Prove the following
Question: (tan + 2) (2 tan + 1) = 5 tan + sec2 Solution: False LHS $=(\tan \theta+2)(2 \tan \theta+1)$ $=2 \tan ^{2} \theta+4 \tan \theta+\tan \theta+2$ $=2\left(\sec ^{2} \theta-1\right)+5 \tan \theta+2$ $\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]$ $=2 \sec ^{2} \theta+5 \tan \theta=R H S$...
Read More →What is the length of each piece?
Question: A cord of length $71 \frac{1}{2} \mathrm{~m}$ has been cut into 26 pieces of equal length. What is the length of each piece? Solution: Length of each piece of the cord $=71 \frac{1}{2} \div 26$ $=\left(71+\frac{1}{2}\right) \div 26$ $=\frac{143}{2} \div 26$ $=\frac{143}{2} \div \frac{26}{1}$ $=\frac{143}{2} \times \frac{1}{26}$ $=\frac{143 \times 1}{2 \times 26}$ $=\frac{143}{52}$ $=\frac{9}{4}$ $=2 \frac{3}{4} \mathrm{~m}$ Hence, the length of each piece of the cord is $2 \frac{3}{4}$...
Read More →Find whether the function is differentiable at x = 1 and x = 2
Question: Find whether the function is differentiable atx= 1 andx= 2 $f(x)=\left\{\begin{array}{cc}x x \leq 1 \\ 2-x 1 \leq x \leq 2 \\ -2+3 x-x^{2} x2\end{array}\right.$ Solution: $f(x)=\left\{\begin{array}{cc}x x \leq 1 \\ 2-x 1 \leq x \leq 2 \\ -2+3 x-x^{2} x2\end{array}\right.$ $\Rightarrow f^{\prime}(x)=\left\{\begin{array}{cc}1 x \leq 1 \\ -1 1 \leq x \leq 2 \\ 3-2 x x2\end{array}\right.$ Now, $\mathrm{LHL}=\lim _{x \rightarrow 1^{-}} f^{\prime}(x)=\lim _{x \rightarrow 1^{-}} 1=1$ $\mathrm...
Read More →If cosA + cos2 A = 1,
Question: If cosA + cos2A = 1, then sin2A + sin4A = 1 Solution: True $\because \quad \cos A+\cos ^{2} A=1$ $\Rightarrow \quad \cos A=1-\cos ^{2} A=\sin ^{2} A \quad\left[\because \sin ^{2} A+\cos ^{2} A=1\right]$ $\Rightarrow \quad \cos ^{2} A=\sin ^{4} A$ $\Rightarrow \quad 1-\sin ^{2} A=\sin ^{4} A$ $\Rightarrow \quad \sin ^{2} A+\sin ^{4} A=1$ $\left[\because \cos ^{2} A=1-\sin ^{2} A\right]$...
Read More →what is the cost of one metre of cloth?
Question: The cost of $3 \frac{1}{2}$ metres of cloth is ₹ $166 \frac{1}{4}$. what is the cost of one metre of cloth? Solution: Cost of $3 \frac{1}{2} \mathrm{~m}$ of cloth $=₹ 166 \frac{1}{4}$ So, the cost of $1 \mathrm{~m}$ of cloth $=\frac{166 \frac{1}{4}}{3 \frac{1}{2}}=\frac{\frac{665}{4}}{\frac{7}{2}}=\frac{665}{4} \times \frac{2}{7}=\frac{95}{2}=₹ 47 \frac{1}{2}$ Hence, the cost of $1 \mathrm{~m}$ of cloth is ₹ $47 \frac{1}{2}$....
Read More →An aeroplane covers 1020 km in an hour.
