A quadratic polynomial whose zeros are 5 and −3, is
Question: A quadratic polynomial whose zeros are 5 and 3, is (a) $x^{2}+2 x-15$ (b) $x^{2}-2 x+15$ (c) $x^{2}-2 x-15$ (d) none of these Solution: (c) $x^{2}-2 x-15$ Here, the zeroes are 5 and $-3$. Let $\alpha=5$ and $\beta$ So, sum of the zeroes, $\alpha+\beta=5+(-3)=2$ Also, product of the zeroes, $\alpha \beta=5 \times(-3)=-15$ The polynomial will be $x^{2}-(\alpha+\beta) x+\alpha \beta$. $\therefore$ The required polynomial is $x^{2}-2 x-15$....
Read More →A common transistor radio set requires 12 V (D.C.) for its operation. The D.C.
Question: A common transistor radio set requires $12 \mathrm{~V}$ (D.C.) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at $220 \mathrm{~V}$ (A.C.) on standard domestic A.C. supply. The number of turns of secondary coil are 24, then the number of turns of primary are ______ Solution: $(440)$ Given Primary voltage, $V_{p}=220 \mathrm{~V}$ Secondary voltage, $v_{s}=12 \mathrm{~V}$ No. of turns in secondary coil is $\mathrm{N}_{\...
Read More →Let S and S' be the foci of an ellipse
Question: Let $S$ and $S^{\prime}$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $\Delta \mathrm{S}^{\prime} \mathrm{BS}$ is a right angled triangle with right angle at $\mathrm{B}$ and area $\left(\Delta \mathrm{S}^{\prime} \mathrm{BS}\right)=8 \mathrm{sq}$. units, hen the length of a latus rectum of the ellipse is :(1) 4(2) $2 \sqrt{2}$(3) $4 \sqrt{2}$(4) 2Correct Option: 1 Solution: $\because \triangle S B S$ is right angled triangle, then $($ Slope of ...
Read More →The sum and product of the zeros of a quadratic polynomial are 3 and −10 respectively.
Question: The sum and product of the zeros of a quadratic polynomial are 3 and 10 respectively. The quadratic polynomial is (a) $x^{2}-3 x+10$ (b) $x^{2}+3 x-10$ (c) $x^{2}-3 x-10$ (d) $x^{2}+3 x+10$ Solution: (c) $x^{2}-3 x-10$ Given: Sum of zeroes, $\alpha+\beta=3$ Also, product of zeroes, $\alpha \beta=-10$ $\therefore$ Required polynomial $=x^{2}-(\alpha+\beta)+\alpha \beta=x^{2}-3 x-10$...
Read More →if
Question: If $2.5 \times 10^{-6} \mathrm{~N}$ average force is exerted by a light wave on a nonreflecting surface of $30 \mathrm{~cm}^{2}$ area during 40 minutes of time span, the energy flux of light just before it falls on the surface is_______ $\mathrm{W} / \mathrm{cm}^{2}$ (Round off to the Nearest Integer) (Assume complete absorption and normal incidence conditions are there) Solution: (25) $F=\frac{I A}{C}$ $I=\frac{F C}{A}=\frac{2.5 \times 10^{-6} \times 3 \times 10^{8}}{30}=25 \mathrm{~W...
Read More →The zeros of the polynomial
Question: The zeros of the polynomial $7 x^{2}-\frac{11}{3} x-\frac{2}{3}$ are (a) $\frac{2}{3}, \frac{-1}{7}$ (b) $\frac{2}{7}, \frac{-1}{3}$ (c) $\frac{-2}{3}, \frac{1}{7}$ (d) none of these Solution: (a) $\frac{2}{3}, \frac{-1}{7}$ Let $f(x)=7 x^{2}-\frac{11}{3} x-\frac{2}{3}=0$ $\Rightarrow 21 x^{2}-11 x-2=0$ $\Rightarrow 21 x^{2}-14 x+3 x-2=0$ $\Rightarrow 7 x(3 x-2)+1(3 x-2)=0$ $\Rightarrow(3 x-2)(7 x+1)=0$ $\Rightarrow x=\frac{2}{3}$ or $x=\frac{-1}{7}$...
