There are two long co-axial solenoids of same length l.
Question: There are two long co-axial solenoids of same length $l$. The inner and outer coils have radii $r_{1}$ and $r_{2}$ and number of turns per unit length $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$, respectively. The ratio of mutual inductance to the self-inductance of the inner-coil is:(1) $\frac{n_{1}}{n_{2}}$(2) $\frac{n_{2}}{n_{1}} \cdot \frac{r_{1}}{r_{2}}$(3) $\frac{n_{2}}{n_{1}} \cdot \frac{r_{2}^{2}}{r_{1}^{2}}$(4) $\frac{n_{2}}{n_{1}}$Correct Option: , 4 Solution: (4) The rate of mutua...
Read More →Let f(1,3)
Question: Let $f(1,3) \rightarrow R$ be a function defined by $f(x)=\frac{x[x]}{1+x^{2}}$, where $[x]$ denotes the greatest integer $\leq x$. Then the range of $f$ is:(1) $\left(\frac{2}{5}, \frac{3}{5}\right) \cup\left(\frac{3}{4}, \frac{4}{5}\right)$(2) $\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right)$(3) $\left(\frac{2}{5}, \frac{4}{5}\right)$(4) $\left(\frac{3}{5}, \frac{4}{5}\right)$Correct Option: , 2 Solution: $f(x) \begin{cases}\frac{x}{x^{2}+1} ; x \in(1,...
Read More →If α, β, γ are the zeroes of the polynomial
Question: If $\alpha, \beta, y$ are the zeroes of the polynomial $p(x)=6 x^{3}+3 x^{2}-5 x+1$, find the value of $\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)$ Solution: Given : $p(x)=6 x^{3}+3 x^{2}-5 x+1$ $=6 x^{2}-(-3) x^{2}+(-5) x-(-1)$ Comparing the polynomial with $x^{3}-x^{2}(\alpha+\beta+\gamma)+x(\alpha \beta+\beta \gamma+\gamma \alpha)-\alpha \beta \gamma$, we get: $\alpha \beta+\beta \gamma+\gamma \alpha=-5$ and $\alpha \beta \gamma=-1$ $\therefore\left(\frac{1}{\alp...
Read More →The self induced emf of a coil is 25 volts.
Question: The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from $10 \mathrm{~A}$ to $25 \mathrm{~A}$ in $1 \mathrm{~s}$, the change in the energy of the inductance is:(1) $740 \mathrm{~J}$(2) $437.5 \mathrm{~J}$(3) $540 \mathrm{~J}$(4) $637.5 \mathrm{~J}$Correct Option: , 2 Solution: (2) According to faraday's law of electromagnetic induc- tion, $\mathrm{e}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ $\mathrm{L} \times \frac{\mathrm{di}}{\mathrm{dt}}=25 \Ri...
Read More →The inverse function of
Question: The inverse function of $f(x)=\frac{8^{2 x}-8^{-2 x}}{8^{2 x}+8^{-2 x}}, x \in(-1,1)$, is(1) $\frac{1}{4} \log _{e}\left(\frac{1+x}{1-x}\right)$(2) $\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1-x}{1+x}\right)$(3) $\frac{1}{4} \log _{e}\left(\frac{1-x}{1+x}\right)$(4) $\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1+x}{1-x}\right)$Correct Option: 1 Solution: $y=\frac{8^{2 x}-8^{-2 x}}{8^{2 x}+8^{-2 x}}$ $\frac{1+y}{1-y}=\frac{8^{2 x}}{8^{-2 x}} \Rightarrow 8^{4 ...
Read More →Show that (x + 2) is a factor of f(x)
Question: Show that $(x+2)$ is a factor of $f(x)=x^{3}+4 x^{2}+x-6$ Solution: Given: $f(x)=x^{3}+4 x^{2}+x-6$ Now, $f(-2)=(-2)^{3}+4(-2)^{2}+(-2)-6$ $=-8+16-2-6$ $=0$ $\therefore(x+2)$ is a factor of $f(x)=x^{3}+4 x^{2}+x-6$....
Read More →Using remainder theorem, find the remainder when p(x)
Question: Using remainder theorem, find the remainder when $p(x)=x^{3}+3 x^{2}-5 x+4$ is divided by $(x-2)$. Solution: Given : $p(x)=x^{3}+3 x^{2}-5 x+4$ Now, $p(2)=2^{3}+3\left(2^{2}\right)-5(2)+4$ $=8+12-10+4$ $=14$...
