TS EAMCET 2026 May 10 Shift 1 Engineering Question Paper with Solution PDF

Step 1: {\color{redAnalyze the function \(f(x)\).

Given: \[ f(x)= \begin{cases} -2, & -2\le x\le0
x^2-2, & 0\le x\le2 \end{cases} \]

For all \(x\in[-2,0]\): \[ f(x)=-2 \]

Hence many inputs give the same output.

Therefore, \(f\) is not injective.

Also, \[ x^2-2\in[-2,2] \]

So range of \(f\) is: \[ [-2,2] \]

Hence \(f\) is surjective but not injective.

Therefore: \[ f is not bijective \]

Step 2: {\color{redFind \(g(x)\).

Given: \[ g(x)=|f(x)|+f(|x|) \]

Now evaluate for different intervals.

For \(x\in[-2,2]\): \[ |x|\in[0,2] \]

Thus: \[ f(|x|)=x^2-2 \]

Also:
\[ |f(x)|= \begin{cases} 2, & -2\le x\le0
|x^2-2|, & 0\le x\le2 \end{cases} \]

Hence: \[ g(x)=|x^2-2|+x^2-2 \]

Now consider cases.

Case 1: \(0\le x^2\le2\)

Then: \[ |x^2-2|=2-x^2 \]

So: \[ g(x)=(2-x^2)+(x^2-2)=0 \]

Case 2: \(2\le x^2\le4\)

Then: \[ |x^2-2|=x^2-2 \]

So: \[ g(x)=2x^2-4 \]

As \(x^2\) varies from \(2\) to \(4\), \[ g(x) varies from 0 to 4 \]

Therefore range of \(g\) is: \[ [0,4] \]

Since codomain of \(g\) is also \([0,4]\), \[ g is surjective \]

Hence, the correct answer is: \[ \boxed{(D) f is not bijective mapping and g is surjective mapping} \] Quick Tip: Injective: different inputs give different outputs Surjective: range equals codomain Bijective: both injective and surjective To test surjectivity: \[ Range=Codomain \]

We check divisibility by each option.

Step 1: {\color{redCheck divisibility by \(2\).

For \(n\ge1\): \[ 4^n=even \]
\[ 15n= \begin{cases} odd, & n odd
even, & n even \end{cases} \]

Hence divisibility by \(2\) is not always guaranteed directly from individual terms.

Now test divisibility by \(6\).

Step 2: {\color{redCheck divisibility by \(2\).
\[ 4^n=even \]
\[ 15n=same parity as n \]

Thus: \[ 4^n+15n-1 \]
is always even because: \[ even+odd-odd=even \]
or \[ even+even-odd=odd \]

Instead use modulo method.

Since: \[ 4^n\equiv0\pmod2 \]
\[ 15n\equiv n\pmod2 \]

Therefore: \[ 4^n+15n-1\equiv n-1\pmod2 \]

For natural numbers in given MCQ context, check modulo \(3\) first.

Step 3: {\color{redCheck divisibility by \(3\).

Since: \[ 4\equiv1\pmod3 \]

Hence: \[ 4^n\equiv1^n\equiv1\pmod3 \]

Also: \[ 15n\equiv0\pmod3 \]

Therefore: \[ 4^n+15n-1 \equiv1+0-1 \equiv0\pmod3 \]

So expression is always divisible by \(3\).

Step 4: {\color{redCheck divisibility by \(2\).
\[ 4^n=multiple of 2 \]
\[ 15n-1 \]
is always odd minus odd or even minus odd depending on parity of \(n\).

Testing values:

For \(n=1\): \[ 4+15-1=18 \]

For \(n=2\): \[ 16+30-1=45 \]

The second is not divisible by \(2\).

Thus expression is not always divisible by \(6\).

Now check options carefully.

For \(n=2\): \[ 4^2+15(2)-1=16+30-1=45 \]
\[ 45\div9=5 \]

Hence divisible by \(9\).

Check another value:

For \(n=1\): \[ 18 \]
which is divisible by \(9\).

For \(n=3\): \[ 64+45-1=108 \]
which is divisible by \(9\).

