Examine the applicability of Mean Value Theorem
Question: Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2. Solution: Mean Value Theorem states that for a function $f:[a, b] \rightarrow \mathbf{R}$, if (a)fis continuous on [a,b] (b)fis differentiable on (a,b) then, there exists some $c \in(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis. (i) $f(x)=[x]$ fo...
Read More →The cable of a uniformly loaded suspension bridge hangs in the form of a parabola.
Question: The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle. Solution: The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken a...
Read More →Express each of the following decimals in the form
Question: Express each of the following decimals in the form $\frac{p}{q}$ : (i) $0.39$ (ii) $0.750$ (iii) $2.15$ (iv) $7.010$ (v) $9.90$ (vi) $1.0001$ Solution: (i) Given decimal is $0.39$ Now we have to convert given decimal number into the $\frac{p}{q}$ form Let $\frac{p}{q}=0.39$ $\Rightarrow \frac{p}{q}=\frac{39}{100}$ (ii) Given decimal is $0.750$ Now we have to convert given decimal number into $\frac{p}{q}$ form Let $\frac{p}{q}=0.750$ $\Rightarrow \frac{p}{q}=\frac{750}{1000}$ $\Rightar...
Read More →An arch is in the form of a parabola with its axis vertical.
Question: An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? Solution: The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the negativey-axis. This can be diagrammatically represented as The equation of the parabola is of the form $x^{2}=-4 a y$ (as it is opening downwards). It can be clearly seen that the parabola passes th...
Read More →If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Question: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. Solution: The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positivex-axis. This can be diagrammatically represented as The equation of the parabola is of the form $y^{2}=4 a x$ (as it is opening to the right). Since the parabola passes through point $A$ ( 5,10 ), $10^{2}=4 a(5)$ $\Rightarrow 100=20 a$ $\Rightarrow a=\...
Read More →Find the equation of the hyperbola satisfying the give conditions: Foci
Question: Find the equation of the hyperbola satisfying the give conditions: Foc $(0, \pm \sqrt{10})$, passing through $(2,3)$ Solution: Foci $(0, \pm \sqrt{10})$, passing through $(2,3)$ Here, the foci are on they-axis. Therefore, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$. Since the foci are $(0, \pm \sqrt{10}), c=\sqrt{10}$. We know that $a^{2}+b^{2}=c^{2}$. $\therefore a^{2}+b^{2}=10$ $\Rightarrow b^{2}=10-a^{2}$ (1) Since the hyperbola passes th...
Read More →Look at several examples of rational numbers in the form
Question: Look at several examples of rational numbers in the form $\frac{p}{q}(q \neq 0)$, where $p$ and $q$ are integers with no common factors other than 1 and having terminating decimal representations. Can you guess what property $q$ must satisfy? Solution: Prime factorization is the process of finding which prime numbers you need to multiply together to get a certain number. So prime factorization of denominators (q)must have only the power of 2or 5 or both....
Read More →Express the following rational numbers as decimals:
Question: Express the following rational numbers as decimals: (i) $\frac{2}{3}$ (ii) $-\frac{4}{9}$ (iii) $\frac{-2}{15}$ (iv) $-\frac{22}{13}$ (v) $\frac{437}{999}$ (vi) $\frac{33}{26}$ Solution: (i) Given rational number is $\frac{2}{3}$ Now we have to express this rational number into decimal form. So we will use long division method (ii) Given rational number is $-\frac{4}{9}$ Now we have to express this rational number into decimal form. So we will use long division method (iii) Given ratio...
Read More →Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0),
Question: Find the equation of the hyperbola satisfying the give conditions: Vertices $(\pm 7,0), e=\frac{4}{3}$ Solution: Vertices $(\pm 7,0), e=\frac{4}{3}$ Here, the vertices are on the $x$-axis. Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Since the vertices are $(\pm 7,0), a=7$. It is given that $e=\frac{4}{3}$ $\therefore \frac{c}{a}=\frac{4}{3}$ $\left[e=\frac{c}{a}\right]$ $\Rightarrow \frac{c}{7}=\frac{4}{3}$ $\Rightarrow c=\frac{2...
Read More →Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0),
Question: Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 4,0)$, the latus rectum is of length 12 Solution: Foci (4, 0), the latus rectum is of length 12. Here, the foci are on thex-axis. Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Since the foci are (4, 0),c= 4. Length of latus rectum = 12 $\Rightarrow \frac{2 b^{2}}{a}=12$ $\Rightarrow b^{2}=6 a$ We know that $a^{2}+b^{2}=c^{2}$. $\therefore a^{2}+6 a=16$ $\...
