Find the quotient and remainder when
Question: Find the quotient and remainder when $f(x)=x^{4}-5 x+6$ is divided by $g(x)=2-x^{2}$ Solution: We can write $f(x)$ as $x^{4}+0 x^{3}+0 x^{2}-5 x+6$ and $g(x)$ as $-x^{2}+2$ Quotient $q(x)=-x^{2}-2$ Remainder $r(x)=-5 x+10$...
Read More →Find the quotient and remainder when:
Question: Find the quotient and remainder when: $f(x)=x^{4}-3 x^{2}+4 x+5$ is divided by $g(x)=x^{2}+1-x$ Solution: Quotient $q(x)=x^{2}+x-3$ Remainder $r(x)=8$...
Read More →if x = 3 tan t and y = 3 sec t,
Question: If $x=3 \tan t$ and $y=3 \sec t$, then the value of $\frac{d^{2} y}{d x^{2}}$ at $\mathrm{t}=\frac{\pi}{4}$, is:(1) $\frac{1}{3 \sqrt{2}}$(2) $\frac{1}{6 \sqrt{2}}$(3) $\frac{3}{2 \sqrt{2}}$(4) $\frac{1}{6}$Correct Option: 2, Solution: $\because \quad x=3 \tan t \Rightarrow \frac{d x}{d t}=3 \sec ^{2} t$ and $y=3 \sec t \Rightarrow \frac{d y}{d t}=3 \sec t \cdot \tan t$ $\because \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t} \quad \therefore \quad \frac{d y}{d x}=\frac{\tan t}{\sec...
Read More →Find the quotient and the remainder when
Question: Find the quotient and the remainder when $f(x)=x^{3}-3 x^{2}+5 x-3$ is divided by $g(x)=x^{2}-2$ Solution: Quotient $q(x)=x-3$ Remainder $r(x)=7 x-9$...
Read More →When the switch S, in the circuit shown, is closed then the valued of current $i$ will be:
Question: When the switch S, in the circuit shown, is closed then the valued of current $i$ will be: (1) $3 \mathrm{~A}$(2) $5 \mathrm{~A}$(3) $4 \mathrm{~A}$(4) $2 \mathrm{~A}$Correct Option: , 2 Solution: Let voltage at $\mathrm{C}=\mathrm{xV}$ From kirchhoff's current law, $\mathrm{KCL}: \mathrm{i}_{1}+\mathrm{i}_{2}=\mathrm{i}$ $\frac{20-x}{2}+\frac{10-x}{4}=\frac{x-0}{2} \Rightarrow x=10$ $\therefore \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{\mathrm{X}}{\mathrm{R}}=\frac{10}{2}=5 \math...
Read More →Find a cubic polynomial with the sum, sum of the product of its
Question: Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes are 5, 2 and 24 respectively. Solution: We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found asx3(Sum of the zeroes)x2+ (sum of the product of the zeroes taking two at a time)x Product of zeroesTherefore, the required polynomial is $x^{3}-5 x^{2}-2 x+24$...
Read More →The derivative of
Question: The derivative of $\tan ^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)$, with respect to $\frac{x}{2}$, where $\left(x \in\left(0, \frac{\pi}{2}\right)\right)$ is :(1) 1(2) $\frac{2}{3}$(3) $\frac{1}{2}$(4) 2Correct Option: , 4 Solution: $f(x)=\tan ^{-1}\left(\frac{\tan x-1}{\tan x+1}\right)$ $=-\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-x\right)\right) \quad\left[\because \frac{\pi}{4}-x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\right]$ So, $f(x)=-\left(\frac{\pi}{4}-x\right)...
Read More →Find a cubic polynomial whose zeroes are
Question: Find a cubic polynomial whose zeroes are $\frac{1}{2}, 1$ and $-3$. Solution: If the zeroes of the cubic polynomial area,bandcthen the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ ........(1) Let $a=\frac{1}{2}, b=1$ and $c=-3$ Substituting the values in (1), we get $x^{3}-\left(\frac{1}{2}+1-3\right) x^{2}+\left(\frac{1}{2}-3-\frac{3}{2}\right) x-\left(\frac{-3}{2}\right)$ $\Rightarrow x^{3}-\left(\frac{-3}{2}\right) x^{2}-4 x+\frac{3}{2}$ $\Rightarrow ...