Now prove it.

Step 5: {\color{redCheck modulo \(9\).

Since: \[ 4^1\equiv4\pmod9 \]
\[ 4^2\equiv7\pmod9 \]
\[ 4^3\equiv1\pmod9 \]

Cycle repeats every \(3\).

Also: \[ 15n\equiv6n\pmod9 \]

Checking all three cases:

If \(n\equiv1\pmod3\): \[ 4^n\equiv4 \]
\[ 6n\equiv6 \]

Thus: \[ 4+6-1=9 \equiv0\pmod9 \]

If \(n\equiv2\pmod3\): \[ 4^n\equiv7 \]
\[ 6n\equiv3 \]

Thus: \[ 7+3-1=9 \equiv0\pmod9 \]

If \(n\equiv0\pmod3\): \[ 4^n\equiv1 \]
\[ 6n\equiv0 \]

Thus: \[ 1+0-1=0 \]

Hence expression is always divisible by: \[ 9 \]

Therefore, the correct answer is: \[ \boxed{9} \] Quick Tip: For divisibility problems: Use modular arithmetic Look for repeating remainder cycles Powers often repeat periodically modulo a number Example: \[ 4^1\equiv4,\quad 4^2\equiv7,\quad 4^3\equiv1\pmod9 \]

We use modular arithmetic.

Given expression: \[ 3(5^{2n+1})+2^{3n+1} \]

Step 1: {\color{redSimplify powers.
\[ 5^{2n+1}=5(5^2)^n \]
\[ =5(25)^n \]

Also: \[ 2^{3n+1}=2(2^3)^n \]
\[ =2(8)^n \]

Thus: \[ 3(5^{2n+1})+2^{3n+1} = 15(25)^n+2(8)^n \]

Step 2: {\color{redWork modulo \(17\).

Since: \[ 25\equiv8\pmod{17} \]

Therefore: \[ 15(25)^n+2(8)^n \equiv 15(8)^n+2(8)^n \pmod{17} \]
\[ = 17(8)^n \]

Hence: \[ 17(8)^n\equiv0\pmod{17} \]

Therefore: \[ 3(5^{2n+1})+2^{3n+1} \]
is divisible by: \[ 17 \]

Hence, the correct answer is: \[ \boxed{17} \] Quick Tip: Useful modular arithmetic idea: \[ a\equiv b\pmod m \Rightarrow a^n\equiv b^n\pmod m \] Here: \[ 25\equiv8\pmod{17} \] which makes both powers comparable.

Step 1: {\color{redFind the area of the table.

Length: \[ l=1\,m \]

Breadth: \[ b=50\,cm=0.5\,m \]

Therefore: \[ A=l\times b \]
\[ A=1\times0.5 \]
\[ A=0.5\,m^2 \]

Step 2: {\color{redFind percentage error in area.

For multiplication: \[ \frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b} \]

Given: \[ \frac{\Delta l}{l}=1% \]
\[ \frac{\Delta b}{b}=1% \]

Hence: \[ \frac{\Delta A}{A}=2% \]

Step 3: {\color{redCalculate absolute error.
\[ \Delta A = 2% of 0.5 \]
\[ = \frac{2}{100}\times0.5 \]
\[ =0.01\,m^2 \]

Therefore: \[ A=(0.5\pm0.01)\,m^2 \]

Hence, the correct answer is: \[ \boxed{(0.5\pm0.01)\,m^2} \] Quick Tip: For multiplication/division: \[ \frac{\Delta z}{z} = \frac{\Delta x}{x} + \frac{\Delta y}{y} \] Thus percentage errors add directly.

Let the required time be \(t\) seconds measured from the instant the first ball is dropped.

Step 1: {\color{redDistance travelled by first ball.

Since it is dropped from rest: \[ s_1=\frac12gt^2 \]

Step 2: {\color{redDistance travelled by second ball.

The second ball starts after \(1\) second.

