A basket contains three types of fruits weighing

Question: A basket contains three types of fruits weighing $19 \frac{1}{3} \mathrm{~kg}$ in all. If $8 \frac{1}{9} \mathrm{~kg}$ of these be apples, $3 \frac{1}{6} \mathrm{~kg}$ be oranges and the rest pears, what is the weight of the pears in the basket? Solution: Weight of pears in the basket = Weight of the basket containing three types of fruits - (Weight of apples + Weight of oranges) $=19 \frac{1}{3}-\left(8 \frac{1}{9}+3 \frac{1}{6}\right)$ Now, $\left(8 \frac{1}{9}+3 \frac{1}{6}\right) \...

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Show that

Question: Show that $f(x)=\left\{\begin{array}{l}12 x-13, \text { if } x \leq 3 \\ 2 x^{2}+5, \text { if } x3\end{array}\right.$ is differentiable at $x=3$. Also, find $f(3)$. Solution: Given: $f(x)= \begin{cases}12 x-13, x \leq 3 \\ 2 x^{2}+5, x3\end{cases}$ We have to show that the given function is differentiable atx= 3. $(\mathrm{LHD}$ at $x=3)=\lim _{x \rightarrow 3^{-}} \frac{f(x)-f(3)}{x-3}$' $=\lim _{x \rightarrow 3} \frac{12 x-13-23}{x-3}$ $=\lim _{x \rightarrow 3} \frac{12 x-36}{x-3}$ ...

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The value of the expression

Question: The value of the expression $\left(\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right)$ is (a) 3 (b) 2 (c) 1 (d) 0 Solution: (b) Given expression, $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}$ $=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}\left(90^{\circ}-22^{\circ}\right)}{\cos ^{2}\left(90^{\ci...

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Show that

Question: Show that $f(x)=x^{1 / 3}$ is not differentiable at $x=0$. Solution: Disclaimer: It might be a wrong question because $f(x)$ is differentiable at $x=0$ Given: $f(x)=x^{\frac{1}{3}}$ We have, $(\mathrm{LHD}$ at $x=0)$ $\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$ $=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}$ $=\lim _{h \rightarrow 0} \frac{(0-h)^{\frac{1}{3}}-0^{\frac{1}{3}}}{-h}$ $=\lim _{h \rightarrow 0} \frac{(-h)^{\frac{1}{3}}}{-h}$ $=\lim _{h \rightarrow 0}(-h)^{\frac{...

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Show that

Question: Show that $f(x)=x^{1 / 3}$ is not differentiable at $x=0$. Solution: Disclaimer: It might be a wrong question because $f(x)$ is differentiable at $x=0$ Given: $f(x)=x^{\frac{1}{3}}$ We have, $(\mathrm{LHD}$ at $x=0)$ $\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$ $=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}$ $=\lim _{h \rightarrow 0} \frac{(0-h)^{\frac{1}{3}}-0^{\frac{1}{3}}}{-h}$ $=\lim _{h \rightarrow 0} \frac{(-h)^{\frac{1}{3}}}{-h}$ $=\lim _{h \rightarrow 0}(-h)^{\frac{...

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A drum full of rice weighs

Question: A drum full of rice weighs $40 \frac{1}{6} \mathrm{~kg}$. If the empty drum weighs $13 \frac{3}{4} \mathrm{~kg}$, find the weight of rice in the drum. Solution: Weight of rice in the drum = Weight of the drum full of rice - Weight of the empty drum $=40 \frac{1}{6}-13 \frac{3}{4}$ $=\left(40+\frac{1}{6}\right)-\left(13+\frac{3}{4}\right)$ $=\frac{241}{6}-\frac{55}{4}$ $=\frac{241}{6}+\left(\right.$ Additive inverse of $\left.\frac{55}{4}\right)$ $=\frac{482-165}{12}$ $=\frac{317}{12}$ ...

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Prove the following

Question: If $\sin \alpha=\frac{1}{2}$ and $\cos \beta=\frac{1}{2}$ then the value of $(\alpha+\beta)$ is (a) $0^{\circ}$ (b) $30^{\circ}$ (C) $60^{\circ}$ (d) $90^{\circ}$ Solution: (d) Given, $\sin \alpha=\frac{1}{2}=\sin 30^{\circ}$ $\left[\because \sin 30^{\circ}=\frac{1}{2}\right]$ $\Rightarrow$ $\alpha=30^{\circ}$ and $\cos \beta=\frac{1}{2}=\cos 60^{\circ}$ $\left[\because \cos 60^{\circ}=\frac{1}{2}\right]$ $\Rightarrow$ $\beta=60^{\circ}$ $\therefore$ $\alpha+\beta=30^{\circ}+60^{\circ}...

