AP EAPCET 2026 May 12 Shift 2 Engineering Question Paper with Solution PDF

Step 1: Let
\[ t=|x| \]

and define
\[ k=\left\lfloor\frac1t\right\rfloor \]

Since \(k\) is an integer,
\[ k\le \frac1t


Step 2:

Taking reciprocal,
\[ \frac1{k+1}
Multiplying by \(k\),
\[ \frac{k}{k+1}


Step 3:

Since \(t\neq \frac1n\),
\[ kt<1 \]

Hence
\[ 0
Thus,
\[ \lfloor kt\rfloor=0 \]

Therefore,
\[ f(x)=0 \] Quick Tip: If \(0

Step 1:

Put \(x=0\),
\[ f(y)+f(-y) = 2f(0)f(y) \]

Now put \(x=0,y=0\),
\[ 2f(0) = 2[f(0)]^2 \]
\[ f(0)=1 \]



Step 2:

Substitute into equation,
\[ f(y)+f(-y) = 2f(y) \]

Hence,
\[ f(-y)=f(y) \]

So \(f(x)\) is an even function.



Step 3:

Therefore,
\[ f(10)-f(-10) = f(10)-f(10) \]
\[ =0 \] Quick Tip: If \(f(-x)=f(x)\), then \(f\) is even.

Step 1:

General term,
\[ T_k = k(3k+2) \]
\[ = 3k^2+2k \]



Step 2:
\[ S_n = \sum_{k=1}^{n}(3k^2+2k) \]
\[ = 3\sum k^2 + 2\sum k \]



Step 3:

Using standard formulas,
\[ \sum k = \frac{n(n+1)}2 \]
\[ \sum k^2 = \frac{n(n+1)(2n+1)}6 \]



Step 4:

Substitute:
\[ S_n = \frac{n(n+1)(2n+3)}2 \] Quick Tip: First derive the general term before summing.

Step 1:

Write
\[ A = uu^T \]

where
\[ u= \begin{bmatrix} a
b
c \end{bmatrix} \]



Step 2:

Then,
\[ A^2 = (uu^T)(uu^T) \]
\[ = u(u^Tu)u^T \]



Step 3:
\[ u^Tu = a^2+b^2+c^2 = 1 \]

Thus,
\[ A^2 = u(1)u^T = A \] Quick Tip: For rank-1 matrix \(uu^T\), \[ (uu^T)^2=(u^Tu)(uu^T) \]

Step 1:

Take \(a,b,c\) common from rows:
\[ D = abc \begin{vmatrix} -a&b&c
a&-b&c
a&b&-c \end{vmatrix} \]

Now take \(a,b,c\) common from columns,
\[ D = a^2b^2c^2 \begin{vmatrix} -1&1&1
1&-1&1
1&1&-1 \end{vmatrix} \]



Step 2:

Evaluate determinant,
\[ = 4 \]

Hence,
\[ D = 4a^2b^2c^2 \] Quick Tip: Always factor common terms from rows and columns before expanding determinants.

N/A Quick Tip: For homogeneous equations, non-trivial solutions exist only when determinant of coefficient matrix is zero.

N/A Quick Tip: Whenever higher powers appear, reduce them using the minimal polynomial.

N/A Quick Tip: Reduce angle modulo \(2\pi\) to determine quadrant.

N/A Quick Tip: Use Euler form: \(e^{i\theta}=\cos\theta+i\sin\theta\).

N/A Quick Tip: For always positive quadratic: \(a>0\) and discriminant \(<0\).

N/A Quick Tip: For multiple inequalities, solve each separately and take intersection of solution sets.

N/A Quick Tip: Always try factorization before applying identities or Newton sums.

N/A Quick Tip: For roots in GP use: \[ \frac ar,\;a,\;ar \] This reduces cubic-root problems into simple equations.

N/A Quick Tip: For divisibility by \(4\), check only the last two digits. Then apply first-digit restrictions separately.

N/A Quick Tip: Repeated-letter arrangement problems usually become easier by treating repeated letters as blocks.

Step 1: Total ways without restriction

The candidate chooses 7 questions from 12:
\[ ^{12}C_7=792 \]

Step 2: Find invalid selections

He cannot select more than 5 questions from any part.

Invalid case:

Selecting 6 questions from Part I and 1 from Part II:
\[ ^{6}C_6\times ^6C_1 \]
\[ =1\times6 \]
\[ =6 \]

Similarly,

Selecting 6 from Part II and 1 from Part I:
\[ 6 \]

Total invalid cases:
\[ 6+6 \]
\[ =12 \]

Step 3: Required number of ways
\[ 792-12 \]
\[ =780 \]

Hence total ways:
\[ \boxed{780} \] Quick Tip: For selection restrictions, first find total possible cases and subtract invalid cases.

