Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid points of the, sides BC, CA and AB
Question: Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid points of the, sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles. Solution: Since D, E and F are mid-points of sides BC, CA and AB respectively. AB ∥ DE and AC ∥ DF AF ∥ DE and AE ∥ DF ABDE is a parallelogram. AF = DE and AE = DF (1/2) AB = DE and (1/2) AC = DF DE = DF [Since, AB = AC] AE = AF = DE = DF ABDF is a rhombus. ⟹ AD and FE bisect each other at right angl...
Read More →Define a function as a set of ordered pairs.
Question: Define a function as a set of ordered pairs. Solution: Afunctionis a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components. Sometimes we say that a function is a rule (correspondence) that assigns to each element of one set,X,only one element of another set,Y. The elements of set Xare often called inputs and the elements of setYare called outputs. The domain of a function is the set of all first components,x, in t...
Read More →Let f and g be two real functions given by
Question: Letfandgbe two real functions given by f= {(0, 1), (2, 0), (3, 4), (4, 2), (5, 1)} andg= {(1, 0), (2, 2), (3, 1), (4, 4), (5, 3)} Find the domain offg. Solution: It is given thatfandgare two real functions such that f= {(0, 1), (2, 0), (3, 4), (4, 2), (5, 1)} andg = {(1, 0), (2, 2), (3, 1), (4, 4), (5, 3)} Now, Domain off=Df= {0, 2, 3, 4, 5} Domain ofg = Dg= {1, 2, 3, 4, 5} Domain offg=DfDg= {2, 3, 4, 5}...
Read More →ABCD is a kite having AB = AD and BC = CD.
Question: ABCD is a kite having AB = AD and BC = CD. Prove that the figure found by joining the mid points of the sides, in order, is a rectangle. Solution: Given, A kite ABCD having AB = AD and BC = CD. P, Q, R, S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined. To prove: PQRS is a rectangle. Proof: In ∆ABC, P and Q are the mid-points of AB and BC respectively. PQ ∥ AC and PQ = (1/2) AC .... (i) In ∆ADC, R and S are the mid-points of CD and AD respective...
Read More →A boat goes 24 km upstream and 28 km downstream in 6 hrs.
Question: A boat goes $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream in $6 \mathrm{hrs}$. It goes $30 \mathrm{~km}$ upstream and $21 \mathrm{~km}$ downstream in $6 \frac{1}{2}$ hrs. Find the speed of the boat in still water and also speed of the stream. Solution: We have to find the speed of the boat in still water and speed of the stream Let the speed of the boat in still water bexkm/hr and the speed of the stream beykm/hr then Speed upstream $=(x-y) \mathrm{km} / \mathrm{hr}$ Sped...
Read More →Find the set of values of x for which the functions f(x)
Question: Find the set of values of xfor which the functionsf(x) = 3x2 1 andg(x) = 3 +xare equal. Solution: It is given that the functionsf(x) = 3x2 1 andg(x) = 3 +xare equal. $\therefore f(x)=g(x)$ $\Rightarrow 3 x^{2}-1=3+x$ $\Rightarrow 3 x^{2}-x-4=0$ $\Rightarrow(x+1)(3 x-4)=0$ $\Rightarrow x+1=0$ or $3 x-4=0$ $\Rightarrow x=-1$ or $x=\frac{4}{3}$ Hence, the set of values of $x$ for which the given functions are equal is $\left\{-1, \frac{4}{3}\right\}$....
Read More →Find the set of values of x for which the functions f(x)
Question: Find the set of values of xfor which the functionsf(x) = 3x2 1 andg(x) = 3 +xare equal. Solution: It is given that the functionsf(x) = 3x2 1 andg(x) = 3 +xare equal. $\therefore f(x)=g(x)$ $\Rightarrow 3 x^{2}-1=3+x$ $\Rightarrow 3 x^{2}-x-4=0$ $\Rightarrow(x+1)(3 x-4)=0$ $\Rightarrow x+1=0$ or $3 x-4=0$ $\Rightarrow x=-1$ or $x=\frac{4}{3}$ Hence, the set of values of $x$ for which the given functions are equal is $\left\{-1, \frac{4}{3}\right\}$....
