Show that
Question: $e^{3 \log x}\left(x^{4}+1\right)^{-1}$ Solution: $e^{3 \log x}\left(x^{4}+1\right)^{-1}=e^{\log x^{3}}\left(x^{4}+1\right)^{-1}=\frac{x^{3}}{\left(x^{4}+1\right)}$ Let $x^{4}+1=t \Rightarrow 4 x^{3} d x=d t$ $\Rightarrow \int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x=\int \frac{x^{3}}{\left(x^{4}+1\right)} d x$ $=\frac{1}{4} \int \frac{d t}{t}$ $=\frac{1}{4} \log |t|+\mathrm{C}$ $=\frac{1}{4} \log \left|x^{4}+1\right|+\mathrm{C}$ $=\frac{1}{4} \log \left(x^{4}+1\right)+\mathrm{C}$...
Read More →If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area.
Question: If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of the median AD, prove that ar(ΔBGC) = 2ar(ΔAGC). Solution: Draw AMBC Since, AD is the median ofΔABC BD = DC ⇒ BD = AM = DC AM ⇒ (1/2)(BD AM) = (1/2)(DC AM) ⇒ ar(ΔABD) = ar(ΔACD) (1) InΔBGC, GD is the median ⇒ar(ΔBGD) = ar(ΔCGD) (2) InΔACD, CG is the median ⇒ar(ΔAGC) = ar(ΔCGD) (3) From (2) and (3) we have, ar(ΔBGD) = ar(ΔAGC) But,ar(ΔBGC) = 2ar(ΔBGD) ⇒ ar(ΔBGC) = 2ar(ΔA...
Read More →The difference between the two acute angles of a right-angled triangle is
Question: The difference between the two acute angles of a right-angled triangle is $\frac{2 \pi}{5}$ radians. Express the angles in degrees. Solution: Given: Difference between two acute angles of a right-angled triangle $=\frac{2 \pi}{5} \mathrm{rad}$ $\because 1 \mathrm{rad}=\left(\frac{180}{\pi}\right)^{\circ}$ $\therefore \frac{2 \pi}{5} \mathrm{rad}=\left(\frac{180}{\pi} \times \frac{2 \pi}{5}\right)^{\circ}$ $=(36 \times 2)^{\circ}$ $=72^{\circ}$...
Read More →Show that
Question: $\cos ^{3} x e^{\log \sin x}$ Solution: $\cos ^{3} x e^{\log \sin x}=\cos ^{3} x \times \sin x$ Let $\cos x=t \Rightarrow-\sin x d x=d t$ $\Rightarrow \int \cos ^{3} x e^{\log \sin x} d x=\int \cos ^{3} x \sin x d x$ $=-\int t \cdot d t$ $=-\frac{t^{4}}{4}+C$ $=-\frac{\cos ^{4} x}{4}+C$...
Read More →If P is any point in the interior of a parallelogram ABCD,
Question: If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram. Solution: Draw DNAB and PMAB Now, ar (∥gmABCD) = AB DN, ar(ΔAPB) = (1/2) (AB PM) Now, PM DN ⇒ AB PM AB DN ⇒ (1/2)(AB PM) (1/2)(AB DN) ⇒ ar(ΔAPB) 1/2 ar(∥gmABCD)...
Read More →In figure, ABC and ABD are two triangles on the base AB.
Question: In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD). Solution: Given that CD is bisected by AB at O To prove:ar(ΔABC) = ar(ΔABD). Construction: Draw CPAB and DQAB. Proof: ar(ΔABC) = 1/2 AB CP (1) ar(ΔABD) = 1/2 AB DQ (2) InΔCPO and ΔDQO CPO = DQO [Each 90] Given that, CO = OD COP = DOQ [Vertically opposite angles are equal] Then,ΔCPO ΔDQO [By AAS condition] CP = DQ (3) [C.P.C.T] Compare equation (1), (2) and ...