Question: An aeroplane covers $1020 \mathrm{~km}$ in an hour. How much distance will it cover in $4 \frac{1}{6}$ hours? Solution: Distance covered by the aeroplane in $4 \frac{1}{6}$ hours $=4 \frac{1}{6} \times 1020$ $=\left(4+\frac{1}{6}\right) \times 1020$ $=\frac{25}{6} \times 1020$ $=\frac{25}{6} \times \frac{1020}{1}$ $=\frac{25 \times 1020}{6 \times 1}$ $=\frac{25500}{6}$ $=4250 \mathrm{~km}$ Therefore, the distance covered by the aeroplane is $4250 \mathrm{~km}$....
Read More →Prove the following
Question: $\sqrt{\left(1-\cos ^{2} \theta\right) \sec ^{2} \theta}=\tan \theta$ Solution: True $\sqrt{\left(1-\cos ^{2} \theta\right) \sec ^{2} \theta}=\sqrt{\sin ^{2} \theta \cdot \sec ^{2} \theta}$$\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$ $=\sqrt{\sin ^{2} \theta \cdot \frac{1}{\cos ^{2} \theta}}=\sqrt{\tan ^{2} \theta}=\tan \theta \quad\left[\because \sec \theta=\frac{1}{\cos \theta}, \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$...
Read More →The value of the expression
Question: The value of the expression (sin 80 cos 80) is negative. Solution: False We know that, sine is increasing when, O9O and cos is decreasing when, O9O. sin80-cos80 0 [positive]...
Read More →Discuss the continuity and differentiability
Question: Discuss the continuity and differentiability of the $f(x)=|x|+|x-1|$ in the interval $(-1,2)$ Solution: Given : $f(x)=|x|+|x-1|$ $|x|=-x$ for $x0$ $|x|=x$ for $x0$ $|x-1|=-(x-1)=-x+1$ for $x-10$ or $x1$ $|x-1|=x-1$ for $x-10$ or $x1$ Now, $f(x)=-x-x+1=-2 x+1 \quad x \in(-1,0)$ Or $f(x)=x-x+1=1 \quad x \in(0,1)$ Or $f(x)=x+x-1=2 x-1 \quad x \in(1,2)$ Now, $\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}-2 x+1=0+1=1$ $\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x...
Read More →One litre of petrol costs ₹
Question: One litre of petrol costs ₹ $63 \frac{3}{4}$. What is the cost of 34 litres of petrol? Solution: Cost of 1 litre of petrol $=$ ₹ $63 \frac{3}{4}$ Cost of 34 litres of petrol $=63 \frac{3}{4} \times 34=\frac{255}{4} \times 34=₹ 2167 \frac{1}{2}$ So, the cost of 34 litres of petrol is ₹ $2167 \frac{1}{2}$....
Read More →The value of the expression
Question: The value of the expression (cos223 sin267) is positive. Solution: False cos223 sin267= (cos23 sin67)(cos23 + sin67) [(a2 b2) = (a b) (a + b)] = [cos23 sin (90 23)] (cos23 + sin67) = (cos23 cos23) (cos23 + sin67) [sin (90 0) = cos 0] = 0.(cos23 + sin67) = 0 which may be either positive or negative....
Read More →Prove the following
Question: $\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=1$ Solution: True $\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=\frac{\tan \left(90^{\circ}-43^{\circ}\right)}{\cot 43^{\circ}}=\frac{\cot 43^{\circ}}{\cot 43^{\circ}}=1$ $\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$...
Read More →Find the area of a square plot of land whose each side measures
Question: Find the area of a square plot of land whose each side measures $8 \frac{1}{2}$ metres. Solution: Area of the square plot $=$ Side $\times$ Side $=(\text { Side })^{2}=a^{2}$ (Because the area of the square is $a^{2}$, where $a$ is the side of the square) $=8 \frac{1}{2} \times 8 \frac{1}{2}$ $=\left(8+\frac{1}{2}\right) \times\left(8+\frac{1}{2}\right)$ $=\frac{17}{2} \times \frac{17}{2}$ $=\frac{17 \times 17}{2 \times 2}$ $=\frac{289}{4}$ $=72 \frac{1}{4} \mathrm{~m}^{2}$ Therefore, ...