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From a rope 11 m long, two pieces of lengths

Question: From a rope $11 \mathrm{~m}$ long, two pieces of lengths $2 \frac{3}{5} \mathrm{~m}$ and $3 \frac{3}{10} \mathrm{~m}$ are cut off. What is the length of the remaining rope? Solution: Length of the rope when two pieces of lengths $2 \frac{3}{5} \mathrm{~m}$ and $3 \frac{3}{10} \mathrm{~m}$ are cut off $=$ Total length of the rope - Length of the two cut off pieces $\therefore 11-\left(2 \frac{3}{5}+3 \frac{3}{10}\right)$ Now, $2 \frac{3}{5}+3 \frac{3}{10} \Rightarrow\left(2+\frac{3}{5}\...

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Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.

Question: Show thatf(x) = |x 2| is continuous but not differentiable atx= 2. Solution: Given: $f(x)=|x-2|= \begin{cases}x-2, x \geq 2 \\ -x+2, x2\end{cases}$ Continuity atx=2: We have, $(\mathrm{LHL}$ at $x=2)$$(\mathrm{LHL}$ at $x=2)$ $=\lim _{x \rightarrow 2^{-}} f(x)$ $=\lim _{h \rightarrow 0} f(2-h)$ $=\lim _{h \rightarrow 0}(-2+h)+2$ $=0$ $(\mathrm{RHL}$ at $x=2)$ $=\lim _{x \rightarrow 2^{+}} f(x)$ $=\lim _{h \rightarrow 0} f(2+h)$ $=\lim _{h \rightarrow 0} 2+h-2$ $=0$ and $f(2)=0$ Thus, $...

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Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.

Question: Show thatf(x) = |x 2| is continuous but not differentiable atx= 2. Solution: Given: $f(x)=|x-2|= \begin{cases}x-2, x \geq 2 \\ -x+2, x2\end{cases}$ Continuity atx=2: We have, $(\mathrm{LHL}$ at $x=2)$$(\mathrm{LHL}$ at $x=2)$ $=\lim _{x \rightarrow 2^{-}} f(x)$ $=\lim _{h \rightarrow 0} f(2-h)$ $=\lim _{h \rightarrow 0}(-2+h)+2$ $=0$ $(\mathrm{RHL}$ at $x=2)$ $=\lim _{x \rightarrow 2^{+}} f(x)$ $=\lim _{h \rightarrow 0} f(2+h)$ $=\lim _{h \rightarrow 0} 2+h-2$ $=0$ and $f(2)=0$ Thus, $...

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If sin A + sin2 A = 1,

Question: If sin A + sin2A = 1, then the value of (cos2A + cos4A) is (a) 1 (b) $\frac{1}{2}$ (c) 2 (d) 3 Solution: (a) Given $\sin A+\sin ^{2} A=1$ $\Rightarrow \quad \sin A=1-\sin ^{2} A=\cos ^{2} A$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$ On squaring both sides, we get $\sin ^{2} A=\cos ^{4} A$ $\Rightarrow$ $1-\cos ^{2} A=\cos ^{4} A$ $\Rightarrow$ $\cos ^{2} A+\cos ^{4} A=1$...

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For any sets A, B and C prove that:

Question: For any sets A, B and C prove that: $A \times(B \cap C)=(A \times B) \cap(A \times C)$ Solution: Given: A, B and C three sets are given. Need to prove: $A \times(B \cap C)=(A \times B) \cap(A \times C)$ Let us consider, $(x, y)^{\in} A \times(B \cap C)$ $\Rightarrow x \in_{A}$ and $y \in_{(B \cap C)}$ $\Rightarrow x \in_{A}$ and $\left(y \in_{B}\right.$ and $\left.y \in_{C}\right)$ $\Rightarrow\left(x \in_{A}\right.$ and $\left.y \in_{B}\right)$ and $\left(x \in_{A}\right.$ and $\left....