Read More →The resistive network shown below is connected to a D.C.
Question: The resistive network shown below is connected to a D.C. source of $16 \mathrm{~V}$. The power consumed by the network is 4 Watt. The value of $R$ is : (1) $6 \Omega$(2) $8 \Omega$(3) $1 \Omega$(4) $16 \Omega$Correct Option: , 2 Solution: (2) Equivalent resistance, $R_{\mathrm{eq}}=\frac{4 R \times 4 R}{4 R+4 R}+R+\frac{6 R \times 12 R}{6 R+12 R}+R$ $=2 R+R+4 R+R$ $=8 R$ Using, $P=\frac{V^{2}}{R_{\mathrm{eq}}}$ $4=\frac{16^{2}}{8 R}$ $\therefore R=\frac{16^{2}}{4 \times 8}=8 \Omega$...
Read More →Find a cubic polynomial whose zeroes are 2, −3 and 4
Question: Find a cubic polynomial whose zeroes are 2, 3 and 4 Solution: If the zeroes of the cubic polynomial area,bandcthen the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ ...........(1) Let $a=2, b=-3$ and $c=4$ Substituting the values in (1), we get $x^{3}-(2-3+4) x^{2}+(-6-12+8) x-(-24)$ $\Rightarrow x^{3}-3 x^{2}-10 x+24$...
Read More →if
Question: If $e^{y}+x y=e$, the ordered pair $\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right)$ at $x=0$ is equal to :(1) $\left(\frac{1}{e},-\frac{1}{e^{2}}\right)$(2) $\left(-\frac{1}{e}, \frac{1}{e^{2}}\right)$(3) $\left(\frac{1}{e}, \frac{1}{e^{2}}\right)$(4) $\left(-\frac{1}{e},-\frac{1}{e^{2}}\right)$Correct Option: , 2 Solution: Given, $e^{y}+x y=e$ ........$\ldots$ (i) Putting $x=0$ in (i), $\Rightarrow e^{y}=e \Rightarrow y=1$ On differentiating (i) w. r. to $x$ $e^{y} \frac{d y}{d...
Read More →Verify that 5, −2 and
Question: Verify that $5,-2$ and $\frac{1}{3}$ are the zeros of the cubic polynomial $p(x)=3 x^{3}-10 x^{2}-27 x+10$ and verify the relation between its zeros and coefficients. Solution: $p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right)$ $p(5)=\left(3 \times 5^{3}-10 \times 5^{2}-27 \times 5+10\right)=(375-250-135+10)=0$ $p(-2)=\left[3 \times\left(-2^{3}\right)-10 \times\left(-2^{2}\right)-27 \times(-2)+10\right]=(-24-40+54+10)=0$ $\left.p\left(\frac{1}{3}\right)=\left\{3 \times\left(\frac{1}{3}\right...
Read More →To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure.
Question: To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained : If $\mathrm{V}_{\mathrm{o}}$ is almost zero, identify the correct statement:(1) The emf of the battery is $1.5 \mathrm{~V}$ and its internal resistance is $1.5 \Omega$(2) The value of the resistance $\mathrm{R}$ is $1.5 \Omega$(3) The potential difference across the battery is $1.5 \mathrm{~...
Read More →If f(1)=1,
Question: If $f(1)=1, f^{\prime}(1)=3$, then the derivative of $f(f(f(x)))+(f(x))^{2}$ at $x=1$ is :(1) 33(2) 12(3) 15(4) 9Correct Option: 1 Solution: Let $g(x)=f(f(f(x)))+(\mathrm{f}(x))^{2}$ Differentiating both sides w.r.t. $x$, we get $g^{\prime}(x)=f^{\prime}(f(f(x))) f^{\prime}(f(x)) f^{\prime}(x)+2 f(x) f^{\prime}(x)$ $g^{\prime}(1)=f^{\prime}(f(f(1))) f^{\prime}(f(1)) f^{\prime}(1)+2 f(1) f^{\prime}(1)$ $=f^{\prime}(f(1)) \mathrm{f}^{\prime}(1) f^{\prime}(1)+2 f(1) f^{\prime}(1)$ $=3 \ti...