Read More →If a pole 6 m high casts a shadow 2√3 m
Question: If a pole 6 m high casts a shadow 23 m long on the ground, then the Suns elevation is (a) 60 (b) 45 (c) 30 (d) 90 Solution: (a) Let $B C=6 \mathrm{~m}$ be the height of the pole and $A B=2 \sqrt{3} \mathrm{~m}$ be the length of the shadow on the ground. let the Sun's makes an angle $\theta$ on the ground. Now, in $\triangle B A C$, $\tan \theta=\frac{B C}{A B}$ $\Rightarrow$ $\tan \theta=\frac{6}{2 \sqrt{3}}=\frac{3}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$ $\Rightarrow$ $\tan \theta...
Read More →Find the area of a rectangular park which is
Question: Find the area of a rectangular park which is $36 \frac{3}{5} \mathrm{~m}$ long and $16 \frac{2}{3} \mathrm{~m}$ broad. Solution: Area of the rectangular park $=$ Length of the park $\times$ Breadth of the park ( $\because$ Area of rectangle $=$ Length $\times$ Breadth) $=36 \frac{3}{5} \times 16 \frac{2}{3}$ $=\left(36+\frac{3}{5}\right) \times\left(16+\frac{2}{3}\right)$ $=\frac{183}{5} \times \frac{50}{3}$ $=\frac{183 \times 50}{5 \times 3}$ $=\frac{9150}{15}$ $=610 \mathrm{~m}^{2}$ ...
Read More →A car is moving at an average speed of
Question: A car is moving at an average speed of $60 \frac{2}{5} \mathrm{~km} / \mathrm{hr}$. How much distance will it cover in $6 \frac{1}{4}$ hours? Solution: Speed $=60 \frac{2}{5} \mathrm{~km} / \mathrm{h}$ Time $=6 \frac{1}{4} \mathrm{~h}$ We know that Speed $=\frac{\text { Distance }}{\text { Time }}$ $\Rightarrow$ Speed $\times$ Time $=$ Distance $\Rightarrow$ Distance $=60 \frac{2}{5} \times 6 \frac{1}{4}$ $\Rightarrow$ Distance $=\frac{302}{5} \times \frac{25}{4}$ $\Rightarrow$ Distanc...
Read More →Show that the function $f$ defined as follows,
Question: Show that the function $f$ defined as follows, is continuous at $x=2$, but not differentiable thereat: $f(x)$$\begin{cases}3 x-2, 0x \leq 1 \\ 2 x^{2}-x, 1x \leq 2 \\ 5 x-4, x2\end{cases}$ Solution: Given: $f(x)= \begin{cases}3 x-2, 0x \leq 1 \\ 2 x^{2}-x, 1x \leq 2 \\ 5 x-4, x2\end{cases}$ First, we will show that $f(x)$ is continuos at $x=2$. We have, $(\mathrm{LHL}$ at $x=2)$ $=\lim _{x \rightarrow 2^{-}} f(x)$ $=\lim _{h \rightarrow 0} f(2-h)$ $=\lim _{h \rightarrow 0} 2(2-h)^{2}-(...
Read More →sin (45° + θ) – cos (45° – θ) is equal to
Question: sin (45 + ) cos (45 ) is equal to (a) 2 cos (b) 0 (c) 2 sin (d) 1 Solution: (b)sin(45 + ) cos(45 ) = cos[90- (45 + )] cos(45- 6) [ cos(90 ) = sin0] = cos (45 0) cos (45 0) = 0...
Read More →Show that the function $f$ defined as follows,
Question: Show that the function $f$ defined as follows, is continuous at $x=2$, but not differentiable thereat: $f(x)$$\begin{cases}3 x-2, 0x \leq 1 \\ 2 x^{2}-x, 1x \leq 2 \\ 5 x-4, x2\end{cases}$ Solution: Given: $f(x)= \begin{cases}3 x-2, 0x \leq 1 \\ 2 x^{2}-x, 1x \leq 2 \\ 5 x-4, x2\end{cases}$ First, we will show that $f(x)$ is continuos at $x=2$. We have, $(\mathrm{LHL}$ at $x=2)$ $=\lim _{x \rightarrow 2^{-}} f(x)$ $=\lim _{h \rightarrow 0} f(2-h)$ $=\lim _{h \rightarrow 0} 2(2-h)^{2}-(...
