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)...
Read More →A sailor goes 8 km downstream in 40 minutes and returns in 1 hours.
Question: A sailor goes 8 km downstream in 40 minutes and returns in 1 hours. Determine the speed of the sailor in still water and the speed of the current. Solution: Let the speed of the sailor in still water bexkm/hr and the speed of the current beykm/hr Speed upstream $=(x-y) \mathrm{km} / \mathrm{hr}$ Speed downstream $=(x+y) \mathrm{km} / \mathrm{hr}$ Now, Time taken to cover $8 \mathrm{~km}$ down stream $=\frac{8}{x+y} h r s$ Time taken to cover $8 \mathrm{~km}$ upstream $=\frac{8}{x-y} h ...
Read More →In a ΔABC median AD is produced to x such that AD = DX. Prove that ABXC is a parallelogram.
Question: In a ΔABC median AD is produced to x such that AD = DX. Prove that ABXC is a parallelogram. Solution: In a quadrilateral ABXC, we have AD = DX [Given] BD = DC [Given] So, diagonals AX and BC bisect each other. Therefore, ABXC is a parallelogram....
Read More →Show that
Question: $\int_{0}^{\pi} \log (1+\cos x) d x$ Solution: Let $I=\int_{0}^{\pi} \log (1+\cos x) d x$ (1) $\Rightarrow I=\int_{1}^{\pi} \log (1+\cos (\pi-x)) d x$ $\Rightarrow I=\int_{0}^{\pi}\log (1-\cos x) d x$ ...(2) Adding (1) and (2), we obtain $2 I=\int_{0}^{\pi}\{\log (1+\cos x)+\log (1-\cos x)\} d x$ $\Rightarrow 2 I=\int_{0}^{\pi} \log \left(1-\cos ^{2} x\right) d x$ $\Rightarrow 2 I=\int_{0}^{\pi} \log \sin ^{2} x d x$ $\Rightarrow 2 I=2 \int_{0}^{\pi} \log \sin x d x$ $\Rightarrow I=\in...
Read More →In a triangle, P, Q and R are the mid points of sides BC, CA and AB respectively.
Question: In a triangle, P, Q and R are the mid points of sides BC, CA and AB respectively. If AC = 21cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ. Solution: InΔABC, R and P are mid points of AB and BC RP ∥ AC, RP = (1/2) AC [By Midpoint Theorem] In a quadrilateral, [A pair of side is parallel and equal] RP ∥ AQ, RP = AQ Therefore, RPQA is a parallelogram ⇒ AR = (1/2) AB = 1/230 = 15 cm AR = QP = 15 cm[Opposite sides are equal] ⇒ RP = (1/2) AC =1/2 21 = 10.5 cm RP ...
Read More →If f(x) =
Question: Iff(x) = 4xx2,x R, then write the value off(a+ 1) f(a 1). Solution: Given: f(x) = 4xx2,x R Now, $f(a+1)=4(a+1)-(a+1)^{2}$ $=4 a+4-\left(a^{2}+1+2 a\right)$ $=4 a+4-a^{2}-1-2 a$ $=2 a-a^{2}+3$ $f(a-1)=4(a-1)-(a-1)^{2}$ $=4 a-4-\left(a^{2}+1-2 a\right)$ $=4 a-4-a^{2}-1+2 a$ $=6 a-a^{2}-5$ Thus, $f(a+1)-f(a-1)=\left(2 a-a^{2}+3\right)-\left(6 a-a^{2}-5\right)$ $=2 a-a^{2}+3-6 a+a^{2}+5$ $=8-4 a$ $=4(2-a)$...
Read More →If f(x) =
Question: Iff(x) = 4xx2,x R, then write the value off(a+ 1) f(a 1). Solution: Given: f(x) = 4xx2,x R Now, $f(a+1)=4(a+1)-(a+1)^{2}$ $=4 a+4-\left(a^{2}+1+2 a\right)$ $=4 a+4-a^{2}-1-2 a$ $=2 a-a^{2}+3$ $f(a-1)=4(a-1)-(a-1)^{2}$ $=4 a-4-\left(a^{2}+1-2 a\right)$ $=4 a-4-a^{2}-1+2 a$ $=6 a-a^{2}-5$ Thus, $f(a+1)-f(a-1)=\left(2 a-a^{2}+3\right)-\left(6 a-a^{2}-5\right)$ $=2 a-a^{2}+3-6 a+a^{2}+5$ $=8-4 a$ $=4(2-a)$...
Read More →Point A and B are 70 km. a part on a highway.
Question: Point A and B are 70 km. a part on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if the travel towards each other, the meet in one hour. Find the speed of the two cars. Solution: We have to find the speed of car Let $X$ and $Y$ be two cars starting from points $A$ and $B$ respectively. Let the speed of car $X$ be $x \mathrm{~km} / \mathrm{hr}$ and that of car $Y$ be $y \mathrm{~km} / \mathrm{...
Read More →In a ΔABC, ∠A = 50°, ∠B = 60° and ∠C = 70°.