Read More →Find the radian measure corresponding to the following degree measures:
Question: Find the radian measure corresponding to the following degree measures: (i) 300 (ii) 35 (iii) 56 (iv) 135 (v) 300 (vi) 7 30' (vii) 125 30' (viii) 47 30' Solution: We have : $180^{\circ}=\pi \mathrm{rad}$ $\therefore 1^{\circ}=\frac{\pi}{180} \mathrm{rad}$ (i) $300^{\circ}$ $=\left(300 \times \frac{\pi}{180}\right)$ $=\frac{5 \pi}{3} \mathrm{rad}$ (ii) $35^{\circ}$ $=35 \times \frac{\pi}{180}$ $=\frac{7 \pi}{36} \mathrm{rad}$ (iii) $-56^{\circ}$ $=-\left(56 \times \frac{\pi}{180}\right)...
Read More →Show that
Question: $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$ Solution: $\therefore \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$ $\Rightarrow 1=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right)$ $\Rightarrow 1=A x^{3}+4 A x+B x^{2}+4 B+C x^{3}+C x+D x^{2}+D$ Equating the coefficients of $x^{3}, x^{2}, x$, and constant term, we obtain $A+C=0$ $B+D=0$ $4 A+C=0$ $4 B+D=1$ On solving these equations, we obtain $A=0, ...
Read More →Find the radian measure corresponding to the following degree measures:
Question: Find the radian measure corresponding to the following degree measures: (i) 300 (ii) 35 (iii) 56 (iv) 135 (v) 300 (vi) 7 30' (vii) 125 30' (viii) 47 30' Solution: We have : $180^{\circ}=\pi \mathrm{rad}$ $\therefore 1^{\circ}=\frac{\pi}{180} \mathrm{rad}$ (i) $300^{\circ}$ $=\left(300 \times \frac{\pi}{180}\right)$ $=\frac{5 \pi}{3} \mathrm{rad}$ (ii) $35^{\circ}$ $=35 \times \frac{\pi}{180}$ $=\frac{7 \pi}{36} \mathrm{rad}$ (iii) $-56^{\circ}$ $=-\left(56 \times \frac{\pi}{180}\right)...
Read More →Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
Question: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(ΔAPB) ar(ΔCPD) = ar(ΔAPD) ar(ΔBPC). Solution: Construction: - Draw BQ AC and DRAC Proof:- L.H.S =ar(ΔAPB) ar(ΔCDP) = (1/2) [(AP BQ)] (1/2 PC DR) =(1/2 PC BQ) (1/2 AP DR) =ar(ΔAPD) ar(ΔBPC). = R.H.S Hence proved....
Read More →In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(ΔADE) = ar(ΔBCF).
Question: In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(ΔADE) = ar(ΔBCF). Solution: Given that ABCD is parallelogram⇒ AD = BC CDEF is parallelogram⇒ DE = CF ABFE is parallelogram⇒ AE = BF Thus, inΔsADF and BCF, we have AD = BC, DE = CF and AE = BF So, by SSS criterion of congruence, we have ΔADE ΔBCF ar(ΔADE) = ar(ΔBCF)...
Read More →A boat goes 12 km upstream and 40 km downstream in 8 hours.
Question: A boat goes 12 km upstream and 40 km downstream in 8 hours. I can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the 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 be II $\mathrm{km} / \mathrm{hr}$ and the speed of the stream be $=\mathrm{km} / \mathrm{hr}$ then Speed upstream $=(x-y) \mathrm{kn} / \mathrm{hr}$ Sped down stream $=...
Read More →Show that
Question: $\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}$ Solution: $\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}$ Let $e^{x}=t \Rightarrow e^{x} d x=d t$ $\Rightarrow \int \frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)} d x=\int \frac{d t}{(t+1)(t+2)}$ $=\int\left[\frac{1}{(t+1)}-\frac{1}{(t+2)}\right] d t$ $=\log |t+1|-\log |t+2|+\mathrm{C}$ $=\log \left|\frac{t+1}{t+2}\right|+\mathrm{C}$ $=\log \left|\frac{1+e^{x}}{2+e^{x}}\right|+\mathrm{C}$...
