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Question: $\frac{3 x-1}{(x-1)(x-2)(x-3)}$ Solution: Let $\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$ $3 x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$ ...(1) Substitutingx= 1, 2, and 3 respectively in equation (1), we obtain A= 1,B= 5, andC= 4 $\therefore \frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}$ $\Rightarrow \int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int\left\{\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\right\} d x$ $=\log |x-1|...

If R is a relation on a finite set having n elements,

Question: If R is a relation on a finite set havingnelements, then the number of relations on A is (a) $2^{n}$ (b) $2^{n^{2}}$ (c) $n^{2}$ (d) $n^{n}$ Solution: (b) $2^{n^{2}}$ Given : A finite set withnelements Its Cartesian product with itself will have n2elements. $\therefore$ Number of relations on $\mathrm{A}=2^{n^{2}}$...

Question: If R is a relation on a finite set havingnelements, then the number of relations on A is (a) $2^{n}$ (b) $2^{n^{2}}$ (c) $n^{2}$ (d) $n^{n}$ Solution: (b) $2^{n^{2}}$ Given : A finite set withnelements Its Cartesian product with itself will have n2elements. $\therefore$ Number of relations on $\mathrm{A}=2^{n^{2}}$...

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Question: If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled. Solution: Given each angle of a triangle less than the sum of the other two X + Y + Z ⇒ X + X X + Y + Z ⇒ 2X 180 [Sum of all the angles of a triangle] ⇒ X 90 SimilarlyY 90 and Z 90 Hence, the triangles are acute angled....

Solve each of the following systems of equations by the method of cross-multiplication :

Question: Solve each of the following systems of equations by the method of cross-multiplication : $5 a x+6 b y=28$ $3 a x+4 b y=18$ Solution: GIVEN: $5 a x+6 b y=28$ $3 a x+4 b y=18$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $5 a x+6 b y-28=0$ $3 a x+4 b y-18=0$ By cross multiplication method we get $\frac{x}{(-18 \times 6 b)-(4 b \times(-28))}=\frac{-y}{(5 a) \times(-18)-((3 a) \times-(28))}=\frac{1}{2...

Can a triangle have:

Question: Can a triangle have: (i) Two right angles? (ii) Two obtuse angles? (iii) Two acute angles? (iv) All angles more than 60? (v) All angles less than 60? (vi) All angles equal to 60? Justify your answer in each case. Solution: (i) No, Two right angles would up to 180. So the third angle becomes zero. This is not possible, so a triangle cannot have two right angles. [Since sum of angles in a triangle is 180] (ii) No, A triangle can't have 2 obtuse angles. Obtuse angle means more than 90 So ...

Show that

Question: $\frac{1}{x^{2}-9}$ Solution: Let $\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}$ $1=A(x-3)+B(x+3)$ Equating the coefficients ofxand constant term, we obtain $A+B=0$ $-3 A+3 B=1$ On solving, we obtain $A=-\frac{1}{6}$ and $B=\frac{1}{6}$ $\therefore \frac{1}{(x+3)(x-3)}=\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}$ $\Rightarrow \int \frac{1}{\left(x^{2}-9\right)} d x=\int\left(\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\right) d x$ $=-\frac{1}{6} \log |x+3|+\frac{1}{6} \log |x-3|+\mathrm{C}$ $=\fr...

If R is a relation from a finite set A having m elements of a finite set B having n elements,

Question: If R is a relation from a finite set A havingmelements of a finite set B havingnelements, then the number of relations from A to B is (a) 2mn (b) 2mn 1 (c) 2mn (d)mn Solution: (a) 2mnGiven: n(A) = m n(B) = n $\therefore n(A \times B)=m n$ Then, the number of relations from $A$ to $B$ is $2^{m n}$....

In a ΔABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that

Question: In a ΔABC, ABC = ACB and the bisectors of ABC and ACB intersect at O such that BOC = 120. Show that A = B = C = 60. Solution: Given, In ΔABC, ABC = ACB Dividing both sides by '2' $\frac{1}{2} \angle \mathrm{ABC}=\frac{1}{2} \angle \mathrm{ACB}$ ⇒ OBC = OCB [ OB, OC bisects B and C] Now, $\angle \mathrm{BOC}=90^{\circ}+\frac{1}{2} \angle \mathrm{A}$ $\Rightarrow 120^{\circ}-90^{\circ}=\frac{1}{2} \angle \mathrm{A}$ ⇒ 30 (2) = A ⇒ A = 60 Now inΔABC A + ABC + ACB = 180 (Sum of all angles ...