Read More →Verify Mean Value Theorem,
Question: Verify Mean Value Theorem, if $f(x)=x^{3}-5 x^{2}-3 x$ in the interval $[a, b]$, where $a=1$ and $b=3$. Find all $c \in(1,3)$ for which $f^{\prime}(c)=0$ Solution: The given function $f$ is $f(x)=x^{3}-5 x^{2}-3 x$ $f$, being a polynomial function, is continuous in $[1,3]$ and is differentiable in $(1,3)$ whose derivative is $3 x^{2}-10 x-3 .$ $f(1)=1^{3}-5 \times 1^{2}-3 \times 1=-7, f(3)=3^{3}-5 \times 3^{2}-3 \times 3=-27$ $\therefore \frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}=\fra...
Read More →Find the equation of the hyperbola satisfying the give conditions:
Question: Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 3 \sqrt{5}, 0)$, the latus rectum is of length $8 .$ Solution: Foci $(\pm 3 \sqrt{5}, 0)$, the latus rectum is of length 8 . Here, the foci are on thex-axis. Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Since the foci are $(\pm 3 \sqrt{5}, 0), c=\pm 3 \sqrt{5}$. Length of latus rectum $=8$ $\Rightarrow \frac{2 b^{2}}{a}=8$ $\Rightarrow b^{2}=4 a$ We know...
Read More →Verify Mean Value Theorem,
Question: Verify Mean Value Theorem, if $f(x)=x^{2}-4 x-3$ in the interval $[a, b]$, where $a=1$ and $b=4$. Solution: The given function is $f(x)=x^{2}-4 x-3$ f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x 4. $f(1)=1^{2}-4 \times 1-3=-6, f(4)=4^{2}-4 \times 4-3=-3$ $\therefore \frac{f(b)-f(a)}{b-a}=\frac{f(4)-f(1)}{4-1}=\frac{-3-(-6)}{3}=\frac{3}{3}=1$ Mean Value Theorem states that there is a point $c \in(1,4)$ such that $f^{\prime...
Read More →Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13),
Question: Find the equation of the hyperbola satisfying the give conditions: Foci $(0, \pm 13)$, the conjugate axis is of length 24 . Solution: Foci $(0, \pm 13)$, the conjugate axis is of length 24 . Here, the foci are on the $y$-axis. Therefore, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$. Since the foci are $(0, \pm 13), c=13$. Since the length of the conjugate axis is $24,2 b=24 \Rightarrow b=12$. We know that $a^{2}+b^{2}=c^{2}$. $\therefore a^{2...
Read More →Express the following rational numbers as decimals:
Question: Express the following rational numbers as decimals: (i) $\frac{42}{100}$ (ii) $\frac{327}{500}$ (iii) $\frac{15}{4}$ Solution: (i) Given rational number is $\frac{42}{100}$ Now we have to express this rational number into decimal form. So we will use long division method as below. (ii) Given rational number is $\frac{327}{500}$ Now we have to express this rational number into decimal form. So we will use long division method as below. (iii) Given rational number is $\frac{15}{4}$ Now w...
Read More →If is a differentiable function and if does not vanish anywhere, then prove that.
Question: If $f:[-5,5] \rightarrow \mathbf{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5)$. Solution: It is given that $f:[-5,5] \rightarrow \mathbf{R}$ is a differentiable function. Since every differentiable function is a continuous function, we obtain (a)fis continuous on [5, 5]. (b)fis differentiable on (5, 5). Therefore, by the Mean Value Theorem, there existsc (5, 5) such that $f^{\prime}(c)=\frac{f(5)-f(-5)}{5-(-5)}$ $\Ri...
Read More →Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the transverse axis is of length 8.
Question: Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 5,0)$, the transverse axis is of length 8 . Solution: Foci (5, 0), the transverse axis is of length 8. Here, the foci are on thex-axis. Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Since the foci are (5, 0),c= 5. Since the length of the transverse axis is 8, 2a= 8 ⇒a= 4. We know that $a^{2}+b^{2}=c^{2}$. $\therefore 4^{2}+b^{2}=5^{2}$ $\Rightarrow b^{2}=...
Read More →Examine if Rolle’s Theorem is applicable to any of the following functions.
Question: Examine if Rolles Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolles Theorem from these examples? (i) $f(x)=[x]$ for $x \in[5,9]$ (ii) $f(x)=[x]$ for $x \in[-2,2]$ (iii) $f(x)=x^{2}-1$ for $x \in[1,2]$ Solution: By Rolle's Theorem, for a function $f:[a, b] \rightarrow \mathbf{R}$, if (a) $f$ is continuous on $[a, b]$ (b) $f$ is differentiable on $(a, b)$ (c) $f(a)=f(b)$ then, there exists some $c \in(a, b)$ such that $f^{\prime}...