Read More →Find a cubic polynomial whose zeros are 3, 5 and −2.
Question: Find a cubic polynomial whose zeros are 3, 5 and 2. Solution: Let $\alpha, \beta$ and $\gamma$ be the zeroes of the required polynomial. Then we have : $\alpha+\beta+\gamma=3+5+(-2)=6$ $\alpha \beta+\beta \gamma+\gamma \alpha=3 \times 5+5 \times(-2)+(-2) \times 3=-1$ and $\alpha \beta \gamma=3 \times 5 \times-2=-30$ Now, $p(x)=x^{3}-x^{2}(\alpha+\beta+\gamma)+x(\alpha \beta+\beta \gamma+\gamma \alpha)-\alpha \beta \gamma$ $=x^{3}-x^{2} \times 6+x \times(-1)-(-30)$ $=x^{3}-6 x^{2}-x+30$...
Read More →if g(x)
Question: If $g(x)=x^{2}+x-1$ and $(g \circ f)(x)=4 x^{2}-10 x+5$, then $f\left(\frac{5}{4}\right)$ is equal to:(1) $\frac{3}{2}$(2) $-\frac{1}{2}$(3) $\frac{1}{2}$(4) $-\frac{3}{2}$Correct Option: , 2 Solution: $(g \circ f)(x)=g(f(x))=f^{2}(x)+f(x)-1$ $g\left(f\left(\frac{5}{4}\right)\right)=4\left(\frac{5}{4}\right)^{2}-10 \cdot \frac{5}{4}+5=-\frac{5}{4}$ $\left[\because g(f(x))=4 x^{2}-10 x+5\right]$ $g\left(f\left(\frac{5}{4}\right)\right)=f^{2}\left(\frac{5}{4}\right)+f\left(\frac{5}{4}\ri...
Read More →A power transmission line feeds input power
Question: A power transmission line feeds input power at $2300 \mathrm{~V}$ to a step down transformer with its primary windings having 4000 turns. The output power is delivered at $230 \mathrm{~V}$ by the transformer. If the current in the primary of the transformer is $5 \mathrm{~A}$ and its efficiency is $90 \%$, the output current would be:(1) $50 \mathrm{~A}$(2) $45 \mathrm{~A}$(3) $35 \mathrm{~A}$(4) $25 \mathrm{~A}$Correct Option: , 2 Solution: (2) Efficiency, $\eta=\frac{P_{\text {out }}...
Read More →For a suitably chosen real constant a
Question: For a suitably chosen real constant a, let a function, $f: \mathrm{R}-\{-\mathrm{a}\} \rightarrow \mathrm{R}$ be defined by $f(x)=\frac{\mathrm{a}-x}{\mathrm{a}+x}$. Further suppose that for any real number $x \neq-a$ and $f(x) \neq-a$, $(f o f)(x)=x$. Then $f\left(-\frac{1}{2}\right)$ is equal to:(1) $\frac{1}{3}$(2) $-\frac{1}{3}$(3) $-3$(4) 3Correct Option: , 4 Solution: $f(f(x))=\frac{a-\left(\frac{a-x}{a+x}\right)}{a+\left(\frac{a-x}{a+x}\right)}=x$ $\Rightarrow \frac{a-a x}{1+x}=...
Read More →The figure shows a square loop L of side 5cm which is connected to a network of resistances.
Question: The figure shows a square loop $\mathrm{L}$ of side $5 \mathrm{~cm}$ which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of $1 \mathrm{~cm} \mathrm{~s}^{-1}$. At some instant, a part of $\mathrm{L}$ is in a uniform magnetic field of 1 $\mathrm{T}$, perpendicular to the plane of the loop. If the resistance of $\mathrm{L}$ is $1.7 \Omega$, the current in the loop at that instant will be close to : (1) $60 \mu \mathrm{A}$(2) $170 \...
Read More →Verify that 2 is a zero of the polynomial
Question: Verify that 2 is a zero of the polynomial $x^{3}+4 x^{2}-3 x-18$ Solution: Let $p(\mathrm{x})=x^{3}+4 x^{2}-3 x-18$ Now, $p(2)=2^{3}+4 \times 2^{2}-3 \times 2-18=0$ $\therefore 2$ is a zero of $p(x)$....