Let R be a relation from a set A to a set B, then

Question: Let R be a relation from a set A to a set B, then (a) R = A B (b) R = A B (c) R A B (d) R B A Solution: (c) R A BIf R is a relation from set A to set B, then R is always a subset of A B....

Solve each of the following systems of equations by the method of cross-multiplication :

Question: Solve each of the following systems of equations by the method of cross-multiplication : $2 a x+3 b y=a+2 b$ $3 a x+2 b y=2 a+b$ Solution: GIVEN: $2 a x+3 b y=a+2 b$ $3 a x+2 b y=2 a+b$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $2 a x+3 b y-(a+2 b)=0$ $3 a x+2 b y-(2 a+b)=0$ By cross multiplication method we get $\frac{x}{(-(2 a+b) \times 3 b)-(2 b \times(-(a+2 b)))}=\frac{-y}{(2 a) \times(-(2 ...

If the set A has p elements, B has q elements,

Question: If the set A haspelements, B hasqelements, then the number of elements in A B is(a)p+q (b)p+q+ 1 (c)pq (d)p2 Solution: (c)pq $n(A \times B)=n(A) \times n(B)$ $=p \times q=p q$...

If the bisectors of the base angles of a triangle enclose an angle of 135°, prove that the triangle is a right angle.

Question: If the bisectors of the base angles of a triangle enclose an angle of 135, prove that the triangle is a right angle. Solution: Given the bisectors of the base angles of a triangle enclose an angle of $135^{\circ}$ i.e., $\angle B O C=135^{\circ}$ But, We know that $\angle \mathrm{BOC}=90^{\circ}+\frac{1}{2} \angle \mathrm{A}$ $\Rightarrow 135^{\circ}=90^{\circ}+\frac{1}{2} \angle \mathrm{A}$ $\Rightarrow \frac{1}{2} \angle \mathrm{A}=135^{\circ}-90^{\circ}$ ⇒ A = 45(2) ⇒ A = 90 Therefo...

Show that

Question: $\frac{x}{(x+1)(x+2)}$ Solution: Let $\frac{x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$ $\Rightarrow x=A(x+2)+B(x+1)$ Equating the coefficients ofxand constant term, we obtain $A+B=1$ $2 A+B=0$ On solving, we obtain $A=-1$ and $B=2$ $\therefore \frac{x}{(x+1)(x+2)}=\frac{-1}{(x+1)}+\frac{2}{(x+2)}$ $\Rightarrow \int \frac{x}{(x+1)(x+2)} d x=\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} d x$ $=-\log |x+1|+2 \log |x+2|+C$ $=\log (x+2)^{2}-\log |x+1|+C$ $=\log \frac{(x+2)^{2}}{(x+1)}+C$...

R is a relation from [11, 12, 13]

Question: R is a relation from [11, 12, 13] to [8, 10, 12] defined byy=x 3. Then, R1is(a) [(8, 11), (10, 13)] (b) [(11, 8), (13, 10)] (c) [(10, 13), (8, 11), (12, 10)] (d) none of these Solution: (a) [(8, 11), (10, 13)]R is a relation from [11, 12, 13] to [8, 10, 12], defined byy=x 3 Now, we have: $11-3=8$ $13-3=10$ So, R = {(13,10),(11,8)} R1= {(10,13),(8,11)}...

The bisectors of base angles of a triangle cannot enclose a right angle in any case.

Question: The bisectors of base angles of a triangle cannot enclose a right angle in any case. Solution: InΔXYZ, Sum of all angles of a triangle is180 i.e., X + Y + Z = 180 Dividing both sides by '2' $\Rightarrow \frac{1}{2} \angle \mathrm{X}+\frac{1}{2} \angle \mathrm{Y}+\frac{1}{2} \angle \mathrm{Z}=180^{\circ}$ $\Rightarrow \frac{1}{2} \angle \mathrm{X}+\angle \mathrm{OYZ}+\angle \mathrm{OYZ}=90^{\circ}[\because \mathrm{OY}, \mathrm{OZ}, \angle \mathrm{Y}$ and $\angle \mathrm{Z}]$ $\Rightarro...

Solve each of the following systems of equations by the method of cross-multiplication :

Question: Solve each of the following systems of equations by the method of cross-multiplication : $a x+b y=\frac{a+b}{2}$ $3 x+5 y=4$ Solution: GIVEN: $a x+b y=\frac{a+b}{2}$ $3 x+5 y=4$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $a x+b y-\frac{a+b}{2}=0$ $3 x+5 y-4=0$ By cross multiplication method we get $\frac{x}{(-4 b)-\left(-\frac{5(a+b)}{2}\right)}=\frac{-y}{(-4 a)-\left(-\frac{3(a+b)}{2}\right)}=\...