Read More →Find the equation of the hyperbola satisfying the give conditions:
Question: Find the equation of the hyperbola satisfying the give conditions: Vertices $(0, \pm 3)$, foci $(0, \pm 5)$ Solution: Vertices $(0, \pm 3)$, foci $(0, \pm 5)$ Here, the vertices are on the $y$-axis. Therefore, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$. Since the vertices are $(0, \pm 3), a=3$. Since the foci are $(0, \pm 5), c=5$. We know that $a^{2}+b^{2}=c^{2}$. $\therefore 3^{2}+b^{2}=5^{2}$ $\Rightarrow b^{2}=25-9=16$ Thus, the equatio...
Read More →Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci (0, ±8)
Question: Find the equation of the hyperbola satisfying the give conditions: Vertices $(0, \pm 5)$, foci $(0, \pm 8)$ Solution: Vertices $(0, \pm 5)$, foci $(0, \pm 8)$ Here, the vertices are on the $y$-axis. Therefore, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$. Since the vertices are $(0, \pm 5), a=5$. Since the foci are $(0, \pm 8), c=8$. We know that $a^{2}+b^{2}=c^{2}$. $\therefore 5^{2}+b^{2}=8^{2}$ $b^{2}=64-25=39$ Thus, the equation of the hy...
Read More →Verify Rolle’s Theorem for the function
Question: Verify Rolle's Theorem for the function $f(x)=x^{2}+2 x-8, x \in[-4,2]$ Solution: The given function, $f(x)=x^{2}+2 x-8$, being a polynomial function, is continuous in $[-4,2]$ and is differentiable in $(-4,2)$. $f(-4)=(-4)^{2}+2 \times(-4)-8=16-8-8=0$ $f(2)=(2)^{2}+2 \times 2-8=4+4-8=0$ $\therefore f(-4)=f(2)=0$ $\Rightarrow$ The value of $f(x)$ at $-4$ and 2 coincides. Rolle's Theorem states that there is a point $c \in(-4,2)$ such that $f^{\prime}(c)=0$ $f(x)=x^{2}+2 x-8$ $\Rightarr...
Read More →Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0)
Question: Find the equation of the hyperbola satisfying the give conditions: Vertices$(\pm 2,0)$, foci $(\pm 3,0)$ Solution: Vertices $(\pm 2,0)$, foci $(\pm 3,0)$ Here, the vertices are on thex-axis. Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Since the vertices are $(\pm 2,0), a=2$. Since the foci are $(\pm 3,0), c=3$. We know that $a^{2}+b^{2}=c^{2}$. $\therefore 2^{2}+b^{2}=3^{2}$ $b^{2}=9-4=5$ Thus, the equation of the hyperbola is $\...
Read More →Discuss briefly the following:
Question: Discuss briefly the following: (a)Greenhouse gases (b)Catalytic converter (c)Ultraviolet B Solution: (a) Greenhouse gases: -Thegreenhouse effect refers to an overall increase in the average temperature of the Earth due to the presence of greenhouse gases. Greenhouse gases mainly consist of carbon dioxide, methane, and water vapour. When solar radiations reach the Earth, some of these radiations are absorbed. These absorbed radiations are released back into the atmosphere. These radiati...
Read More →If, show that
Question: If $y=\left(\tan ^{-1} x\right)^{2}$, show that $\left(x^{2}+1\right)^{2} y_{2}+2 x\left(x^{2}+1\right) y_{1}=2$ Solution: The given relationship is $y=\left(\tan ^{-1} x\right)^{2}$ Then, $y_{1}=2 \tan ^{-1} x \frac{d}{d x}\left(\tan ^{-1} x\right)$ $\Rightarrow y_{1}=2 \tan ^{-1} x \cdot \frac{1}{1+x^{2}}$ $\Rightarrow\left(1+x^{2}\right) y_{1}=2 \tan ^{-1} x$ Again differentiating with respect to $x$ on both the sides, we obtain $\left(1+x^{2}\right) y_{2}+2 x y_{1}=2\left(\frac{1}{...
Read More →What initiatives were taken for reducing vehicular air pollution in Delhi?
Question: What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi? Solution: Delhi has been categorized as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi. Various steps have been taken to improve the quality of air in Delhi. (a)Introduction of CNG (Compressed Natural Gas):By the orderof the supreme court of India, CNG-powered vehicles were introduced at the ...
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