Read More →If f(x+y) =
Question: If $f(x+y)=f(x) f(y)$ and $\sum_{x=1}^{\infty} f(x)=2, x, y \in \mathrm{N}$, where $\mathrm{N}$ is the set of all natural numbers, then the value of $\frac{f(4)}{f(2)}$ is :(1) $\frac{2}{3}$(2) $\frac{1}{9}$(3) $\frac{1}{3}$(4) $\frac{4}{9}$Correct Option: , 4 Solution: Let $f(1)=k$, then $f(2)=f(1+1)=k^{2}$ $f(3)=f(2+1)=k^{3}$ $\sum_{x=1}^{\infty} f(x)=2 \Rightarrow k+k^{2}+k^{3}+\ldots \ldots \infty=2$ $\Rightarrow \frac{k}{1-k}=2 \Rightarrow k=\frac{2}{3}$ Now, $\frac{f(4)}{f(2)}=\f...
Read More →If the zeroes of the polynomial
Question: If the zeroes of the polynomial $x^{3}-3 x^{2}+x+1$ are $(a-b), a$ and $(a+b)$, find the values of $a$ and $b$. Solution: The given polynomial $=x^{3}-3 x^{2}+x+1$ and its roots are $(a-b), a$ and $(a+b)$. Comparing the given polynomial with $A x^{3}+B x^{2}+C x+D$, we have : $A=1, B=-3, C=1$ and $D=1$ Now, $(a-b)+a+(a+b)=\frac{-B}{A}$ $=3 a=-\frac{-3}{1}$ $=a=1$ Also, $(a-b) \times a \times(a+b)=\frac{-D}{A}$ $=a\left(a^{2}-b^{2}\right)=\frac{-1}{1}$ $=1\left(1^{2}-b^{2}\right)=-1$ $=...
Read More →Let A={a, b, c} and B={1,2,3,4}.
Question: Let $A=\{a, b, c\}$ and $B=\{1,2,3,4\}$. Then the number of elements in the set $C=\{f: A \rightarrow B \mid 2 \in f(A)$ and $f$ is not one-one $\}$ is Solution: The desired functions will contain either one element or two elements in its codomain of which ' 2 ' always belongs to $f(A)$. $\therefore$ The set $B$ can be $\{2\},\{1,2\},\{2,3\},\{2,4\}$ Total number of functions $=1+\left(2^{3}-2\right) 3=19$....
Read More →A thin ring of 10 cm radius carries a uniformly distributed charge.
Question: A thin ring of $10 \mathrm{~cm}$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40 \pi \mathrm{rad} \mathrm{s}^{-1}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8 \times 10^{-9} \mathrm{~T}$, then the charge carried by the ring is close to $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{~N} / \mathrm{A}^{2}\right)$.(1) $2 \times 10^{-6} \mathrm{C}$(2) $3 \times 10^{-5} \mathrm{C}$(3) $4 \times 10^{-5...
Read More →Suppose a differentiable function
Question: Suppose a differentiable function $f(x)$ satisfies the identity $f(x+y)=f(x)+f(y)+x y^{2}+x^{2} y$, for all real $x$ and $y$. If $\lim _{x \rightarrow 0} \frac{f(x)}{x}=1$, then $f^{\prime}(3)$ is equal to_______. Solution: $f(x+y)=f(x)+f(y)+x y^{2}+x^{2} y$ Differentiate w.r.t. $x$ : $f^{\prime}(x+y)=f^{\prime}(x)+0+y^{2}+2 x y$ Put $y=-x$ $f^{\prime}(0)=f^{\prime}(x)+x^{2}-2 x^{2}$....(i) $\because \lim _{x \rightarrow 0} \frac{f(x)}{x}=1 \Rightarrow f(0)=0$ $\therefore f^{\prime}(0)...
Read More →Find a quadratic polynomial whose zeros are 2 and −5.
Question: Find a quadratic polynomial whose zeros are 2 and 5. Solution: It is given that the two roots of the polynomial are 2 and5. Let $\alpha=2$ and $\beta=-5$ Now, sum of the zeroes, $\alpha+\beta=2+(-5)=-3$ Product of the zeroes, $\alpha \beta=2 \times-5=-10$ $\therefore$ Required polynomial $=x^{2}-(\alpha+\beta) x+\alpha \beta$ $=x^{2}-(-3) x+(-10)$ $=x^{2}+3 x-10$...
