plotting the two points $A(0,-3)$ and $B(6,0)$ equation ...
Read More →Find the value
Question: Find the value $125 x^{3}-27 y^{3}-225 x^{2} y+135 x y^{2}$ Solution: $=(5 x)^{3}-(3 y)^{3}-3(5 x)^{2}(3 y)+3(5 x)(3 y)^{2}$ $\left[\therefore a^{3}-b^{3}-3 a^{2} b+3 a b^{2}=(a-b)^{3}\right]$ $=(5 x-3 y)^{3}$ $=(5 x-3 y)(5 x-3 y)(5 x-3 y)$ $\therefore 125 x^{3}-27 y^{3}-225 x^{2} y+135 x y^{2}$ $=(5 x-3 y)(5 x-3 y)(5 x-3 y)$...
Read More →The two equal sides of an isosceles triangle with fixed base b are decreasing
Question: The two equal sides of an isosceles triangle with fixed basebare decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base? Solution: Let ΔABC be isosceles where BC is the base of fixed lengthb. Let the length of the two equal sides of ΔABC bea. Draw $A D \perp B C$. Now, in ΔADC, by applying the Pythagoras theorem, we have: $\mathrm{AD}=\sqrt{a^{2}-\frac{b^{2}}{4}}$ $\therefore$ Area of triangle $(A)=\frac{1}{2} b \sqrt{a...
Read More →If A is any set, prove that:
Question: If $A$ is any set, prove that: $A \subseteq \phi \Leftrightarrow A=\phi .$ Solution: To prove: $A \subseteq \phi \Leftrightarrow A=\phi$ Proof: Let: $A \subseteq \phi$ IfAis a subset of an empty set, thenAis the empty set. $\therefore A=\phi$ Now, let $A=\phi$ This means thatAis an empty set. We know that every set is a subset of itself. $\therefore A \subseteq \phi$ Thus, we have: $A \subseteq \phi \Leftrightarrow A=\phi$...
Read More →Find the value
Question: Find the value $64 a^{3}+125 b^{3}+240 a^{2} b+300 a b^{2}$ Solution: $=(4 a)^{3}+(5 b)^{3}+3(4 a)^{2}(5 b)+3(4 a)(5 b)^{2}$ $\left[\therefore a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$ $=(4 a+5 b)^{3}$ $=(4 a+5 b)(4 a+5 b)(4 a+5 b)$ $\therefore 64 a^{3}+125 b^{3}+240 a^{2} b+300 a b^{2}$ $=(4 a+5 b)(4 a+5 b)(4 a+5 b)$...
Read More →What is the total number of proper subsets of a set consisting of n elements?
Question: What is the total number of proper subsets of a set consisting ofnelements? Solution: We know that the total number of subsets of a finite set consisting ofnelements is 2n.Therefore, the total number of proper subsets of a set consisting ofnelements is 2n-1....
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