Find the angle of elevation of the sun
Question: Find the angle of elevation of the sun when the shadow of a pole h m high is 3 h m long. Solution: Let the angle of elevation of the sun is . Given, height of pole = h Now, in $\triangle A B C$, $\tan \theta=\frac{A B}{B C}=\frac{h}{\sqrt{3} h}$ $\Rightarrow$$\tan \theta=\frac{1}{\sqrt{3}}=\tan 30^{\circ} \Rightarrow \theta=30^{\circ}$ Hence, the angle of elevation of the Sun is $30^{\circ}$....
Read More →If A and B be two sets such that
Question: If $A$ and $B$ be two sets such that $n(A)=3, n(B)=4$ and $n(A \cap B)=2$ then find. (i) $n(A \times B)$ (ii) $n(B \times A)$ (iii) $n(A \times B) \cap(B \times A)$ Solution: Given: $n(A)=3, n(B)=4$ and $n(A \cap B)=2$ (i) $n(A \times B)=n(A) \times n(B)$ $\Rightarrow n(A \times B)=3 \times 4$ $\Rightarrow n(A \times B)=12$ (ii) $n(B \times A)=n(B) \times n(A)$ $\Rightarrow n(B \times A)=4 \times 3$ $\Rightarrow \mathrm{n}(\mathrm{B} \times \mathrm{A})=12$ (iii) $n((A \times B) \cap(B ...
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Question: If $f(x)=x^{3}+7 x^{2}+8 x-9$, find $f(4)$ Solution: Given: $f(x)=x^{3}+7 x^{2}+8 x-9$ Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $f$ at $x$ is given by: $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^{3}+7(x+h)^{2}+8(x+h)-9-x^{3}-7 x^{2}-8 x+9}{h}$ $\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{x^{3}+h^{3}+3 x^{2} h+3 x h^{2}+7 x^{2}+7 h^{2}+14 x...
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Question: Tick (✓) the correct answer The sum of two numbers is $\frac{-4}{3}$. If one of the numbers is $-5$, what is the other? (a) $\frac{-11}{3}$ (b) $\frac{11}{3}$ (C) $\frac{-19}{3}$ (d) $\frac{19}{3}$ Solution: (b) $\frac{11}{3}$ Let the other number bex. Now, $x+(-5)=\frac{-4}{3}$ $\Rightarrow x=\frac{-4}{3}+($ Additive inverse of $-5)$ $\Rightarrow x=\frac{-4}{3}+5$ $=\frac{-4}{3}+\frac{5}{1}$ $=\frac{(-4)+15}{3}$ $=\frac{11}{3}$...
Read More →tanθ + tan (90° – θ) = sec θ
Question: tan + tan (90 ) = sec sec (90 ) Solution: LHS $=\tan \theta+\tan \left(90^{\circ}-\theta\right)$ $\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$ $=\tan \theta+\cot \theta=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$ $=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}$$\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$ $=\frac{1}{\sin \theta \cos \th...
Read More →If for the function
Question: If for the function $\Phi(x)=\lambda x^{2}+7 x-4, \Phi^{\prime}(5)=97$, find $\lambda$. Solution: Given: $\phi(x)=\lambda x^{2}+7 x-4$ Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $\phi$ at $x$ is given by: $\phi^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\phi(x+h)-\phi(x)}{h}$ $\Rightarrow \phi^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\lambda(x+h)^{2}+7(x+h)-4-\lambda x^{2}-7 x+4}{h}$ $\Rightarrow \phi^{\prime}(x)=\lim _{h \rightarrow...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer $\left(\frac{2}{3}+\frac{-4}{5}+\frac{7}{15}+\frac{-11}{20}\right)=?$ (a) $\frac{-1}{5}$ (b) $\frac{-4}{15}$ (c) $\frac{-13}{60}$ (d) $\frac{-7}{30}$ Solution: (c) $\frac{-13}{60}$ Using the commutative and associative laws, we can arrange the terms in any suitable manner. Using this rearrangement property, we have: $\frac{2}{3}+\frac{-4}{5}+\frac{7}{15}+\frac{-11}{20}=\left(\frac{2}{3}+\frac{7}{15}\right)+\left(\frac{-4}{5}+\frac{-11}{20}\right)$ $=\frac{(1...