Question: In a ΔABC, A = 50, B = 60 and C = 70. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle. Solution: InΔABC, D and E are mid points of AB and BC. By Mid point theorem, DE ∥ AC, DE = (1/2) AC F is the midpoint of AC Then,DE = (1/2) AC = CF In a Quadrilateral DECF DE ∥ AC, DE = CF Hence DECF is a parallelogram. C = D = 70 [Opposite sides of a parallelogram] Similarly BEFD is a parallelogram,B = F = 60 ADEF is a parallelogram,A = ...
Read More →In ΔABC, D, E and F are, respectively the mid points of BC, CA and AB.
Question: In ΔABC, D, E and F are, respectively the mid points of BC, CA and AB. If the lengths of sides AB, BC and CA are 7cm, 8 cm and 9 cm, respectively, find the perimeter of ΔDEF. Solution: Given that, AB = 7 cm, BC = 8 cm, AC = 9 cm In ∆ABC, F and E are the mid points of AB and AC. EF = 1/2 BC Similarly DF = 1/2 AC and DE = 1/2 AB Perimeter of ∆DEF = DE + EF + DF = (1/2) AB + (1/2) BC + (1/2) AC = 1/2 7 + 1/2 8 + 1/2 9 = 3.5 + 4 + 4.5 = 1/2 cm Perimeter of ΔDEF = 1/2 cm...
Read More →Write the domain and range of the function f(x)
Question: Write the domain and range of the function $f(x)=\frac{x-2}{2-x}$. Solution: Given: $f(x)=\frac{x-2}{2-x}$ Domain (f) : Clearly,f(x) is defined for allxsatisfying: if 2-x 0 ⇒x 2. Hence, domain (f) =R-{2} Range off: Letf(x) =y $\Rightarrow \frac{x-2}{2-x}=y$ $\Rightarrow x-2=y(2-x)$ $\Rightarrow x-2=-y(x-2)$ $\Rightarrow y=-1$ Hence, range $(f)=\{-1\} .$...
Read More →Write the domain and range of the function f(x)
Question: Write the domain and range of the function $f(x)=\frac{x-2}{2-x}$. Solution: Given: $f(x)=\frac{x-2}{2-x}$ Domain (f) : Clearly,f(x) is defined for allxsatisfying: if 2-x 0 ⇒x 2. Hence, domain (f) =R-{2} Range off: Letf(x) =y $\Rightarrow \frac{x-2}{2-x}=y$ $\Rightarrow x-2=y(2-x)$ $\Rightarrow x-2=-y(x-2)$ $\Rightarrow y=-1$ Hence, range $(f)=\{-1\} .$...
Read More →If f(x)
Question: If $f(x)=1-\frac{1}{x}$, then write the value of $f\left(f\left(\frac{1}{x}\right)\right)$. Solution: Given: $f(x)=1-\frac{1}{x}$ Now, $f\left(\frac{1}{x}\right)=1-\frac{1}{\frac{1}{x}}=1-x$ $\Rightarrow f\left(f\left(\frac{1}{x}\right)\right)=f(1-x)$ Again, If $f(x)=1-\frac{1}{x}$ Thus, $f(1-x)=1-\frac{1}{1-x}$ $=\frac{1-x-1}{1-x}$ $=\frac{-x}{1-x}$ $=\frac{-x}{-(x-1)}$ $=\frac{x}{x-1}$...
Read More →If f(x)
Question: If $f(x)=1-\frac{1}{x}$, then write the value of $f\left(f\left(\frac{1}{x}\right)\right)$. Solution: Given: $f(x)=1-\frac{1}{x}$ Now, $f\left(\frac{1}{x}\right)=1-\frac{1}{\frac{1}{x}}=1-x$ $\Rightarrow f\left(f\left(\frac{1}{x}\right)\right)=f(1-x)$ Again, If $f(x)=1-\frac{1}{x}$ Thus, $f(1-x)=1-\frac{1}{1-x}$ $=\frac{1-x-1}{1-x}$ $=\frac{-x}{1-x}$ $=\frac{-x}{-(x-1)}$ $=\frac{x}{x-1}$...
Read More →ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F.
Question: ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC. Solution: Draw a parallelogram ABCD with AC and BD intersecting at O. Produce AD to E such that DE = DC Join EC and produce it to meet AB produced at F. In ΔDCE, DCE = DEC ... (i) [In a triangle, equal sides have equal angles] AB ∥ CD [Opposite sides of the parallelogram are parallel] AE ∥ CD [AB lies on AF] AF∥CDand EF is the Transversal. DCE = BFC ... (ii) [Pair of...
Read More →Point A and B are 70 km. a part on a highway. A car starts from A and another car starts from B simultaneously.
Question: Point A and B are 70 km. a part on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if the travel towards each other, the meet in one hour. Find the speed of the two cars. Solution: We have to find the speed of car Let $X$ and $Y$ be two cars starting from points $A$ and $B$ respectively. Let the speed of car $X$ be $x \mathrm{~km} / \mathrm{hr}$ and that of car $Y$ be $y \mathrm{~km} / \mathrm{...