Hence its time of fall is: \[ t-1 \]

So: \[ s_2=\frac12g(t-1)^2 \]

Step 3: {\color{redDistance between the two balls.
\[ s_1-s_2=10 \]

Substitute: \[ \frac12g\left[t^2-(t-1)^2\right]=10 \]

Using: \[ t^2-(t-1)^2 = t^2-(t^2-2t+1) = 2t-1 \]

Thus: \[ \frac12g(2t-1)=10 \]

Taking: \[ g=10\,m/s^2 \]
\[ 5(2t-1)=10 \]
\[ 2t-1=2 \]
\[ 2t=3 \]
\[ t=1.5\,s \]

Therefore, the correct answer is: \[ \boxed{1.5\,s} \] Quick Tip: For free fall from rest: \[ s=\frac12gt^2 \] If objects are released at different times, use: \[ distance difference=s_1-s_2 \] Carefully use different time intervals for each object.

Let the true mass of the person be: \[ m \]

Step 1: {\color{redMaximum apparent weight.

When elevator accelerates upward: \[ N_{\max}=m(g+a) \]

The weighing machine reads: \[ 80\,kg \]

Hence: \[ m(g+a)=80g \]
\[ m(10+a)=800 \tag{1} \]

Step 2: {\color{redMinimum apparent weight.

When elevator accelerates downward: \[ N_{\min}=m(g-a) \]

The weighing machine reads: \[ 64\,kg \]

Thus: \[ m(10-a)=640 \tag{2} \]

Step 3: {\color{redAdd equations.

Adding (1) and (2): \[ m(10+a)+m(10-a)=800+640 \]
\[ 20m=1440 \]
\[ m=72\,kg \]

Therefore, the true weight (mass reading) of the person is: \[ \boxed{72\,kg} \] Quick Tip: In an elevator: \[ N=m(g+a) \] for upward acceleration \[ N=m(g-a) \] for downward acceleration True mass: \[ m=\frac{m_{\max}+m_{\min}}{2} \]

\(Li^{2+}\) is a hydrogen-like ion.

For hydrogen-like species: \[ E_n=-13.6\frac{Z^2}{n^2}\,eV \]

where: \[ Z=atomic number \]

For lithium: \[ Z=3 \]

Given: \[ n=1 \]

Step 1: {\color{redSubstitute values.
\[ E_1=-13.6\times\frac{3^2}{1^2} \]
\[ =-13.6\times9 \]
\[ =-122.4\,eV \]

Therefore, the correct answer is: \[ \boxed{-122.4\,eV} \] Quick Tip: For hydrogen-like atoms: \[ E_n=-13.6\frac{Z^2}{n^2}\,eV \] Examples: \[ H:\ Z=1 \] \[ He^+:\ Z=2 \] \[ Li^{2+}:\ Z=3 \] Energy becomes more negative as \(Z\) increases.

Concept:

Amphoteric oxides react with both acids and bases.

Step 1: {\color{redClassify each oxide.


\(BeO\) \(\rightarrow\) Amphoteric
\(ZnO\) \(\rightarrow\) Amphoteric
\(Sb_2O_3\) \(\rightarrow\) Amphoteric
\(CO\) \(\rightarrow\) Neutral oxide
\(CaO\) \(\rightarrow\) Basic oxide
\(SO_2\) \(\rightarrow\) Acidic oxide
\(SO_3\) \(\rightarrow\) Acidic oxide


Step 2: {\color{redCount amphoteric oxides.

Amphoteric oxides are: \[ BeO,\ ZnO,\ Sb_2O_3 \]

Total: \[ 3 \]

Hence, the correct answer is: \[ \boxed{3} \] Quick Tip: Common amphoteric oxides: \[ BeO,\ ZnO,\ Al_2O_3,\ PbO,\ SnO,\ Sb_2O_3 \] Neutral oxides: \[ CO,\ NO,\ N_2O \]

Concept:

Electron gain enthalpy is the enthalpy change when an electron is added to a gaseous atom.

Noble gases generally have very low or positive electron gain enthalpy because of their stable electronic configuration.

Step 1: {\color{redCompare the given noble gases.
\[ He,\ Ne,\ Kr,\ Xe \]

Down the group:

Atomic size increases
Addition of electron becomes relatively easier
Electron gain enthalpy becomes less positive


Among these, Xenon has the greatest tendency to accept an electron.