Read More →Let the length of the latus rectum of an ellipse with its
Question: Let the length of the latus rectum of an ellipse with its major axis along $x$-axis and centre at the origin, be 8 . If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?(1) $(4 \sqrt{2}, 2 \sqrt{2})$(2) $(4 \sqrt{3}, 2 \sqrt{2})$(3) $(4 \sqrt{3}, 2 \sqrt{3})$(4) $(4 \sqrt{2}, 2 \sqrt{3})$Correct Option: , 2 Solution: Let the ellipse be $\frac{x^{2}}{a^{2}} \frac{y^{2}}{b^{2}}=1$ Then, $\frac{2 b^{2...
Read More →The zeros of the polynomial
Question: The zeros of the polynomial $x^{2}+\frac{1}{6} x-2$ are (a) $-3,4$ (b) $\frac{-3}{2}, \frac{4}{3}$ (c) $\frac{-4}{3}, \frac{3}{2}$ (d) none of these Solution: (b) $\frac{-3}{2}, \frac{4}{3}$ Let $f(x)=x^{2}+\frac{1}{6} x-2=0$ $=6 x^{2}+x-12=0$ $=6 x^{2}+9 x-8 x-12=0$ $=3 x(2 x+3)-4(2 x+3)=0$ $=(2 x+3)(3 x-4)=0$ $\therefore x=\frac{-3}{2}$ or $x=\frac{4}{3}$...
Read More →The zeros of the polynomial
Question: The zeros of the polynomial $4 x^{2}+5 \sqrt{2} x-3$ are (a) $-3 \sqrt{2}, \sqrt{2}$ (b) $-3 \sqrt{2}, \frac{\sqrt{2}}{2}$ (c) $\frac{-3 \sqrt{2}}{2}, \frac{\sqrt{2}}{4}$ (d) none of these Solution: (c) $-\frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$ Let $f(x)=4 x^{2}+5 \sqrt{2} x-3=0$ $=4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3=0$ $=2 \sqrt{2} x(\sqrt{2} x+3)-1(\sqrt{2} x+3)=0$ $=(\sqrt{2} x+3)(2 \sqrt{2} x-1)=0$ $=x=-\frac{3}{\sqrt{2}}$ or $x=\frac{1}{2 \sqrt{2}}$ $=x=-\frac{3}{\sqrt{2}}$ or $x=\f...
Read More →A conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure
Question: A conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure Two resistors $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ are connected across the ends of the rails. There is\ a uniform magnetic field $\overrightarrow{\mathrm{B}}$ pointing into the page. An external agent pulls the bar to the left at a constant speed $v$ The correct statement about the directions of induced currents $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$ flowing through $\mathrm{R}_{1}$...
Read More →If tangents are drawn to the ellipse
Question: If tangents are drawn to the ellipse $x^{2}+2 y^{2}=2$ at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :(1) $\frac{1}{4 x^{2}}+\frac{1}{2 y^{2}}=1$(2) $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$(3) $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$(4) $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$Correct Option: , 3 Solution: Given the equation of ellipse, $\frac{x^{2}}{(\sqrt{2})^{2}}+y^{2}=1$ $\frac{\sqrt{2} \co...
Read More →The zeros of the polynomial
Question: The zeros of the polynomial $x^{2}-\sqrt{2} x-12$ are (a) $\sqrt{2},-\sqrt{2}$ (b) $3 \sqrt{2},-2 \sqrt{2}$ (c) $-3 \sqrt{2}, 2 \sqrt{2}$ (d) $3 \sqrt{2}, 2 \sqrt{2}$ Solution: (b) $3 \sqrt{2},-2 \sqrt{2}$ Let $f(x)=x^{2}-\sqrt{2} x-12=0$ $=x^{2}-3 \sqrt{2} x+2 \sqrt{2} x-12=0$ $=x(x-3 \sqrt{2})+2 \sqrt{2}(x-3 \sqrt{2})=0$ $=(x-3 \sqrt{2})(x+2 \sqrt{2})=0$ $=x=3 \sqrt{2}$ or $x=-2 \sqrt{2}$...