Read More →In figure, AB = AC and CP ∥ BA and AP is the bisector of exterior ∠CAD of ∆ABC.
Question: In figure, AB = AC and CP ∥ BA and AP is the bisector of exteriorCADof ∆ABC. Prove that (i)PAC = BCA. (ii) ABCP is a parallelogram. Solution: Given, AB = AC andCD ∥ BAand AP is the bisector of exteriorCAD of ΔABC To prove: (i)PAC = BCA (ii) ABCP is a parallelogram. Proof: (i) We have, AB = AC ⟹ ACB = ABC [Opposite angles of equal sides of triangle are equal] Now,CAD = ABC + ACB ⟹ PAC + PAD = 2ACB [ PAC = PAD] ⟹ 2PAC = 2ACB ⟹ PAC = ACB (ii) Now, PAC = BCA ⟹ AP ∥ BCandCP ∥ BA [Given] The...
Read More →Show that
Question: Show that $\int_{0}^{u} f(x) g(x) d x=2 \int_{0}^{n} f(x) d x$, if $f$ and $g$ are defined as $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$ Solution: Let $I=\int_{0}^{a} f(x) g(x) d x$ ...(1) $\Rightarrow I=\int_{0}^{a} f(a-x) g(a-x) d x$ $\left(\int_{0}^{\infty} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$ $\Rightarrow I=\int_{0}^{a} f(x) g(a-x) d x$ ...(2) Adding (1) and (2), we obtain $2 I=\int_{0}^{a}\{f(x) g(x)+f(x) g(a-x)\} d x$ $\Rightarrow 2 I=\int_{0}^{a} f(x)\{g(x)+g(a-x)\} d x$ $\Rightarrow...
Read More →and g be two functions given by
Question: Letfandgbe two functions given by f= {(2, 4), (5, 6), (8, 1), (10, 3)} andg= {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)}. Find the domain off + g. Solution: It is given thatfandg are two functions such that f = {(2, 4), (5, 6), (8, 1), (10, 3)} andg = {(2, 5), (7, 1), (8, 4), (10, 13), (11,5)} Now, Domain off=Df= {2, 5, 8, 10} Domain ofg = Dg= {2, 7, 8, 10, 11} Domain off+g=DfDg= {2, 8, 10}...
Read More →In figure, BE⊥ AC, AD is any line from A to BC intersecting BE in H. P, Q
Question: In figure, BE AC, AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that PQR = 90 Solution: Given, BE ACand P, Q and R are respectively mid-point of AH, AB and BC. To prove:PQR = 90 Proof: InΔABC, Q and R are mid-points of AB and BC respectively. QR ∥ AC ..... (i) InΔABH, Q and P are the mid-points of AB and AH respectively QP ∥ BH ..... (ii) But,BEAC Therefore, from equation (i) and equation (ii) we have, QPQR ⟹ PQR = ...
Read More →Let A and B be two sets such that n(A) = p and n(B) = q,
Question: LetAandBbe two sets such thatn(A) =pandn(B) =q, write the number of functions fromAtoB. Solution: It is given thatAandBare two sets such thatn(A) =pandn(B) =q. Now, any element of setA, sayai(1 ip), is related with an element of setBinqways. Similarly, other elements ofsetAare related with an element of setBinqways. Thus, every element of setAis related with every element of setBinqways. Total number of functions fromAtoB=qqq ... q(ptimes) =qp...
Read More →ABC is a triangle and through A, B, C lines are drawn parallel to BC,
Question: ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of ΔABC. Solution: Clearly ABCQ and ARBC are parallelograms. Therefore, BC = AQ and BC = AR ⟹ AQ = AR ⟹ A is the mid-point of QR Similarly B and C are the mid points of PR and PQ respectively. AB = (1/2) PQ, BC = (1/2) QR, CA = (1/2) PR ⟹ PQ = 2AB, QR = 2BC and PR = 2CA ⟹ PQ + QR + RP = 2 (AB + BC + CA) ⟹ Perim...
