Ask & Answer: India's Largest Education Community

The given area of the circle is $x^2+y^2=16-------(1)$ is a circle with centre (0,0) and radius, $\pi =4$ and the parabola is $y^2=6x$ -------------(2)

Solving (1) and (2) for x and y.

$\begin{array}{l}{x}^2+6x=16\\ =x^2+6x-16=0\\ =x^2+8x-2x-16=0\\ =x\left(x+8\right)-2\left(x+8\right)=0\\ =\left(x+8\right)\left(x-2\right)=0\\ =x=-8\&x=2\end{array}$

For, $x=-8,y^2=6\left(-8\right)=-48$

Which is not possible.

For, $x=2,y^2=6\left(2\right)=12$

$y=\pm \mathrm{2\surd 3}$

$areaOACBO=2\times \left\{area\left(OADO\right)+area\left(ACDA\right)\right\}$

38. (C) Tm

In the periodic table, actinides lie from atomic number 90 to 103. 

Thulium is not an actinide; it is a lanthanide. 

The given curve is $y=x\left|x\right|$

$=\left\{\left(x.xifx\ge 0\right)\left(x.\left(-x\right)ifx\le 0\right)\right\}=\left\{\left(x^{2}if,x\ge 0\right)\left(-x^{2}if,x\le 0\right)\right\}$

Which is in the form of a parabola nad the lines are $x=-1\&x-axis$

At $x=1>0,y=1^2=1$

At $$$x=-1<0,y=-1^2=-1$

Shaded area of the Ist quadrant

$\begin{array}{l}={\displaystyle \underset{0}{\overset{1}{\int }}ydx}={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}\\ ={\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{3}\end{array}$

Shaded area of the IInd quadrant

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{0}{\int }}ydx}={\displaystyle \underset{-1}{\overset{0}{\int }}{x}^{2}dx}\\ ={\left[\frac{-x^3}{-3}\right]}_{-1}^0\\ =\frac{1}{3}\end{array}$

$\therefore$ Total area of the enclosed region $=\frac{1}{3}+\frac{1}{3}$

$=\frac{2}{3}uni{t}^2$

$\therefore$ Option (c) is correct.

37. (D) Mn²⁺ acts as auto catalyst

Mn²⁺  formed in the reaction acts as an autocatalyst. Auto catalyst is a compound that is one of the products that are formed in the reaction itself catalyses the reaction.

Reaction: 2MnO₄^-  +  16H⁺  +  5C₂O₄^2- → 2Mn₂ +  +  10CO₂ +  8H₂O

55.  (ii) Both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation: Hydrometallurgy is used to extract copper from low-grade ore. Hydrometallurgy entails dissolving the ore in a suitable reagent and then precipitating it. In this method, more electropositive metal is used, allowing pure metal to be displaced.

Given is $y=x^3$ and the ines $x=-2\&x=1$

For $y=x^3$

$\begin{array}{l}area\left(OAB\right)={\displaystyle \underset{0}{\overset{1}{\int }}ydx}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\\ ={\left[\frac{x^4}{4}\right]}_0^1=\frac{1}{4}\\ area\left(ODC\right)={\displaystyle \underset{-2}{\overset{0}{\int }}ydx}\\ ={\displaystyle \underset{-2}{\overset{0}{\int }}{x}^{3}dx}\\ =\left|{\left[\frac{x^4}{4}\right]}_{-2}^0\right|=\left|\left[\frac{0^4}{4}-\frac{{\left(-2\right)}^4}{4}\right]\right|=4\end{array}$

$\therefore$ Total area of the bounded region $=\frac{1}{4}+4$

$=\frac{17}{4}uni{t}^2$

UCEOU is a constituent of Osmania University. Thus, the institute is expected to follow the fee refund policy framed by the university. As per the university's policy, students who withdraw their admissions are entitled to a refund of fees. However, the amount of refund depends on the following policy:

36.  (A)i, ii

Disproportionation/redox reaction is a reaction in which one compound is oxidised as well as reduced at the same time.