Read More →If one zero of the polynomial
Question: If one zero of the polynomial $\left(a^{2}+9\right) x^{2}+13 x+6 a$ is the reciprocal of the other, find the value of $a$. Solution: $(a+9) x^{2}-13 x+6 a=0$ Here, $A=\left(a^{2}+9\right), B=13$ and $C=6 a$ Let $\alpha$ and $\frac{1}{\alpha}$ be the two zeroes. Then, product of the zeroes $=\frac{C}{A}$ $=\alpha \cdot \frac{1}{\alpha}=\frac{6 a}{a^{2}+9}$ $=1=\frac{6 a}{a^{2}+9}$ $=a^{2}+9=6 a$ $=a^{2}-6 a+9=0$ $=a^{2}-2 \times a \times 3+3^{2}=0$ $=(a-3)^{2}=0$ $=a-3=0$ $=a=3$...
Read More →Let [t] denote the greatest integer
Question: Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x,[x]^{2}+2[x+2]-7=0$ has:(1) exactly two solutions(2) exactly four integral solutions(3) no integral solution(4) infinitely many solutionsCorrect Option: , 4 Solution: The given equation $[x]^{2}+2[x]+4-7=0$ $\Rightarrow[x]^{2}+2[x]-3=0$ $\Rightarrow[x]^{2}+3[x]-[x]-3=0$ $\Rightarrow([x]+3)([x]-1)=0 \Rightarrow[x]=1$ or $-3$ $\Rightarrow x \in[-3,-2) \cup[1,2)$ $\therefore$ The equation has infinitely many solutions...
Read More →Two coils 'P' and ' Q ' are separated by some distance.
Question: Two coils 'P' and ' $Q$ ' are separated by some distance. When a current of $3 A$ flows through coil ' $P$ ', a magnetic flux of $10^{-3} \mathrm{~Wb}$ passes through ' $\mathrm{Q}$ '. No current is passed through ' $Q$ '. When no current passes through ' $P$ ' and a current of 2 A passes through ' $Q$ ', the flux through ' $P$ ' is:(1) $6.67 \times 10^{-4} \mathrm{~Wb}$(2) $3.67 \times 10^{-3} \mathrm{~Wb}$(3) $6.67 \times 10^{-3} \mathrm{~Wb}$(4) $3.67 \times 10^{-4} \mathrm{~Wb}$Cor...
Read More →Find the zeros of the polynomial
Question: Find the zeros of the polynomial $x^{2}+2 x-195$ Solution: Here, $p(x)=x^{2}+2 x-195$ Let $p(x)=0$ $=\mathrm{x}^{2}+(15-13) \mathrm{x}-195=0$ $=\mathrm{x}^{2}+15 \mathrm{x}-13 \mathrm{x}-195=0$ $=\mathrm{x}(\mathrm{x}+15)-13(\mathrm{x}+15)=0$ $=(\mathrm{x}+15)(\mathrm{x}-13)=0$ $=\mathrm{x}=-15,13$ Hence, the zeroes are $-15$ and 13 ....
Read More →Let f:
Question: Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function which satisfies $f(x+y)=f(x)+f(y), \forall x, y \in \mathbf{R}$. If $f(1)=2$ and $g(n)=\sum_{k=1}^{(n-1)} f(k), n \in \mathbf{N}$, then the value of $n$, for which $g(n)=20$, is :(1) 5(2) 20(3) 4(4) 9Correct Option: 1 Solution: Given: $f(x+y)=f(x)+f(y), \forall x, y \in R, f(1)=2$ $\Rightarrow f(2)=f(1)+f(1)=2+2=4$ $f(3)=f(1)+f(2)=2+4=6$ $f(n-1)=2(n-1)$ Now, $g(n)=\sum_{k=1}^{n-1} f(k)$ $=f(1)+f(2)+f(3)+\ldots . f(n-1)$ $=2+4+6+\...
Read More →A very long solenoid of radius R
Question: A very long solenoid of radius $R$ is carrying current $\mathrm{I}(\mathrm{t})=\mathrm{kte}^{-\alpha t}(k0)$, as a function of time $(t \geq 0)$. Counter clockwise current is taken to be positive. A circular conducting coil of radius $2 R$ is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by:Correct Option: 1 Solution: (1) $Q=B A$ $=\left(\mu_{0} n_{i}\right) A$ $=\mu_{...
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