If the normal at an end of a latus rectum
Question: If the normal at an end of a latus rectum of an ellipse passes through an extermity of the minor axis, then the eccentricity e of the ellipse satisfies:(1) $e^{4}+2 e^{2}-1=0$(2) $\mathrm{e}^{2}+\mathrm{e}-1=0$(3) $e^{4}+e^{2}-1=0$(4) $\mathrm{e}^{2}+2 \mathrm{e}-1=0$Correct Option: , 3 Solution: Normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $\left(a e, \frac{b^{2}}{a}\right)$ is $\frac{a^{2} x}{a e}-\frac{b^{2} y}{b^{2} / a}=a^{2}-b^{2}$ $\Rightarrow x-e y=\frac...
Read More →If α, β are the zeros of a polynomial such that
Question: If $\alpha, \beta$ are the zeros of a polynomial such that $\alpha+\beta=6$ and $\alpha \beta=4$ the write the polynomial. Solution: If the zeroes of the quadratic polynomial areandthen the quadratic polynomial can be found as $x^{2}-(\alpha+\beta) x+\alpha \beta$ ..........(1) Substituting the values in (1), we get $x^{2}-6 x+4$...
Read More →In the experimental set up of metre bridge shown in the figure,
Question: In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of $40 \mathrm{~cm}$ from A. If a $10 \Omega$ resistor is connected in series with $\mathrm{R}_{1}$, the null point shifts by $10 \mathrm{~cm}$. The resistance that should be connected in parallel with $\left(\mathrm{R}_{1}+10\right) \Omega$ such that the null point shifts back to its initial position is : (1) $20 \Omega$(2) $40 \Omega$(3) $60 \Omega$(4) $30 \Omega$Correct Option: , ...
Read More →Which of the following points lies on the locus
Question: Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ from any of its foci?(1) $(-2, \sqrt{3})$(2) $(-1, \sqrt{2})$(3) $(-1, \sqrt{3})$(4) $(1,2)$Correct Option: 3, Solution: We know that the locus of the feet of the perpendicular draw from foci to any tangent of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is the auxiliary circle $x^{2}+y^{2}=a^{2}$ $\therefore$ Auxiliary circl...
Read More →Find the zeros of the polynomial
Question: Find the zeros of the polynomial $x^{2}-3 x-m(m+3)$ Solution: $f(x)=x^{2}-3 x-m(m+3)$ By adding and subtractingmx, we get $f(x)=x^{2}-m x-3 x+m x-m(m+3)$ $=x[x-(m+3)]+m[x-(m+3)]$ $=[x-(m+3)](x+m)$ $f(x)=0 \Rightarrow[x-(m+3)](x+m)=0$ $\Rightarrow[x-(m+3)]=0$ or $(x+m)=0$ $\Rightarrow x=m+3$ or $x=-m$ So, the zeros off(x) are mandm + 3....
Read More →If the point P on the curve,
Question: If the point $P$ on the curve, $4 x^{2}+5 y^{2}=20$ is farthest from the point $\mathrm{Q}(0,-4)$, then $\mathrm{PQ}^{2}$ is equals to :(1) 36(2) 48(3) 21(4) 29Correct Option: 1, Solution: Ellipse $\equiv \frac{x^{2}}{5}+\frac{y^{2}}{4}=1$ Let a point on ellipse be $(\sqrt{5} \cos \theta, 2 \sin \theta)$ $\therefore P Q^{2}=(\sqrt{5} \cos \theta)^{2}+(-4-2 \sin \theta)^{2}$ $=5 \cos ^{2} \theta+4 \sin ^{2} \theta+16+16 \sin \theta$ $=21+16 \sin \theta-\sin ^{2} \theta$ $=21+64-(\sin \t...
Read More →In the circuit shown, the potential difference between A and B is:
Question: In the circuit shown, the potential difference between $\mathrm{A}$ and $B$ is: (1) $1 \mathrm{~V}$(2) $2 \mathrm{~V}$(3) $3 \mathrm{~V}$(4) $6 \mathrm{~V}$Correct Option: 2, Solution: (2) Given, $E_{1}=1 V, E_{2}=2 V, E_{3}=3 V, r_{1}=1 \Omega$, $r_{2}=1 \Omega$ and $r_{3}=1 \Omega$ $\mathrm{V}_{\mathrm{AB}}=\mathrm{V}_{\mathrm{CD}}=\frac{\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}+\frac{\mathrm{E}_{3}}{\mathrm{r}_{3}}}{\frac{1}{\mathrm{r}_{1}}+\frac{1}...