(A) Cu⁺  ?  Cu²⁺  +  Cu 

  ( + 1)       ( + 2)        (0) 

(B) 3MnO₄^- +  4H⁺  ?  2MnO₄^- +  MnO₂ +  2H₂O 

  ( + 6)                          ( + 7)           ( + 4) 

54.  (ii) Both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation: The zone refining method is very useful for producing high-purity semiconductors and other metals, such as germanium.

35. (B)+3

Out of all +3 is the most common oxidation state for all the lanthanide elements.

The given equation of curve $y^2\le 4x$ i.e,  $y^2=4x$ - (1) is a parabola and

$4x^2+4y^2\le 9\Rightarrow x^2+y^2\le \frac{9}{4}\Rightarrow x^2+y^2={\left(\frac{3}{2}\right)}^2$ - (2) is a circle

With centre (0,0)and radius 

Solving (1) and (2) for x and y,                                 

$\begin{array}{l}{x}^2+4x=\frac{9}{4}\\ =x^2+4x-\frac{9}{4}=0\\ =4x^2+16x-9=0\end{array}$

Yes, there are many Textile Design colleges in India that offer certificate courses. Some of them are mentioned below along with their tuition fees:

Disclaimer: This information is sourced from the official website/ sanctioning body and may vary.

The given equation of the lines are

$\begin{array}{l}2x+y=4------(1)\\ 3x-2y=6-----(2)\\ x-3y+5=0----(3)\end{array}$

$\therefore$ Area of $?ABC=area(PCBQ)-area(?APC)-area(?AQB)$

$\begin{array}{l}={\displaystyle \underset{1}{\overset{4}{\int }}{y}_{(3)}dx-}{\displaystyle \underset{1}{\overset{2}{\int }}{y}_{(1)}dx-{\displaystyle \underset{2}{\overset{4}{\int }}{y}_{(2)}dx-}}\\ ={\displaystyle \underset{1}{\overset{4}{\int }}\frac{\left(x+5\right)}{3}dx-{\displaystyle \underset{1}{\overset{2}{\int }}\left(4-2x\right)dx-{\displaystyle \underset{2}{\overset{4}{\int }}\left(\frac{3x-6}{2}dx\right)}}}\end{array}$

$\therefore$ The point of intersection of the circle and the parabola is . $A\left(\frac{1}{2},\right)\&C\left(\frac{1}{2},-\right)$

Taking in first quadrant

Area of $(OABO)=area(OADO)+area(BADB)$

You can find the TS inter syllabus for 1st and 2nd year on the official website, tgbie.cgg.gov.in. On the home page, there is the syllabus tab on the left, click on it and you'll find subject-wise syllabus for TS Intermediate exams. However, the you'll have to find syllabus for different subjects there as they are provided in a simple format. 

You can find the updated TS inter syllabus 2026 subject-wise by clicking on the link provided here. Click and download the syllabus and start preparing for your exams. The syllabus mentions all the topics that you need to prepare to score well in the exams. 

Yes, there are many BDes in Textile Design colleges in India. Some of them are mentioned below along with their tuition fees:

Disclaimer: This information is sourced from the official website/ sanctioning body and may vary.

52. (i) Both assertion and reason are true and reason is the correct explanation of assertion.

Explanation: Van Arkel method is generally used to obtain pure forms of Zirconium (Zr) and Titanium (Ti).

The metal iodide is heated at 1800 K and is decomposed on a tungsten filament. The pure metal is dropped on the tungsten filament. This proves that ZrI4 is volatile and decomposes at 1800 K.

34. (B) 3d⁵

Magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of magnetic moment.

3d⁷ =  3 unpaired electrons 

3d⁵ =  5 unpaired electrons

3d⁸ =  2 unpaired electrons 

3d² =   2 unpaired electrons 

Out of all, 3d⁵ has five unpaired electrons that is maximum and hence it has the highest magnetic moment.