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Find 12 rational numbers between −1 and 2.

Question: Find 12 rational numbers between 1 and 2. Solution: We may write: $-1=\frac{-10}{10}$ and $2=\frac{20}{10}$ Rational numbers between $-1$ and 2 : $\frac{-9}{10}, \frac{-8}{10}, \frac{-7}{10}, \frac{-6}{10}, \frac{-5}{10}, \frac{-4}{10}, \ldots, \frac{14}{10}, \frac{15}{10}, \frac{16}{10}, \frac{17}{10}, \frac{18}{10}$ and $\frac{19}{10}$ We can take any 12 numbers out of these....

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If ΔABC is right angled at C,

Question: If ΔABC is right angled at C, then the value of cos (A + B) is (a) 0 (b) 1 (c) $\frac{1}{2}$ (d) $\frac{\sqrt{3}}{2}$ Solution: (a) We know that, in $\triangle A B C$, sum of three angles $=180^{\circ}$ i.e., $\quad \angle A+\angle B+\angle C=180^{\circ}$ But right angled at $C$ i.e., $\angle C=90^{\circ}$ [given] But right angled at $C$ i.e., $\angle C=90^{\circ}$ $\angle A+\angle B+90^{\circ}=180^{\circ}$ $\Rightarrow$ $A+B=90^{\circ}$ $[\because \angle A=A]$ $\therefore$ $\cos (A+B)...

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Find 10 rational numbers between

Question: Find 10 rational numbers between $\frac{-3}{4}$ and $\frac{5}{6}$. Solution: LCM of 4 and 6 is 12 . Now, $\frac{-3}{4}=\frac{-3 \times 3}{4 \times 3}=\frac{-9}{12}$ And, $\frac{5}{6}=\frac{5 \times 2}{6 \times 2}=\frac{10}{12}$ Rational numbers lying between $\frac{-3}{4}$ and $\frac{5}{6}$ : $\frac{-8}{12}, \frac{-7}{12}, \frac{-6}{12}, \frac{-5}{12}, \frac{-4}{12}, \ldots \frac{1}{12}, \frac{2}{12}, \frac{3}{12}, \frac{4}{12}, \frac{5}{12}, \frac{6}{12}, \frac{7}{12}, \frac{8}{12}, \...

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Find three rational numbers between

Question: Find three rational numbers between $\frac{2}{3}$ and $\frac{3}{4}$. Solution: Rational number between $\frac{2}{3}$ and $\frac{3}{4}$ : $\frac{1}{2}\left(\frac{2}{3}+\frac{3}{4}\right)$ $=\frac{1}{2}\left(\frac{8+9}{12}\right)$ $=\frac{17}{24}$ We know, $\frac{2}{3}\frac{17}{24}\frac{3}{4}$ Rational number between $\frac{2}{3}$ and $\frac{17}{24}$ : $\frac{1}{2}\left(\frac{2}{3}+\frac{17}{24}\right)$ $=\frac{1}{2}\left(\frac{16+17}{24}\right)$ $=\frac{1}{2}\left(\frac{33}{24}\right)$ ...

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If cos 9α = sin α and 9α < 90°

Question: If cos 9 = sin and 9 90 ,then the value of tan 5 is (a) $\frac{1}{\sqrt{3}}$ (b) $\frac{\sqrt{3}}{1}$ (C) 7 (d) 0 Solution: (c) Given. $\cos 9 \alpha=\sin \alpha$ and $9 \alpha90^{\circ}$ i.e., acute angle. $\sin \left(90^{\circ}-9 \alpha\right)=\sin \alpha$$\left[\because \cos A=\sin \left(90^{\circ}-A\right)\right]$ $\Rightarrow \quad 90^{\circ}-9 \alpha=\alpha$ $\Rightarrow \quad 10 \alpha=90^{\circ}$ $\begin{array}{ll}\Rightarrow \alpha=9^{\circ}\end{array}$ $\therefore$$\tan 5 \al...

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Find three rational numbers between 4 and 5.