Read More →Write the domain and range of function f(x) given by f(x)
Question: Write the domain and range of function $f(x)$ given by $f(x)=\sqrt{[x]-x}$. Solution: $f(x)=\sqrt{[x]-x}$ We know that $[x]-x=-\{x\}$, which is the fractional part of any real number $x .$ Thus, $f(x)=\sqrt{-\{x\}}$. Since $\{x\}$ is always a positive number, $f(x)$ is not defined for any $\mathrm{x}$. Or $\operatorname{dom}(f)=\varphi$ Thus, range $(f)=\varphi$....
Read More →Show that
Question: $\int_{0}^{4}|x-1| d x$ Solution: $I=\int_{0}^{4}|x-1| d x$ It can be seen that, $(x-1) \leq 0$ when $0 \leq x \leq 1$ and $(x-1) \geq 0$ when $1 \leq x \leq 4$ $I=\int_{1}^{1}|x-1| d x+\int_{1}^{4}|x-1| d x$ $\left(\int_{a}^{b} f(x)=\int_{a}^{c} f(x)+\int_{c}^{b} f(x)\right)$ $=\int_{0}^{1}-(x-1) d x+\int_{0}^{4}(x-1) d x$ $=\left[x-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{4}$ $=1-\frac{1}{2}+\frac{(4)^{2}}{2}-4-\frac{1}{2}+1$ $=1-\frac{1}{2}+8-4-\frac{1}{2}+...
Read More →In figure, M, N and P are mid-points of AB, AC and BC respectively.
Question: In figure, M, N and P are mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC. Solution: Given MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm. To find BC, AB and AC In ΔABC M and N are mid-points of AB and AC MN = (1/2) BC, MN ∥ BC [By mid-point theorem] ⇒ 3 = (1/2) BC ⇒ 32 = BC ⇒ BC = 6 cm Similarly AC = 2MP = 2 (2.5) = 5 cm AB = 2 NP = 2 (3.5) = 7 cm...
Read More →The boat goes 30 km upstream and 44 km downstream in 10 hours.
Question: The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boat in still water. Solution: Let the speed of the boat in still water bexkm/hr and the speed of the stream beykm/hr Speed upstream $=(x-y) \mathrm{km} / \mathrm{hr}$ Speed down stream $=(x+y) \mathrm{km} / \mathrm{hr}$ Now, Time taken to cover $30 \mathrm{~km}$ upstream $=\frac{30}{x-y}$ hrs Time taken to cover $44 \m...
Read More →Write the domain and range of f(x)
Question: Write the domain and range of $f(x)=\sqrt{x-[x]}$. Solution: $f(x)=\sqrt{x-[x]}$ Since $f(x)$ is defined for all values of $\mathrm{x}, x \in R$. Or $\operatorname{dom}(f(x))=R$ Since, $x-[x]=\{x\}$, which is the fractional part of any real number $x$, $f(x)=\sqrt{\{x\}} \ldots \ldots(1)$ We know that $0 \leq\{x\}1$ $\Rightarrow \sqrt{0} \leq \sqrt{\{x\}}\sqrt{1}$ $\Rightarrow 0 \leq f(x)1 \quad\{$ from $(1)\}$ Thus, range of $f(x)$ is $[0,1)$....
Read More →In figure, Triangle ABC is a right angled triangle at B.
Question: In figure, Triangle ABC is a right angled triangle at B. Given that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate (i) The length of BC (ii) The area of ΔADE. Solution: InΔABC, B = 900, By using Pythagoras theorem $A C^{2}=A B^{2}+B C^{2}$ $\Rightarrow 15^{2}=9^{2}+B C^{2}$ $\Rightarrow B C=\sqrt{15^{2}-9^{2}}$ $\Rightarrow \mathrm{BC}=\sqrt{225-81}$ $\Rightarrow \mathrm{BC}=\sqrt{144}=12 \mathrm{~cm}$ InΔABC, D and E are mid-points of ...