Read More →The zeros of the polynomial
Question: The zeros of the polynomial $x^{2}-2 x-3$ are (a) 3, 1(b) 3, 1(c) 3, 1(d) 3, 1 Solution: (c) $3,-1$ Let $f(x)=x^{2}-2 x-3=0$ $=x^{2}-3 x+x-3=0$ $=x(x-3)+1(x-3)=0$ $=(x-3)(x+1)=0$ $=x=3$ or $x=-1$...
Read More →An ellipse, with foci at (0,2) and (0,-2)
Question: An ellipse, with foci at $(0,2)$ and $(0,-2)$ and minor axis of length 4, passes through which of the following points ?(1) $(\sqrt{2}, 2)$(2) $(2, \sqrt{2})$(3) $(2,2 \sqrt{2})$(4) $(1,2 \sqrt{2})$Correct Option: 1 Solution: Let the equation of ellipse : $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Given that length of minor axis is 4 i.e. $a=4$. Also given be $=2$ $\because a^{2}=b^{2}\left(1-e^{2}\right) \Rightarrow 4=b^{2}-4 \Rightarrow b=2 \sqrt{2}$ Hence, equation of ellipse will ...
Read More →Which of the following is not a polynomial?
Question: Which of the following is not a polynomial? (a) $\sqrt{3} x^{2}-2 \sqrt{3} x+5$ (b) $9 x^{2}-4 x+\sqrt{2}$ (c) $\frac{3}{2} x^{3}+6 x^{2}-\frac{1}{\sqrt{2}} x-8$ (d) $x+\frac{3}{x}$ Solution: (d) $x+\frac{3}{x}$ is not a polynomial. It is because in the second term, the degree ofxis 1 and an expression with a negative degree is not a polynomial....
Read More →Which of the following is a polynomial?
Question: Which of the following is a polynomial? (a) $x^{2}-5 x+4 \sqrt{x}+3$ (b) $x^{3 / 2}-x+x^{1 / 2}+1$ (c) $\sqrt{x}+\frac{1}{\sqrt{x}}$ (d) $\sqrt{2} x^{2}-3 \sqrt{3} x+\sqrt{6}$ Solution: (d) is the correct option. A polynomial in $x$ of degree $n$ is an expression of the form $p(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x_{n}$, where $a_{n} \neq 0$....
Read More →If α and β are the zeroes of a polynomial f(x)
Question: If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x)=6 x^{2}+x-2$, find the value of $\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)$ Solution: By using the relationship between the zeroes of the quadratic ploynomial.We have, Sum of zeroes $=\frac{-(\text { coefficient } \text { of } x)}{\text { coefficent } \text { of } x^{2}}$ and Product of zeroes $=\frac{\text { constant term }}{\text { coefficent of } x^{2}}$ $\therefore \alpha+\beta=\frac{-1}{6}$ and $\alpha \beta...
Read More →If α and β are the zeroes of a polynomial f(x)
Question: If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x)=x^{2}-5 x+k$, such that $\alpha-\beta=1$, find the value of $k$. 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 $=\frac{\text { constant term }}{\text { coefficent of } x^{2}}$ $\therefore \alpha+\beta=\frac{-(-5)}{1}$ and $\alpha \beta=\frac{k}{1}$ $\Rightarrow \alpha+\beta=5...
Read More →If the normal to the ellipse
Question: If the normal to the ellipse $3 x^{2}+4 y^{2}=12$ at a point $\mathrm{P}$ on it is parallel to the line, $2 x+y=4$ and the tangent to the ellipse at $\mathrm{P}$ passes through $\mathrm{Q}(4,4)$ then $\mathrm{PQ}$ is equal to :(1) $\frac{5 \sqrt{5}}{2}$(2) $\frac{\sqrt{61}}{2}$(3) $\frac{\sqrt{221}}{2}$(4) $\frac{\sqrt{157}}{2}$Correct Option: 1, Solution: Slope of tangent on the line $2 x+y=4$ at point $P$ is $\frac{1}{2}$. Given ellipse is, $3 x^{2}+4 y^{2}=12 \Rightarrow \frac{x^{2}...