Read More →Find the cost of
Question: Find the cost of $3 \frac{2}{5}$ metres of cloth at Rs $63 \frac{3}{4}$ per metre. Solution: Cost of $1 \mathrm{~m}$ of cloth $=₹ 63 \frac{3}{4}$ So, cost of $3 \frac{2}{5} \mathrm{~m}$ of cloth $=63 \frac{3}{4} \times 3 \frac{2}{5}$ $=\frac{255}{4} \times \frac{17}{5}$ $=₹ 216 \frac{3}{4}$ So, the cost of $3 \frac{2}{5} \mathrm{~m}$ of cloth is ₹ $216 \frac{3}{4}$....
Read More →Prove the following
Question: If $\sin \theta-\cos \theta=0$, then the value of $\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$ is (a) 1 (b) $\frac{3}{4}$ (c) $\frac{1}{2}$ (d) $\frac{1}{4}$ Solution: (c) Given, $\sin \theta-\cos \theta=0$ $\Rightarrow$ $\sin \theta=\cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=1$ $\Rightarrow$ $\tan \theta=1$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\tan 45^{\circ}=1\right]$ $\Rightarrow$ $\tan \theta=\tan 45^{\circ}$ $\therefore$ $...
Read More →For any sets A, B and C prove that:
Question: For any sets A, B and C prove that: $A \times(B-C)=(A \times B)-(A \times C)$ Solution: Given: A, B and C three sets are given. Need to prove: $A \times(B-C)=(A \times B)-(A \times C)$ Let us consider, $(x, y)^{\in} A \times(B-C)$ $\Rightarrow x \in_{A}$ and $y \in(B-C)$ $\Rightarrow x^{\in} A$ and $\left(y \in_{B}\right.$ and $\left.y \notin C\right)$ $\Rightarrow\left(x^{\in} A\right.$ and $\left.y \in B\right)$ and $\left(x \in_{A}\right.$ and $\left.y \notin C\right)$ $\Rightarrow(...
Read More →On one day a rickshaw puller earned Rs 160. Out of his earnings he spent Rs
Question: On one day a rickshaw puller earned Rs 160 . Out of his earnings he spent Rs $26 \frac{3}{5}$ on tea and snacks, Rs $50 \frac{1}{2}$ on food and Rs $16 \frac{2}{5}$ on repairs of the rickshaw. How much did he save on that day? Solution: Total earning $=$ ₹ 160 Money spent on tea and snacks $=₹ 26 \frac{3}{5}$ Money spent on food $=₹ 50 \frac{1}{2}$ Money spent on repairs $=₹ 16 \frac{2}{5}$ Let the savings be ₹x. Money spent on tea and snacks + Money spent on food + Money spent on repa...
Read More →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 $x=3)=\lim _{x \rightarrow 3^{-}} \frac{f(x)-f(3)}{x-3}$' $=\lim _{x \rightarrow 3} \frac{12 x-13-23}{x-3}$ $=\lim _{x \rightarrow 3} \frac{12 x-36}{x-3}$ ...
Read More →Prove the following
Question: If $4 \tan \theta=3$, then $\left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right)$ is equal to (a) $\frac{2}{3}$ (b) $\frac{1}{3}$ (c) $\frac{1}{2}$ (d) $\frac{3}{4}$ Solution: (c) Given, $4 \tan \theta=3$ $\Rightarrow$ $\tan \theta=\frac{3}{4}$...(i) $\therefore$ $\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}=\frac{4 \frac{\sin \theta}{\cos \theta}-1}{4 \frac{\sin \theta}{\cos \theta}+1}$ [divide by $\cos \theta$ in both numerator and denominator] $=\f...
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