Read More →if 2y =
Question: If $2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^{2}, x \in\left(0, \frac{\pi}{6}\right)$ then $\frac{d y}{d x}$ is equal to :(1) $\frac{\pi}{6}-x$(2) $x-\frac{\pi}{6}$(3) $\frac{\pi}{3}-x$(4) $2 x-\frac{\pi}{3}$Correct Option: , 2 Solution: $2 y=\left[\cot ^{-1}\left(\frac{\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x}{\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x}\right)\right]^{2}$ $\Rightarrow 2 y=\left[\cot \left(\frac{\cos \left(\fra...
Read More →Space between two concentric conducting spheres
Question: Space between two concentric conducting spheres of radii $a$ and $b(ba)$ is filled with a medium of resistivity $p$. The resistance between the two spheres will be :(1) $\frac{p}{4 \pi}\left(\frac{1}{a}-\frac{1}{b}\right)$(2) $\frac{p}{2 \pi}\left(\frac{1}{a}-\frac{1}{b}\right)$(3) $\frac{p}{2 \pi}\left(\frac{1}{a}+\frac{1}{b}\right)$(4) $\frac{p}{4 \pi}\left(\frac{1}{a}+\frac{1}{b}\right)$Correct Option: 1 Solution: (1) $d R=\frac{(\rho)(d x)}{4 \pi x^{2}}$ $R=\int d R$ $\int d R=\rho...
Read More →Verity that 3, −2, 1 are the zeros of the cubic polynomial
Question: Verity that $3,-2,1$ are the zeros of the cubic polynomial $p(x)=x^{3}-2 x^{2}-5 x+6$ and verify the relation between its zeros and coefficients. Solution: The given polynomial is $p(x)=\left(x^{3}-2 x^{2}-5 x+6\right)$ $\therefore p(3)=\left(3^{3}-2 \times 3^{2}-5 \times 3+6\right)=(27-18-15+6)=0$ $p(-2)=\left[\left(-2^{3}\right)-2 \times\left(-2^{2}\right)-5 \times(-2)+6\right]=(-8-8+10+6)=0$ $p(1)=\left(1^{3}-2 \times 1^{2}-5 \times 1+6\right)=(1-2-5+6)=0$ $\therefore 3,-2$ and 1 ar...
Read More →In an experiment, the resistance of a material is plotted as a function of temperature (in some range).
Question: In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may canclude that:(1) $R(T)=\frac{R_{0}}{T^{2}}$(2) $R(T)=R_{0} \mathrm{e}^{-T_{0}^{2} / T^{2}}$(3) $\mathrm{R}(\mathrm{T})=\mathrm{R}_{0} \mathrm{e}^{-\mathrm{T}^{2} / \mathrm{T}_{0}^{2}}$(4) $\mathrm{R}(\mathrm{T})=\mathrm{R}_{0} \mathrm{e}^{\mathrm{T}^{2} / \mathrm{T}_{0}^{2}}$Correct Option: 2, Solution: (2) Equation of straight ...
Read More →If the zeros of the polynomial
Question: If the zeros of the polynomial $f(x)=x^{3}-3 x^{2}+x+1$ are $(a-b), a$ and $(a+b)$, Find $a$ and $b$. Solution: By using the relationship between the zeroes of the cubic ploynomial. We have, Sum of zeroes $=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficent of } x^{3}}$ $\therefore a-b+a+a+b=\frac{-(-3)}{1}$ $\Rightarrow 3 a=3$ $\Rightarrow a=1$ Now, Product of zeros $=\frac{-(\text { constant } t e r m)}{\text { coefficent of } x^{3}}$ $\therefore(a-b)(a)(a+b)=\fra...