Read More →Prove the following
Question: $1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}=\operatorname{cosec} \alpha$ Solution: $\mathrm{LHS}=1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}=1+\frac{\cos ^{2} \alpha / \sin ^{2} \alpha}{1+1 / \sin \alpha}$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right.$ and $\left.\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$ $=1+\frac{\cos ^{2} \alpha}{\sin \alpha(1+\sin \alpha)}=\frac{\sin \alpha(1+\sin \alpha)+\cos ^{2} \alpha}{\sin \alpha(1...
Read More →Show that the derivative of the function f given by
Question: Show that the derivative of the functionfgiven by $f(x)=2 x^{3}-9 x^{2}+12 x+9$, at $x=1$ and $x=2$ are equal. Solution: Given: $f(x)=2 x^{3}-9 x^{2}+12 x+9$ Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $f$ at $x$ is given by: $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h-f(x)}{h}$ $\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{2(x+h)^{3}-9(x+h)^{2}+12(x+h)+9-2 x^{3}+9 x^{2}-12 x-9}{h}$ $\Rightarrow f^{\prime}(x)=\lim _{...
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Question: Tick (✓) the correct answer What should be added to $\frac{7}{12}$ to get $\frac{-4}{15} ?$ (a) $\frac{17}{20}$ (b) $\frac{-17}{20}$ (c) $\frac{7}{20}$ (d) $\frac{-7}{20}$ Solution: (b) $\frac{-17}{20}$ Let the required number bex. Now, $\frac{7}{12}+x=\frac{-4}{15}$ $\Rightarrow x=\left(\frac{-4}{15}+\frac{-7}{12}\right)$ $=\frac{4 \times(-4)+5 \times(-7)}{60}$ $=\frac{(-16)+(-35)}{60}$ $=\frac{-51}{60}$ $=\frac{-17}{20}$...
Read More →Prove the following
Question: (3 + 1) (3 cot 30) = tan360 2sin60 Solution: RHS $=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}=(\sqrt{3})^{3}-2 \frac{\sqrt{3}}{3}=3 \sqrt{3}-\sqrt{3}=2 \sqrt{3}$ $\mathrm{LHS}=(\sqrt{3}+1)\left(3-\cot 30^{\circ}\right)=(\sqrt{3}+1)(3-\sqrt{3})$ $\left[\because \tan 60^{\circ}=\sqrt{3} \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right.$ and $\left.=(\sqrt{3}+1) \sqrt{3}(\sqrt{3}-1) \cot 30^{\circ}=\sqrt{3}\right]$ $\left.=\sqrt{3}(\sqrt{3})^{2}-1\right)=\sqrt{3}(3-1)=2 \sqrt{3}$ $\therefore \quad L ...
Read More →prove that
Question: If $A \times B \subseteq C \times D$ and $A \times B \neq \phi$, prove that $A \subseteq C$ and $B \subseteq D .$ Solution: Given: $A \times B \subseteq C \times D$ and $A \times B \neq \phi$ Need to prove: $A \subseteq C$ and $B \subseteq D$ Let us consider, $(x, y)^{\in}(A \times B) \cdots(1)$ $\Rightarrow(x, y)^{\in}(C \times D)[\operatorname{as} A \times B \subseteq C \times D] \cdots(2)$ From (1) we can say that, $x \in_{A}$ and $y \in_{B \cdots}$ (a) From (2) we can say that, $x^...
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Question: Tick (✓) the correct answer $\left(\frac{31}{-4}+\frac{-5}{8}\right)=?$ (a) $\frac{67}{8}$ (b) $\frac{57}{8}$ (c) $\frac{-57}{8}$ (d) $\frac{-67}{8}$ Solution: (d) $\frac{-67}{8}$ $\frac{31}{-4}=\frac{-31}{4}$ We have: $\left(\frac{31}{-4}+\frac{-5}{8}\right)=\left(\frac{-31}{4}+\frac{-5}{8}\right)$ LCM of 4 and 8 is 8 , that is, $(4 \times 1 \times 2)$. $\left(\frac{-31}{4}+\frac{-5}{8}\right)=\frac{2 \times(-31)+1 \times(-5)}{8}$ $=\frac{(-62)+(-5)}{8}$ $=\frac{-67}{8}$...