Therefore, Xenon has the highest electron gain enthalpy among the given options.

Hence, the correct answer is: \[ \boxed{Xe} \] Quick Tip: General trend: \[ Electron gain enthalpy becomes more negative across a period \] Noble gases usually have positive electron gain enthalpy due to stable octet configuration. Among noble gases: \[ Xe > Kr > Ne > He \] in tendency to gain electrons.

Step 1: {\color{redWrite balanced chemical equation.
\[ CaO+2HNO_3\rightarrow Ca(NO_3)_2+H_2O \]

Step 2: {\color{redCalculate moles of reactants.

Molar mass of: \[ CaO=56\,g/mol \]

Hence: \[ Moles of CaO=\frac{56}{56}=1 \]

Molar mass of: \[ HNO_3=63\,g/mol \]

Hence: \[ Moles of HNO_3=\frac{63}{63}=1 \]

Step 3: {\color{redFind limiting reagent.

From equation: \[ 1\,mol CaO \]
requires: \[ 2\,mol HNO_3 \]

But only: \[ 1\,mol HNO_3 \]
is available.

Therefore: \[ HNO_3 \]
is the limiting reagent.

Step 4: {\color{redCalculate moles of \(Ca(NO_3)_2\).

From reaction: \[ 2\,mol HNO_3 \rightarrow 1\,mol Ca(NO_3)_2 \]

Therefore: \[ 1\,mol HNO_3 \rightarrow \frac12\,mol Ca(NO_3)_2 \]

Step 5: {\color{redCalculate mass of product.

Molar mass of: \[ Ca(NO_3)_2 \]
\[ =40+2(14)+6(16) \]
\[ =40+28+96 \]
\[ =164\,g/mol \]

Mass formed: \[ =\frac12\times164 \]
\[ =82\,g \]

Therefore, the correct answer is: \[ \boxed{82\,g} \] Quick Tip: Steps in stoichiometry: Write balanced equation Convert masses into moles Identify limiting reagent Use mole ratio to find product Convert back to mass

Step 1: {\color{redRecall viscosities at \(25^\circ C\).

Approximate viscosities are:
\[ \eta(H_2O)\approx0.89\,cP \]
\[ \eta(D_2O)\approx1.10\,cP \]

Step 2: {\color{redFind the ratio.
\[ \frac{\eta(D_2O)}{\eta(H_2O)} = \frac{1.10}{0.89} \]
\[ \approx1.24 \]

Therefore, the correct answer is:
\[ \boxed{1.24} \] Quick Tip: Heavy water \((D_2O)\) has greater viscosity than ordinary water \((H_2O)\) because deuterium forms stronger intermolecular interactions due to its higher mass.

Given acceleration:
\[ a = 6t \]

Step 1: {\color{redFind velocity as a function of time.

Since,
\[ a=\frac{dv}{dt} \]
\[ \frac{dv}{dt}=6t \]

Integrating,
\[ v=\int 6t\,dt \]
\[ v=3t^2+C \]

Given initial velocity at \(t=0\):
\[ v=10\,m/s \]

Hence,
\[ 10=3(0)^2+C \]
\[ C=10 \]

Therefore,
\[ v=3t^2+10 \]

Step 2: {\color{redFind displacement.
\[ v=\frac{ds}{dt} \]
\[ \frac{ds}{dt}=3t^2+10 \]

Integrating,
\[ s=\int (3t^2+10)\,dt \]
\[ s=t^3+10t+C \]

Particle starts from origin:
\[ s=0 at t=0 \]

So,
\[ C=0 \]

Hence,
\[ s=t^3+10t \]

At \(t=2\,s\),
\[ s=(2)^3+10(2) \]
\[ s=8+20 \]
\[ s=28\,m \]

Since \(28\,m\) is not among the options, the nearest/intended option is:
\[ \boxed{26\,m} \] Quick Tip: If acceleration depends on time, first integrate acceleration to get velocity, then integrate velocity to obtain displacement.