Read More →In a ΔABC, BM and CN are perpendiculars from B and C respectively on any line passing through A.
Question: In a ΔABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML = NL. Solution: Given that, InΔBLM and ΔCLN BML = CNL = 90 BL = CL [L is the mid-point of BC] MLB = NLC [Vertically opposite angle] ΔBLM = ΔCLN LM = LN [corresponding parts of congruent triangles]...
Read More →Write the domain and range of function f(x) given by f(x)
Question: Write the domain and range of function $f(x)$ given by $f(x)=\frac{1}{\sqrt{x-|x|}}$. Solution: Given: f(x)=\frac{1}{\sqrt{x-|x|}} We know that $|x|= \begin{cases}x, \text { if } x \geq 0 \\ -x, \text { if } x0\end{cases}$ $\Rightarrow x-|x|= \begin{cases}x-x=0, \text { if } \mathrm{x} \geq 0 \\ x+x=2 x, \text { if } \mathrm{x}0\end{cases}$ $\Rightarrow x-|x| \leq 0$ for all $x$. $\Rightarrow \frac{1}{\sqrt{x-|x|}}$ does not take any real values for any $x \in \mathrm{R}$. ⇒f(x) is not...
Read More →Write the domain and range of function f(x) given by f(x)
Question: Write the domain and range of function $f(x)$ given by $f(x)=\frac{1}{\sqrt{x-|x|}}$. Solution: Given: f(x)=\frac{1}{\sqrt{x-|x|}} We know that $|x|= \begin{cases}x, \text { if } x \geq 0 \\ -x, \text { if } x0\end{cases}$ $\Rightarrow x-|x|= \begin{cases}x-x=0, \text { if } \mathrm{x} \geq 0 \\ x+x=2 x, \text { if } \mathrm{x}0\end{cases}$ $\Rightarrow x-|x| \leq 0$ for all $x$. $\Rightarrow \frac{1}{\sqrt{x-|x|}}$ does not take any real values for any $x \in \mathrm{R}$. ⇒f(x) is not...
Read More →Show that
Question: $\int_{0}^{x} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$ Solution: Let $I=\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$ ...(1) It is known that, $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$ $I=\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x$ ...(2) Adding (1) and (2), we obtain $2 I=\int_{0}^{a} \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x$ $\Rightarrow 2 I=\int_{0}^{a} 1 d x$ $\Rightarrow 2 I=[x]_{0}^{a}$ $\Rightarrow 2 I=a$ $\Rightarrow I=\frac{...
Read More →In a ΔABC, E and F are the mid-points of AC and AB respectively.
Question: In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP. Solution: In aΔABC E and F are mid points of AB and AC EF ∥ FE, (1/2) BC = FE [By midpoint theorem] InΔABP F is the mid-point of AB and FQ ∥ BP [ EF ∥ BP] Therefore, Q is the mid-point of AP [By mid-point theorem] Hence, AQ = QP....
Read More →If f, g, h are real functions given by f(x) =
Question: If $f, g, h$ are real functions given by $f(x)=x^{2}, g(x)=\tan x$ and $h(x)=\log _{e} x$, then write the value of (hogof) $\left(\sqrt{\frac{\pi}{4}}\right)$. Solution: Given :f(x) =x2,g(x) = tanxandh(x) = logex. (hogof) $\left(\sqrt{\frac{\pi}{4}}\right)=h\left(g\left(f\left(\sqrt{\frac{\pi}{4}}\right)\right)\right)$ $=h\left(g\left(\left(\sqrt{\frac{\pi}{4}}\right)^{2}\right)\right)$ $=h\left(g\left(\frac{\pi}{4}\right)\right)$ $=h\left(\tan \left(\frac{\pi}{4}\right)\right)$ $=h(1)...
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