Read More →If f is defined by
Question: If $f$ is defined by $f(x)=x^{2}-4 x+7$, show that $f^{\prime}(5)=2 f^{\prime}\left(\frac{7}{2}\right)$ Solution: Given: $f(x)=x^{2}-4 x+7$ Clearly, $f(x)$ being a polynomial function, is everywhere differentiable. The derivative of $f$ at $x$ is given by: $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^{2}-4(x+h)+7-\left(x^{2}-4 x+7\right)}{h}$ $\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{x^{2}+h^...
Read More →Prove the following
Question: $(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \beta)=\sec \alpha+\operatorname{cosec} \beta$ Solution: $\mathrm{LHS}=(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)$ $=(\sin \alpha+\cos \alpha)\left(\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha}\right)$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$ $=(\sin \alpha+\cos \alpha)\left(\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{\sin \alpha \...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer $\left(3+\frac{5}{-7}\right)=?$ (a) $\frac{-16}{7}$ (b) $\frac{16}{7}$ (c) $\frac{-26}{7}$ (d) $\frac{-8}{7}$ Solution: (b) $\frac{16}{7}$ $3=\frac{3}{1}$ and $\frac{5}{-7}=\frac{-5}{7}$ Now, we have: $\left(3+\frac{5}{-7}\right)=\left(\frac{3}{1}+\frac{-5}{7}\right)$ LCM of 1 and 7 is 7 $\left(\frac{3}{1}+\frac{-5}{7}\right)=\frac{7 \times 3+1 \times(-5)}{7}$ $=\frac{21+(-5)}{7}$ $=\frac{16}{7}$...
Read More →If f is defined by
Question: If $f$ is defined by $f(x)=x^{2}$, find $f(2)$. Solution: Given: $f(x)=x^{2}$ We know a polynomial function is everywhere differentiable. Therefore $f(x)$ is differentiable at $x=2$. $f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$ $\Rightarrow f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{(2+h)^{2}-2^{2}}{h}$ $\Rightarrow f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{\left(4+h^{2}+4 h\right)-4}{h}$ $\Rightarrow f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{h(h+4)}{h}$ $\Rightar...
Read More →Prove the following
Question: If $\tan \mathrm{A}=\frac{3}{4}, t h e n \sin \mathrm{A} \cos \mathrm{A}=\frac{12}{25}$ Solution: Given, $\tan A=\frac{3}{4}=\frac{P}{B}=\frac{\text { Perpendicular }}{\text { Base }}$ Let $P=3 k$ and $B=4 k$ By Pythagoras theorem, $H^{2}=P^{2}+B^{2}=(3 k)^{2}+(4 k)^{2}$ $=9 k^{2}+16 k^{2}=25 k^{2}$ $\begin{array}{lll}\Rightarrow H=5 k \text { [since, side cannot be negative] }\end{array}$ $\therefore \quad \sin A=\frac{P}{H}=\frac{3 k}{5 k}=\frac{3}{5}$ and $\cos A=\frac{B}{H}=\frac{4...
Read More →Solve this
Question: (i) If $A \subseteq B$, prove that $A \times C \subseteq B \times C$ for any set $C$. (ii) If $A \subseteq B$ and $C \subseteq D$ then prove that $A \times C \subseteq B \times D$. Solution: (i) Given: A\subseteq B Need to prove: $A \times C \subseteq B \times C$ Let us consider, $(x, y) \in(A \times C)$ That means, $x \in A$ and $y \in C$ Here given, $A \subseteq B$ That means, $x$ will surely be in the set $B$ as $A$ is the subset of $B$ and $x \in A$. So, we can write $x^{\in} B$ Th...