Question: Find three rational numbers between 4 and 5. Solution: Rational number between 4 and 5 : $\frac{1}{2}(4+5)$ $=\frac{9}{2}$ Rational number between 4 and $\frac{9}{2}$ : $\frac{1}{2}\left(4+\frac{9}{2}\right)$ $=\frac{1}{2}\left(\frac{8+9}{2}\right)$ $=\frac{1}{2}\left(\frac{17}{2}\right)$ $=\frac{17}{4}$ Rational number between $\frac{9}{2}$ and 5 : $\frac{1}{2}\left(\frac{9}{2}+5\right)$ $=\frac{1}{2}\left(\frac{9+10}{2}\right)$ $=\frac{19}{4}$ We know, $4\frac{17}{4}\frac{9}{2}\frac{...

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Find two rational numbers between −3 and −2.

Question: Find two rational numbers between 3 and 2. Solution: Required number $=\frac{1}{2} \times(-3-2)$ $=\frac{1}{2}(-5)$ $=\frac{-5}{2}$ We know, $-3\frac{-5}{2}-2$ Rational number between $-3$ and $\frac{-5}{2}=\frac{1}{2} \times\left(-3-\frac{5}{2}\right)$ $=\frac{1}{2}\left(\frac{-6-5}{2}\right)$ $=\frac{1}{2} \times \frac{-11}{2}$ $=\frac{-11}{4}$ Thus, the required numbers are $\frac{-5}{2}$ and $\frac{-11}{4}$....

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Find a rational number between

Question: Find a rational number between $\frac{-1}{3}$ and $\frac{1}{2}$. Solution: Required number $=\frac{1}{2} \times\left(\frac{-1}{3}+\frac{1}{2}\right)$ $=\frac{1}{2} \times\left(\frac{-2+3}{6}\right)$ $=\frac{1}{2} \times \frac{1}{6}$ $=\frac{1}{12}$...

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The value of

Question: The value of $\left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right)$ is (a) 0 (b) 1 (c) 2 (d) $\frac{1}{2}$ Solution: (b)tan1-tan2-tan3 tan 89 = tan1-tan2-tan3 tan44 . tan 45 . tan 46 tan 87-tan 88tan 89 = tan 1- tan2 - tan 3 tan 44 . (1)- tan (90 44) tan (90 3) tan (90 -2)- tan (90 -1) ( tan 45 = 1) = tan1-tan2-tan3. tan44 (1) . cot 44. cot3-cot2-cot1 $\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$ $=\tan 1^{\circ} \cdot \tan 2^{\circ...

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Find a rational number between 2 and 3.

Question: Find a rational number between 2 and 3. Solution: Required Number $=\frac{1}{2} \times(2+3)$ $=\frac{5}{2}$...

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Find a rational number between

Question: Find a rational number between $\frac{1}{4}$ and $\frac{1}{3}$. Solution: Required number $=\frac{1}{2}\left(\frac{1}{4}+\frac{1}{3}\right)$ $=\frac{1}{2}\left(\frac{3+4}{12}\right)$ $=\left(\frac{1}{2} \times \frac{7}{12}\right)$ $=\frac{7}{24}$...

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For any sets A, B and C prove that:

Question: For any sets A, B and C prove that: $A \times(B \cup C)=(A \times B) \cup(A \times C)$ Solution: Given: A, B and C three sets are given. Need to prove: $A \times(B \cup C)=(A \times B) \cup(A \times C)$ Let us consider, $(x, y)^{\in} A \times(B \cup C)$ $\Rightarrow x^{\in} A$ and $y \in(B \cup C)$ $\Rightarrow x^{\in}_{A}$ and $\left(y \in_{B}\right.$ or $\left.y \in_{C}\right)$ $\Rightarrow\left(x^{\in}_{A}\right.$ and $\left.y \in_{B}\right)$ or $\left(x^{\in}_{A}\right.$ and $\left...

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Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find:

Question: Let A = {3, 1}, B = {1, 3) and C = {3, 5). Find: (i) $A \times B$ (ii) $(A \times B) \times C$ (iii) $B \times C$ (iv) $A \times(B \times C)$ Solution: (i) Given: $A=\{-3,-1\}$ and $B=\{1,3\}$ To find: A B By the definition of the Cartesian product Given two non empty sets P and Q. The Cartesian product P Q is the set of all ordered pairs of elements from P and Q, .i.e. $P \times Q=\{(p, q): p \in P, q \in Q\}$ Here, $A=\{-3,-1\}$ and $B=\{1,3\} .$ So, $A \times B=\{-3,-1\} \times\{1,3...

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