Read More →Find the zeros of the quadratic polynomial
Question: Find the zeros of the quadratic polynomial $f(x)=4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}$ Solution: To find the zeros of the quadratic polynomial we will equatef(x) to 0 $\therefore f(x)=0$ $\Rightarrow 4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}=0$ $\Rightarrow 4 \sqrt{3} x^{2}+8 x-3 x-2 \sqrt{3}=0$ $\Rightarrow 4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0$ $\Rightarrow(\sqrt{3} x+2)(4 x-\sqrt{3})=0$ $\Rightarrow(\sqrt{3} x+2)=0$ or $(4 x-\sqrt{3})=0$ $\Rightarrow x=-\frac{2}{\sqrt{3}}$ or $x=\frac{\sqrt{3...
Read More →The tangent and normal to the ellipse
Question: The tangent and normal to the ellipse $3 x^{2}+5 y^{2}=32$ at the point $\mathrm{P}(2,2)$ meet the $\mathrm{x}$-axis at $\mathrm{Q}$ and $\mathrm{R}$, respectively. Then the area (in sq. units) of the triangle $P Q R$ is : (1) $\frac{34}{15}$(2) $\frac{14}{3}$(3) $\frac{16}{3}$(4) $\frac{68}{15}$Correct Option: , 4 Solution: $3 x^{2}+5 y^{2}=32 \Rightarrow \frac{3 x^{2}}{32}+\frac{5 y^{2}}{32}=1$ Tangent on the ellipse at $\mathrm{P}$ is $\frac{3(2) x}{32}+\frac{5(2) y}{32}=1 \Rightarr...
Read More →Write the zeros of the quadratic polynomial f(x)
Question: Write the zeros of the quadratic polynomial $f(x)=6 x^{2}-3$ Solution: To find the zeros of the quadratic polynomial we will equatef(x) to 0 $\therefore f(x)=0$ $\Rightarrow 6 x^{2}-3=0$ $\Rightarrow 3\left(2 x^{2}-1\right)=0$ $\Rightarrow 2 x^{2}-1=0$ $\Rightarrow 2 x^{2}=1$ $\Rightarrow x^{2}=\frac{1}{2}$ $\Rightarrow x=\pm \frac{1}{\sqrt{2}}$ Hence, the zeros of the quadratic polynomial $f(x)=6 x^{2}-3$ are $\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$....
Read More →An ideal battery of 4 V and resistance R
Question: An ideal battery of $4 \mathrm{~V}$ and resistance $\mathrm{R}$ are connected in series in the primary circuit of a potentionmeter of length $1 \mathrm{~m}$ and resistance $5 \Omega$. The value of $\mathrm{R}$, to give a potential difference of $5 \mathrm{mV}$ across $10 \mathrm{~cm}$ of potentiometer wire is:(1) $490 \Omega$(2) $480 \Omega$(3) $395 \Omega$(4) $495 \Omega$Correct Option: , 3 Solution: Current flowing through the circuit (I) is given by $I=\left(\frac{4}{R+5}\right) A$ ...
Read More →The sum of the zero and the product of zero of a quadratic polynomial are
Question: The sum of the zero and the product of zero of a quadratic polynomial are $\frac{-1}{2}$ and $-3$ respectively, write the polynomial. Solution: We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula $x^{2}-$ (Sum of the zeros) $x+$ Product of zeros $\Rightarrow x^{2}-\left(-\frac{1}{2}\right) x+(-3)$ $\Rightarrow x^{2}+\frac{1}{2} x-3$ Hence, the required polynomial is $x^{2}+\frac{1}{2} x-3$....
Read More →State division algorithm for polynomials.
Question: State division algorithm for polynomials. Solution: Iff(x) andg(x) are two polynomials such that degree off(x) is greater than degree ofg(x) whereg(x) 0, then there exists unique polynomialsq(x) andr(x) such that f(x) =g(x) q(x) +r(x), wherer(x) = 0 or degree ofr(x) degree ofg(x)....
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