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Question: Tick (✓) the correct answer $\left(\frac{7}{-26}+\frac{16}{39}\right)=?$ (a) $\frac{11}{78}$ (b) $\frac{-11}{78}$ (c) $\frac{11}{39}$ (d) $\frac{-11}{39}$ Solution: $\frac{7}{-26}=\frac{-7}{26}$ Now, we have: $\left(\frac{7}{-26}+\frac{16}{39}\right)=\left(\frac{-7}{26}+\frac{16}{39}\right)$ LCM of 26 and 39 is 1014 , that is, $(29 \times 1 \times 36)$. (a) $\frac{11}{78}$ $\left(\frac{-7}{26}+\frac{16}{39}\right)=\frac{39 \times(-7)+26 \times 16}{1014}$ $=\frac{(-273)+416}{1014}$ $=\f...
Read More →Find the values of a and b,
Question: Find the values of $a$ and $b$, if the function $f$ defined by $f(x)=\left\{\begin{array}{cl}x^{2}+3 x+a, x \leqslant 1 \\ b x+2, x1\end{array}\right.$ is differentiable at $x=1 .$ Solution: Given thatf(x) is differentiable atx= 1. Therefore,f(x) is continuous atx= 1. $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$ $\Rightarrow \lim _{x \rightarrow 1}\left(x^{2}+3 x+a\right)=\lim _{x \rightarrow 1}(b x+2)=1+3+a$ $\Rightarrow 1+3+a=b+2$ $\Rightarrow a-b+2=0$ ......
Read More →Find the values of a and b,
Question: Find the values of $a$ and $b$, if the function $f$ defined by $f(x)=\left\{\begin{array}{cl}x^{2}+3 x+a, x \leqslant 1 \\ b x+2, x1\end{array}\right.$ is differentiable at $x=1 .$ Solution: Given thatf(x) is differentiable atx= 1. Therefore,f(x) is continuous atx= 1. $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$ $\Rightarrow \lim _{x \rightarrow 1}\left(x^{2}+3 x+a\right)=\lim _{x \rightarrow 1}(b x+2)=1+3+a$ $\Rightarrow 1+3+a=b+2$ $\Rightarrow a-b+2=0$ ......
Read More →Prove the following
Question: $\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2 \operatorname{cosec} A$ Solution: $\mathrm{LHS}=\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=\frac{\tan A(1-\sec A-1-\sec A)}{(1+\sec A)(1-\sec A)}$ $=\frac{\tan A(-2 \sec A)}{\left(1-\sec ^{2} A\right)}=\frac{2 \tan A \cdot \sec A}{\left(\sec ^{2} A-1\right)}$ $\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$ $=\frac{2 \tan A \cdot \sec A}{\tan ^{2} A}$ $\left[\because \sec ^{2} A-\tan ^{2} A=1\right]\left[\because \sec \theta=\frac{1}...
Read More →If A and B are nonempty sets, prove that
Question: If A and B are nonempty sets, prove that $A \times B=B \times A \Leftrightarrow A=B$ Solution: Given: A = B, where A and B are nonempty sets. Need to prove: $A \times B=B \times A$ Let us consider, $(x, y)^{\in}(A \times B)$ That means, $x \in_{A}$ and $y \in_{B}$ As given in the problem A = B, we can write, $\Rightarrow x^{\in}_{B}$ and $y^{\in} A$ $\Rightarrow(x, y)^{\in}(B \times A)$ That means, $(A \times B) \subseteq(B \times A) \cdots(1)$ Similarly we can prove, $\Rightarrow(B \t...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer $\left(\frac{8}{-15}+\frac{4}{-3}\right)=?$ (a) $\frac{28}{15}$ (b) $\frac{-28}{15}$ (c) $\frac{-4}{5}$ (d) $\frac{-4}{15}$ Solution: (b) $\frac{-28}{15}$ $\frac{8}{-15}=\frac{-8}{15}$ and $\frac{4}{-3}=\frac{-4}{3}$ Now, we have: $\left(\frac{8}{-15}+\frac{4}{-3}\right)=\left(\frac{-8}{15}+\frac{-4}{3}\right)$ LCM of 15 and 3 is $(3 \times 5 \times 1)$, that is, 15 $\frac{-8}{15}+\frac{-4}{3}=\frac{1 \times(-8)+5 \times(-4)}{15}$ $=\frac{(-8)+(-20)}{15}$ $=...
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