Read More →If the co-ordinates of two points
Question: If the co-ordinates of two points $\mathrm{A}$ and $\mathrm{B}$ are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively and $P$ is any point on the conic, $9 x^{2}+16 y^{2}=144$, then $\mathrm{PA}+\mathrm{PB}$ is equal to :(1) 16(2) 8(3) 6(4) 9Correct Option: , 2 Solution: Ellipse : $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$, $a=4, b=3, c=\sqrt{16-9}=\sqrt{7}$ $\therefore(\pm \sqrt{7}, 0)$ are the foci of given ellipse. So for any point $P$ on it; $P A+P B=2 a$ $\Rightarrow P A+P B=2(4)=8$...
Read More →In a Wheatstone bridge (see fig.),
Question: In a Wheatstone bridge (see fig.), Resistances $P$ and $Q$ are approximately equal. When $\mathrm{R}=400 \Omega$, the bridge is balanced. On interchanging $P$ and $Q$, the value of $R$, for balance, is $405 \Omega$. The value of $\mathrm{X}$ is close to: (1) $401.5 \mathrm{ohm}$(2) $404.5 \mathrm{ohm}$(3) $403.5 \mathrm{ohm}$(4) $402.5 \mathrm{ohm}$Correct Option: , 4 Solution: (4)...
Read More →Let x=4 be a directrix to an ellipse whose centre is at the
Question: Let $x=4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$. If $P(1, \beta), \beta0$ is a point on this ellipse, then the equation of the normal to it at $P$ is :(1) $4 x-3 y=2$(2) $8 x-2 y=5$(3) $7 x-4 y=1$(4) $4 x-2 y=1$Correct Option: , 4 Solution: $\frac{a}{e}=4 \Rightarrow a=4 \times \frac{1}{2}=2$ Now, $b^{2}=a^{2}\left(1-e^{2}\right)$ $\Rightarrow b^{2}=4\left(1-\frac{1}{4}\right)=4 \times \frac{3}{4}=3$ So, equation $\frac{x^{2}}{...
Read More →Find the zeros of the polynomial
Question: Find the zeros of the polynomial $x^{2}+x-p(p+1)$ Solution: $f(x)=x^{2}+x-p(p+1)$ By adding and subtractingpx, we get $f(x)=x^{2}+p x+x-p x-p(p+1)$ $=x^{2}+(p+1) x-p x-p(p+1)$ $=x[x+(p+1)]-p[x+(p+1)]$ $=[x+(p+1)](x-p)$ $f(x)=0$ $\Rightarrow[x+(p+1)](x-p)=0$ $\Rightarrow[x+(p+1)]=0$ or $(x-p)=0$ $\Rightarrow x=-(p+1)$ or $x=p$ So, the zeros off(x) are (p+ 1) andp....
Read More →Two equal resistances when connected in series to a battery,
Question: Two equal resistances when connected in series to a battery, consume electric power of $60 \mathrm{~W}$. If these resistance are now connected in parallel combination to the same battery, the electric power consumed will be :(1) $60 \mathrm{~W}$(2) $240 \mathrm{~W}$(3) $120 \mathrm{~W}$(4) $30 \mathrm{~W}$Correct Option: , 2 Solution: (2) When two resistances are connected in series, $R_{e q}=2 R$ Power consumed, $P=\frac{\varepsilon^{2}}{R_{e q}}=\frac{\varepsilon^{2}}{2 R}$ In parall...