Read More →ABCD is a rhombus, EAFB is a straight line such that EA = AB = BF.
Question: ABCD is a rhombus, EAFB is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles. Solution: We know that the diagonals of a rhombus are perpendicular bisector of each other. OA = OC, OB = OD, and AOD = COD = 90 AndAOB = COB = 90 InΔBDE, A and O are mid-points of BE and BD respectively. OA ∥ DE OC ∥ DG InΔCFA, B and O are mid-points of AF and AC respectively. OB ∥ CF OD ∥ GC Thus, in quadrilateral DOGC, we have OC ∥ DG and OD ∥ GC ⟹ DOCG is a pa...
Read More →Let f(x) =
Question: Let $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then write the value of a satisfying $f(f(x))=x$ for all $x \neq-1$. Solution: Given: $f(x)=\frac{\alpha x}{x+1}, x \neq-1$ Since $f(f(x))=x$ $\frac{\alpha\left(\frac{\alpha x}{x+1}\right)}{\frac{\alpha x}{x+1}+1}=x$ $\Rightarrow \frac{\alpha^{2} x}{\alpha x+x+1}=x$ $\Rightarrow \alpha^{2} x-\alpha x^{2}-\left(x^{2}+x\right)=0$ Solving | the quadratic lequation in $\alpha:$ $\alpha=\frac{x^{2} \pm \sqrt{x^{4}+4 x\left(x^{2}+x\right)}}{2 x}$ $\...
Read More →ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively,
Question: ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square. Solution: We have, AE = BF = CG = DH = x (say) BE = CF = DG = AH = y (say) In ΔAEH and ΔBEF, we have AE = BF A = B And AH = BE So, by SAS congruency criterion, we have ΔAEH ≃ ΔBFE ⇒ 1 = 2 and 3 = 4 But1 + 3 = 90 and 2 + A = 90 ⇒ 1 + 3 + 2 + A = 90 + 90 ⇒ 1 + 4 + 1 + 4 = 180 ⇒ 2(1 + 4) = 180 ⇒ 1 + 4 = 90 HEF = 90 Similarly we haveF = G = H = 90 Hence, E...
Read More →P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD.
Question: P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ. Solution: We know that, Diagonals of a parallelogram bisect each other. Therefore, OA = OC and OB = OD Since P and Q are point of intersection of BD. Therefore, BP = PQ = QD Now, OB = OD are BP = QD ⟹ OB - BP = OD - QD ⟹ OP = OQ Thus in quadrilateral APCQ, we have OA = OC and OP = OQ Diagonals of Quadrilateral APCQ bisect each other. Therefore...
Read More →Show that
Question: $\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$ Solution: Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$ ...(1) $\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x$ $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$ $\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x}...
Read More →Write the range of the function f(x) =
Question: Write the range of the functionf(x) =ex[x],x R. Solution: $f(x)=e^{x-[x]}, x \in \mathrm{R}$ We know that $x-[x]=\{x\}$, which is the fractional part of any number $x$. Thus, $f(x)=e^{\{x\}}$ Also, $0 \leq\{x\}1$ $\Rightarrow e^{0} \leq e^{\{x\}}e^{1}$ $\Rightarrow 1 \leq f(x)e$ Thus range of $f(x)$ is $[1, e)$....
Read More →Write the range of the function f(x) = cos [x],
Question: Write the range of the function $1(x)=\cos [x]$, where $\frac{-\pi}{2}x\frac{\pi}{2}$. Solution: Since $f(x)=\cos [x]$, where $\frac{-\pi}{2}x\frac{\pi}{2}$ $-\frac{\pi}{2}x\frac{\pi}{2}$ $\Rightarrow-1.57x1.57$ $\Rightarrow[x] \in\{-1,0,1,2\}$ Thus, $\cos [x]=\{\cos (-1), \cos 0, \cos 1, \cos 2\}$. Range of $f(x)=\{\cos 1,1, \cos 2\}$....
Read More →The sides AB and CD of a parallelogram ABCD are bisected at E and F.
Question: The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram. Solution: Since ABCD is a parallelogram AB ∥ DC and AB = DC ⇒ EB ∥ DF and (1/2) AB = (1/2) DC ⇒ EB ∥ DF and EB = DF EBFD is a parallelogram....
Read More →Show that
Question: $\int_{0}^{2 \pi} \cos ^{5} x d x$ Solution: Let $I=\int_{0}^{2 \pi} \cos ^{5} x d x$ ...(1) $\cos ^{5}(2 \pi-x)=\cos ^{5} x$ It is known that, $\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x$, if $f(2 a-x)=f(x)$ $=0$ if $f(2 a-x)=-f(x)$ $\therefore I=2 \int_{0}^{\pi} \cos ^{5} x d x$ $\Rightarrow I=2(0)=0$ $\left[\cos ^{5}(\pi-x)=-\cos ^{5} x\right]$...
Read More →