Show that

Question: $\int \frac{d x}{\sqrt{9 x-4 x^{2}}}$ equals A. $\frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+\mathrm{C}$ B. $\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+\mathrm{C}$ C. $\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+\mathrm{C}$ D. $\frac{1}{2} \sin ^{-1}\left(\frac{9 x-8}{9}\right)+\mathrm{C}$ Solution: $\int \frac{d x}{\sqrt{9 x-4 x^{2}}}$ $=\int \frac{1}{\sqrt{-4\left(x^{2}-\frac{9}{4} x\right)}} d x$ $=\int \frac{1}{-4\left(x^{2}-\frac{9}{4} x+\frac{81}{64}-\frac{8...

Let R be a relation on N defined by x + 2y = 8.

Question: Let R be a relation on N defined byx+ 2y= 8. The domain of R is(a) [2, 4, 8] (b) [2, 4, 6, 8] (c) [2, 4, 6] (d) [1, 2, 3, 4] Solution: (c) {2, 4, 6} x+ 2y= 8 ⇒x= 8-2y Fory= 1,x= 6 y= 2,x= 4 y= 3,x= 2 Then R = {(2,3),(4,2),(6,1)} Domain of R = {2,4,6}...

ABC is a triangle in which ∠A = 720°, the internal bisectors of angles B and C meet in O.

Question: ABC is a triangle in which A = 720, the internal bisectors of angles B and C meet in O. Find the magnitude of BOC. Solution: Given, ABC is a triangle whereA = 72and the internal bisector of angles B and C meeting O. InΔABC, A + B + C = 180 ⇒ 72 + B + C = 180 ⇒ B + C = 180 72 Dividing both sides by '2' ⇒ B/2 + C/2 = 108/2 ⇒ OBC + OCB = 54 Now,In ΔBOC ⇒ OBC + OCB + BOC = 180 ⇒ 540 + BOC = 180 ⇒ BOC = 180 54=126 BOC = 126...

If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right angle triangle.

Question: If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right angle triangle. Solution: If one angle of a triangle is equal to the sum of the other two angles ⇒ B = A + C InΔABC, Sum of all angles of a triangle is180 ⇒ A + B + C = 180 ⇒ B + B = 180[ B = A + C] ⇒ 2B = 180 ⇒ B = 180/2 ⇒ B = 90 Therefore, ABC is a right angled triangle....

Solve each of the following systems of equations by the method of cross-multiplication :

Question: Solve each of the following systems of equations by the method of cross-multiplication : $\frac{2}{x}+\frac{3}{y}=13$ $\frac{5}{x}-\frac{4}{y}=-2$ where $x \neq 0$ and $y \neq 0$ Solution: GIVEN: $\frac{2}{x}+\frac{3}{y}=13$ $\frac{5}{x}-\frac{4}{y}=-2$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation Rewriting the equation again $\frac{2}{x}+\frac{3}{y}-13=0$ $\frac{5}{x}-\frac{4}{y}+2=0$ Taking $u=\...

Show that the function

Question: $\int \frac{d x}{x^{2}+2 x+2}$ equals A. $x \tan ^{-1}(x+1)+C$ B. $\tan ^{-1}(x+1)+C$ C. $(x+1) \tan ^{-1} x+\mathrm{C}$ D. $\tan ^{-1} x+\mathrm{C}$ Solution: $\int \frac{d x}{x^{2}+2 x+2}=\int \frac{d x}{\left(x^{2}+2 x+1\right)+1}$ $=\int \frac{1}{(x+1)^{2}+(1)^{2}} d x$ $=\left[\tan ^{-1}(x+1)\right]+\mathrm{C}$ Hence, the correct answer is B....

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°.

Question: Two angles of a triangle are equal and the third angle is greater than each of those angles by30. Determine all the angles of the triangle. Solution: Given that, Two angles of a triangle are equal and the third angle is greater than each of those angles by30. Let x, x, x + 30be the angles of a triangle We know that, Sum of all angles in a triangle is180 x + x + x +30=180 3x +30=180 3x =180 30 3x =150 x =50 Therefore, the three angles are 50, 50, 80....

A relation

Question: A relation $\phi$ from $\mathrm{C}$ to $\mathrm{R}$ is defined by $x \phi y \Leftrightarrow|x|=y$. Which one is correct? (a) $(2+3 i) \phi 13$ (b) $3 \phi(-3)$ (c) $(1+i) \phi 2$ (d) $i \phi 1$ Solution: (d) $i \phi 1$ We have $|i|=\sqrt{1^{2}+0^{2}}=1$ Thus, $i \phi 1$ satisfies $x \phi y \Leftrightarrow|x|=y .$...