Read More →If one zero of the polynomial
Question: If one zero of the polynomial $x^{2}-4 x+1$ is $2+\sqrt{3}$. Write the other zero. Solution: Let the other zeroes of $x^{2}-4 x+1$ be $a$. By using the relationship between the zeroes of the quadratic ploynomial. We have, Sum of zeroes $=\frac{-(\text { coefficient of } x)}{\text { coefficent of } x^{2}}$ $\therefore 2+\sqrt{3}+a=\frac{-(-4)}{1}$ $\Rightarrow a=2-\sqrt{3}$ Hence, the other zeroes of $x^{2}-4 x+1$ is $2-\sqrt{3}$....
Read More →If its eccentricity is the maximum value of the function,
Question: Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(ab)$ be a given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, $\phi(t)=\frac{5}{12}+t-t^{2}$, then $a^{2}+b^{2}$ is equal to:(1) 145(2) 116(3) 126(4) 135Correct Option: , 3 Solution: The given ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(ab)$ Length of latus rectum $=\frac{2 b^{2}}{a}$ $\Rightarrow \frac{2 b^{2}}{a}=10 \Rightarrow b^{2}=5 a$......(i) Now $\phi(t)=\frac{5}{...
Read More →The resistance of the meter bridge $mathrm{AB}$
Question: The resistance of the meter bridge $\mathrm{AB}$ in given figure is $4 \Omega$. With a cell of emf $\varepsilon=0.5 \mathrm{~V}$ and rheostat resistance $\mathrm{R}_{\mathrm{h}}=$ $2 \Omega$ the null point is obtained at some point $\mathrm{J}$. When the cell is replaced by another one of $\mathrm{emf} \varepsilon=\varepsilon_{2}$ the same null point $\mathrm{J}$ is found for $\mathrm{R}_{\mathrm{h}}=6 \Omega$. The emf $\varepsilon_{2}$ is, (1) $0.4 \mathrm{~V}$(2) $0.3 \mathrm{~V}$(3)...
Read More →Find all the zeros of the polynomial
Question: Find all the zeros of the polynomial $\left(2 x^{4}-11 x^{3}+7 x^{2}+13 x\right)$, it being given that two if its zeros are $3+\sqrt{2}$ and $3-\sqrt{2}$. Solution: The given polynomial is $f(x)=2 x^{4}-11 x^{3}+7 x^{2}+13 x-7$. Since $(3+\sqrt{2})$ and $(3-\sqrt{2})$ are the zeroes of $f(x)$, it follows that each one of $(x+3+\sqrt{2})$ and $(x+3-\sqrt{2})$ is a factor of $f(x)$. Consequently, $[x-(3+\sqrt{2})][x-(3-\sqrt{2})]=[(x-3)-\sqrt{2}][(x-3)+\sqrt{2}]$ $=\left[(x-3)^{2}-2\righ...
Read More →Let e1 and e2 be the eccentricities of the ellipse,
Question: Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse, $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b5)$ and the hyperbola, $\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$ respectively satisfying $e_{1} e_{2}=1$. If $\alpha$ and $\beta$ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $(\alpha, \beta)$ is equal to :(1) $(8,12)$(2) $\left(\frac{20}{3}, 12\right)$(3) $\left(\frac{24}{5}, 10\right)$(4) $(8,10)$Correct Option: ,...
Read More →Let e1 and e2 be the eccentricities of the ellipse,
Question: Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse, $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b5)$ and the hyperbola, $\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$ respectively satisfying $e_{1} e_{2}=1$. If $\alpha$ and $\beta$ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $(\alpha, \beta)$ is equal to :(1) $(8,12)$(2) $\left(\frac{20}{3}, 12\right)$(3) $\left(\frac{24}{5}, 10\right)$(4) $(8,10)$Correct Option: ,...
Read More →Find all the zeros of
Question: Find all the zeros of $2 x^{4}-3 x^{3}-3 x^{2}+6 x-2$ if it is given that two of its zeros are 1 and $\frac{1}{2}$. Solution: Let $f(x)=2 x^{4}-3 x^{3}-3 x^{2}+6 x-2$ It is given that 1 and $\frac{1}{2}$ are two zeroes of $f(x)$. Thus, $f(x)$ is completely divisible by $(x-1)$ and $\left(x-\frac{1}{2}\right)$. Therefore, one factor of $f(x)$ is $(x-1)\left(x-\frac{1}{2}\right)$ $\Rightarrow$ one factor of $f(x)$ is $\left(x^{2}-\frac{3}{2} x+\frac{1}{2}\right)$ We get another factor of...