Read More →Find the degree measure corresponding to the following radian measures:
Question: Find the degree measure corresponding to the following radian measures: (i) $\frac{9 \pi}{5}$ (ii) $-\frac{5 \pi}{6}$ (iii) $\left(\frac{18 \pi}{5}\right)$ (iv) $(-3)^{\mathrm{C}}$ (v) $11^{\mathrm{c}}$ (vi) $1^{\mathrm{c}}$ Solution: We have : $\pi \operatorname{rad}=180^{\circ}$ $\therefore 1 \mathrm{rad}=\left(\frac{180}{\pi}\right)^{\circ}$ (i) $\frac{18 \pi}{5}=\left(\frac{180}{\pi} \times \frac{9 \pi}{5}\right)^{\circ}$ $=(36 \times 9)^{\circ}$ $=324^{\circ}$ (ii) $-\frac{5 \pi}{...
Read More →Find the degree measure corresponding to the following radian measures:
Question: Find the degree measure corresponding to the following radian measures: (i) $\frac{9 \pi}{5}$ (ii) $-\frac{5 \pi}{6}$ (iii) $\left(\frac{18 \pi}{5}\right)$ (iv) $(-3)^{\mathrm{C}}$ (v) $11^{\mathrm{c}}$ (vi) $1^{\mathrm{c}}$ Solution: We have : $\pi \operatorname{rad}=180^{\circ}$ $\therefore 1 \mathrm{rad}=\left(\frac{180}{\pi}\right)^{\circ}$ (i) $\frac{18 \pi}{5}=\left(\frac{180}{\pi} \times \frac{9 \pi}{5}\right)^{\circ}$ $=(36 \times 9)^{\circ}$ $=324^{\circ}$ (ii) $-\frac{5 \pi}{...
Read More →Find the degree measure corresponding to the following radian measures:
Question: Find the degree measure corresponding to the following radian measures: (i) $\frac{9 \pi}{5}$ (ii) $-\frac{5 \pi}{6}$ (iii) $\left(\frac{18 \pi}{5}\right)$ (iv) $(-3)^{\mathrm{C}}$ (v) $11^{\mathrm{c}}$ (vi) $1^{\mathrm{c}}$ Solution: We have : $\pi \operatorname{rad}=180^{\circ}$ $\therefore 1 \mathrm{rad}=\left(\frac{180}{\pi}\right)^{\circ}$ (i) $\frac{18 \pi}{5}=\left(\frac{180}{\pi} \times \frac{9 \pi}{5}\right)^{\circ}$ $=(36 \times 9)^{\circ}$ $=324^{\circ}$ (ii) $-\frac{5 \pi}{...
Read More →Show that
Question: $\frac{x^{3}}{\sqrt{1-x^{8}}}$ Solution: $\frac{x^{3}}{\sqrt{1-x^{8}}}$ Let $x^{4}=t \Rightarrow 4 x^{3} d x=d t$ $\Rightarrow \int \frac{x^{3}}{\sqrt{1-x^{8}}} d x=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^{2}}}$ $=\frac{1}{4} \sin ^{-1} t+C$ $=\frac{1}{4} \sin ^{-1}\left(x^{4}\right)+C$...
Read More →Show that
Question: $\frac{1}{\cos (x+a) \cos (x+b)}$ Solution: $\frac{1}{\cos (x+a) \cos (x+b)}$ Multiplying and dividing by $\sin (a-b)$, we obtain $\frac{1}{\sin (a-b)}\left[\frac{\sin (a-b)}{\cos (x+a) \cos (x+b)}\right]$ $=\frac{1}{\sin (a-b)}\left[\frac{\sin [(x+a)-(x+b)]}{\cos (x+a) \cos (x+b)}\right]$ $=\frac{1}{\sin (a-b)}\left[\frac{\sin (x+a) \cdot \cos (x+b)-\cos (x+a) \sin (x+b)}{\cos (x+a) \cos (x+b)}\right]$ $=\frac{1}{\sin (a-b)}\left[\frac{\sin (x+a)}{\cos (x+a)}-\frac{\sin (x+b)}{\cos (x...