Read More →if x=2
Question: If $x=2 \sin \theta-\sin 2 \theta$ and $y=2 \cos \theta-\cos 2 \theta, \theta \in[0,2 \pi]$, then $\frac{d^{2} y}{d x^{2}}$ at $\theta=\pi$ is :(1) $\frac{3}{4}$(2) $\frac{3}{8}$(3) $\frac{3}{2}$(4) $-\frac{3}{4}$Correct Option: , 2 Solution: It is given that $x=2 \sin \theta-\sin 2 \theta$$\ldots$ (i) $y=2 \cos \theta-\cos 2 \theta$$\ldots$ (ii) Differentiating (i) w.r.t. $\theta$, we get $\frac{d x}{d \theta}=2 \cos \theta-2 \cos 2 \theta$ Differentiating (ii) w.r.t. $\theta$; we get...
Read More →If α and β are the zeroes of a polynomial
Question: If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x)=x^{2}+x-2$, find the value of $\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)$ Solution: By using the relationship between the zeroes of the quadratic ploynomial.We have, Sum of zeroes $=\frac{-(\text { coefficient of } x)}{\text { coefficent of } x^{2}}$ and Product of zeroes $=\frac{\text { constant term }}{\text { coefficent of } x^{2}}$ $\Rightarrow \alpha+\beta=\frac{-1}{1}$ and $\alpha \beta=\frac{-2}{1}$ $\Rightarrow \a...
Read More →A current of 5 A passes through a copper conductor
Question: A current of $5 \mathrm{~A}$ passes through a copper conductor (resistivity) $=1.7 \times 10^{-8} \Omega \mathrm{m}$ ) of radius of cross-section $5 \mathrm{~mm}$. Find the mobility of the charges if their drift velocity is $1.1 \times 10^{-3} \mathrm{~m} / \mathrm{s}$.(1) $1.8 \mathrm{~m}^{2} / \mathrm{V} \mathrm{s}$(2) $1.5 \mathrm{~m}^{2} / \mathrm{V}_{\mathrm{S}}$(3) $1.3 \mathrm{~m}^{2} / \mathrm{Vs}$(4) $1.0 \mathrm{~m}^{2} / \mathrm{Vs}$Correct Option: , 4 Solution: (4) Charge m...
Read More →If α and β are the zeroes of a polynomial
Question: If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x)=5 x^{2}-7 x+1$, find the value of $\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$ Solution: By using the relationship between the zeroes of the quadratic ploynomial.We have, Sum of zeroes $=\frac{-(\text { coefficient of } x)}{\text { coefficent of } x^{2}}$ and Product of zeroes $=\frac{\text { constant term }}{\text { coefficent of } x^{2}}$ $\therefore \alpha+\beta=\frac{-(-7)}{5}$ and $\alpha \beta=\frac{1}{5}$ $\Rightarr...
Read More →Let f and g be differentiable functions on
Question: Let $f$ and $g$ be differentiable functions on $\mathbf{R}$ such that fog is the identity function. If for some $a, b \in \mathbf{R}, g^{\prime}(a)=5$ and $g$ $(a)=b$, then $f^{\prime}(b)$ is equal to:(1) $\frac{1}{5}$(2) 1(3) 5(4) $\frac{2}{5}$Correct Option: 1, Solution: It is given that functions $f$ and $g$ are differentiable and fog is identity function. $\therefore \quad(f o g)(x)=x \Rightarrow f(g(x))=x$ Differentiating both sides, we get $f^{\prime}(g(x)) \cdot g^{\prime}(x)=1$...
Read More →Obtain all other zeros of
Question: Obtain all other zeros of $\left(x^{4}+4 x^{3}-2 x^{2}-20 x-15\right)$ if two of its zeros are $\sqrt{5}$ and $-\sqrt{5}$. Solution: The given polynomial is $f(x)=x^{4}+4 x^{3}-2 x^{2}-20 x-15$. Since $(x-\sqrt{5})$ and $(x+\sqrt{5})$ are the zeroes of $f(x)$, it follows that each one of $(x-\sqrt{5})$ and $(x+\sqrt{5})$ is a factor of $f(x)$. Consequently, $(x-\sqrt{5})(x+\sqrt{5})=\left(x^{2}-5\right)$ is a factor of $f(x)$. On dividing $f(x)$ by $\left(x^{2}-5\right)$, we get: $\the...
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