Read More →For some
Question: For some $\theta \in\left(0, \frac{\pi}{2}\right)$, if the eccentricity of the hyperbola, $x^{2}-y^{2} \sec ^{2} \theta=10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec ^{2} \theta+y^{2}=5$, then the length of the latus rectum of the ellipse, is :(1) $2 \sqrt{6}$(2) $\sqrt{30}$(3) $\frac{2 \sqrt{5}}{3}$(4) $\frac{4 \sqrt{5}}{3}$Correct Option: , 4 Solution: Hyperbola : $\frac{x^{2}}{10}-\frac{y^{2}}{10 \cos ^{2} \theta}=1 \Rightarrow e_{1}=\sqrt{1+\cos ^{2} \theta}$...
Read More →The Wheatstone bridge shown in Fig.
Question: The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as $R_{1}$ has the colour code (Orange, Red, Brown). The resistors $R_{2}$ and $R_{4}$ are $80 \Omega$ and $40 \Omega$, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as $R_{3}$, would be: (1) Brown, Blue, Brown(2) Brown, Blue, Black(3) Red, Green, Brown(4) Grey, Black, BrownCorrect Option: 1 Solution: (1)...
Read More →One zero of the polynomial
Question: One zero of the polynomial $3 x^{3}+16 x^{2}+15 x-18$ is $\frac{2}{3}$. Find the other zeros of the polynomial. Solution: Let $f(x)=3 x^{3}+16 x^{2}+15 x-18$ It is given that one of its zeroes is $\frac{2}{3}$. Therefore, one factor of $f(x)$ is $\left(x-\frac{2}{3}\right)$ We get another factor of $f(x)$ by dividing it with $\left(x-\frac{2}{3}\right)$. On division, we get the quotient $3 x^{2}+18 x+27$ $\Rightarrow f(x)=\left(x-\frac{2}{3}\right)\left(3 x^{2}+18 x+27\right)$ $=\left(...
Read More →A current of 2mA was passed through an unknown resistor which dissipated a power of 4.4 W.
Question: A current of $2 \mathrm{~mA}$ was passed through an unknown resistor which dissipated a power of $4.4 \mathrm{~W}$. Dissipated power when an ideal power supply of $11 \mathrm{~V}$ is connected across it is:(1) $11 \times 10^{-5} \mathrm{~W}$(2) $11 \times 10^{-3} \mathrm{~W}$(3) $11 \times 10^{-4} \mathrm{~W}$(4) $11 \times 10^{5} \mathrm{~W}$Correct Option: 1 Solution: (1) Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $4.4=4 \times 10^{-6} \times \mathrm{R}$ $\Rightarrow \mathrm{R}=1....
Read More →The actual value of resistance R,
Question: The actual value of resistance $R$, shown in the figure is $30 \Omega$. This is measured in an experiment as shown using the standard formula $R=\frac{V}{I}$, where $V$ and $I$ are the reading of the voltmeter and ammeter, respectively. If the measured value of $\mathrm{R}$ is $5 \%$ less, then the internal resistance of the voltmeter is: (1) $600 \Omega$(2) $570 \Omega$(3) $35 \Omega$(4) $350 \Omega$Correct Option: , 2 Solution: (2) using, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{...
Read More →Find all the zeros of
Question: Find all the zeros of $2 x^{4}-13 x^{3}+19 x^{2}+7 x-3$ if two of its zeros are $(2+\sqrt{3})$ and $(2-\sqrt{3})$. Solution: Let $f(x)=2 x^{4}-13 x^{3}+19 x^{2}+7 x-3$ It is given that $(2+\sqrt{3})$ and $(2-\sqrt{3})$ are two zeroes of $f(x)$ Thus, $f(x)$ is completely divisible by $(x-2-\sqrt{3})$ and $(x-2+\sqrt{3})$. Therefore, one factor of $f(x)$ is $\left((x-2)^{2}-3\right)$ $\Rightarrow$ one factor of $f(x)$ is $\left(x^{2}-4 x+1\right)$ We get another factor of $f(x)$ by divid...
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