Read More →Show that
Question: $\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}$ Solution: $\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}=\frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{4} x-\cos ^{4} x\right)}{\sin ^{2} x+\cos ^{2} x-\sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x}$ $=\frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)}{\left(\sin ^{2} x-\sin ^{2} x \cos ^{2} x\right)+\left(\cos ^{2} x-\sin ^{2} x \cos ^...
Read More →Show that
Question: $\frac{\cos x}{\sqrt{4-\sin ^{2} x}}$ Solution: $\frac{\cos x}{\sqrt{4-\sin ^{2} x}}$ Let $\sin x=t \Rightarrow \cos x d x=d t$ $\Rightarrow \int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x=\int \frac{d t}{\sqrt{(2)^{2}-(t)^{2}}}$ $=\sin ^{-1}\left(\frac{t}{2}\right)+\mathrm{C}$ $=\sin ^{-1}\left(\frac{\sin x}{2}\right)+\mathrm{C}$...
Read More →Places A and B are 80 km apart from each other on a highway.
Question: Places A and B are 80 km apart from each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find the speeds of the cars. Solution: Let x and y be two cars starting from points A and B respectively. Let the speed of the car X be x km/hr and that of the car Y be y km/hr. Case I: When two cars move in the same directions: Suppose two...
Read More →Show that
Question: $\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}$ Solution: $\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}=\frac{e^{4 \log x}\left(e^{\log x}-1\right)}{e^{2 \log x}\left(e^{\log x}-1\right)}$ $=e^{2 \log x}$ $=e^{\log x^{2}}$ $=x^{2}$ $\therefore \int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x=\int x^{2} d x=\frac{x^{3}}{3}+\mathrm{C}$...
Read More →Show that
Question: $\frac{\sin x}{\sin (x-a)}$ Solution: $\frac{\sin x}{\sin (x-a)}$ Let $x-a=t \Rightarrow d x=d t$ $\int \frac{\sin x}{\sin (x-a)} d x=\int \frac{\sin (t+a)}{\sin t} d t$ $=\int \frac{\sin t \cos a+\cos t \sin a}{\sin t} d t$ $=\int(\cos a+\cot t \sin a) d t$ $=t \cos a+\sin a \log |\sin t|+C_{1}$ $=(x-a) \cos a+\sin a \log |\sin (x-a)|+\mathrm{C}_{1}$ $=x \cos a+\sin a \log |\sin (x-a)|-a \cos a+\mathrm{C}_{1}$ $=\sin a \log |\sin (x-a)|+x \cos a+\mathrm{C}$...
Read More →Express the function f :
Question: Express the functionf:XRgiven byf(x) =x3+ 1 as set of ordered pairs, whereX= {1, 0, 3, 9, 7}. [NCERT EXEMPLAR] Solution: The functionf:XRis definedbyf(x) =x3+ 1, whereX= {1, 0, 3, 9, 7}. Now, $f(-1)=(-1)^{3}+1=0$ $f(0)=0^{3}+1=1$ $f(3)=3^{3}+1=28$ $f(7)=7^{3}+1=344$ $f(9)=9^{3}+1=730$ So, $f=\{(x, f(x)): x \in X\}=\{(-1,0),(0,1),(3,28),(7,344),(9,730)\}$...
Read More →In figure, ABCD is a trapezium in which AB ∥ DC. Prove that ar(ΔAOD) = ar(ΔBOC)
Question: In figure, ABCD is a trapezium in which AB ∥ DC. Prove that ar(ΔAOD) = ar(ΔBOC) Solution: Given: ABCD is a trapezium in which AB ∥ DC To prove:ar(ΔAOD) = ar(ΔBOC) Proof: Since,ΔADCandΔBDCare on the same base DC and between same parallels AB and DC Then, ar(ΔADC) = ar(ΔBDC) ⇒ ar(ΔAOD) + ar(ΔDOC) = ar(ΔBOC) + ar(ΔDOC) ⇒ ar(ΔAOD) = ar(ΔBOC)...
Read More →