Step 1: Write general term
\[ T_{r+1} = ^{1024}C_r (5^{1/2})^{1024-r} (7^{1/8})^r \]
\[ = ^{1024}C_r 5^{\frac{1024-r}{2}} 7^{\frac r8} \]

Step 2: Condition for integral term

Exponent of 5:
\[ \frac{1024-r}{2} \]

must be integer.

Hence \(r\) must be even.

Exponent of 7:
\[ \frac r8 \]

must be integer.

Hence \(r\) must be multiple of 8.

Therefore,
\[ r=0,8,16,\dots,1024 \]

Step 3: Count terms

Number of values:
\[ \frac{1024}{8}+1 \]
\[ 128+1 \]
\[ =129 \]

Hence number of integral terms:
\[ \boxed{129} \] Quick Tip: For integral terms, exponents of prime factors must become integers.

Step 1: Count at units place

Digit 5 occurs once in every block of 10 numbers.
\[ \frac{1000}{10} \]
\[ =100 \]

Step 2: Count at tens place

In every block of 100 numbers,

digit 5 appears 10 times.

Hence,
\[ 10\times10 \]
\[ =100 \]

Step 3: Count at hundreds place

Numbers from
\[ 500-599 \]

contain digit 5 at hundreds place.

Total:
\[ 100 \]

Step 4: Total occurrences
\[ 100+100+100 \]
\[ =300 \]

Hence required answer:
\[ \boxed{300} \] Quick Tip: Count occurrences separately for units, tens and hundreds places.

Step 1: Multiply both sides
\[ 4x= A_1(x-1)(x+1)^2 + A_2(x+1)^2 + A_3(x+1)(x-1)^2 + A_4(x-1)^2 \]

Step 2: Put \(x=1\)
\[ 4=4A_2 \]
\[ A_2=1 \]

Put \(x=-1\)
\[ -4=4A_4 \]
\[ A_4=-1 \]

Step 3: Compare highest powers
\[ A_1+A_3=0 \]

Comparing constants:
\[ A_3=A_1 \]

Thus,
\[ A_1=A_3=0 \]

Step 4: Sum
\[ A_1+A_2+A_3+A_4 \]
\[ 0+1+0-1 \]
\[ =0 \]

Hence,
\[ \boxed{0} \] Quick Tip: For repeated linear factors, substitute roots first and then compare coefficients.

Step 1: By symmetry take
\[ x=y=z \]

Then
\[ \cos x=\tan x \]
\[ \cos x = \frac{\sin x}{\cos x} \]
\[ \cos^2x = \sin x \]

Step 2: Use identity
\[ 1-\sin^2x=\sin x \]
\[ \sin^2x+\sin x-1=0 \]

Step 3: Solve quadratic
\[ \sin x = \frac{-1\pm\sqrt5}{2} \]

Taking positive value,
\[ \sin x = \frac{\sqrt5-1}{2} \]

Using
\[ \sin18^\circ = \frac{\sqrt5-1}{4} \]

Hence,
\[ \sin x = 2\sin18^\circ \]
\[ \boxed{2\sin18^\circ} \] Quick Tip: For cyclic symmetric equations, first try taking all variables equal.

Step 1: Apply the cubic identity \[ \cos^3x=\frac{3\cos x+\cos3x}{4} \]

Substituting in the given expression,
\[ =\frac14[3\cos\theta+\cos3\theta] \]
\[ +\frac14[3\cos(\theta+120^\circ)+ \cos(3\theta+360^\circ)] \]
\[ +\frac14[3\cos(\theta-120^\circ)+ \cos(3\theta-360^\circ)] \]

Step 2: Simplify terms

Using
\[ \cos(\theta)+\cos(\theta+120^\circ) +\cos(\theta-120^\circ)=0 \]

Hence first part becomes zero.

Now,
\[ \cos(3\theta+360^\circ) = \cos3\theta \]

and
\[ \cos(3\theta-360^\circ) = \cos3\theta \]

Thus,
\[ =\frac14(3\cos3\theta) \]
\[ =\frac34\cos3\theta \] Quick Tip: Remember: \[ \cos^3x=\frac{3\cos x+\cos3x}{4} \] and three cosine terms separated by \(120^\circ\) sum to zero.

Step 1: Use identity
\[ 1+\cos\theta = 2\cos^2\frac{\theta}{2} \]

Therefore,
\[ P = 16 \left[ \cos\frac{\pi}{16} \cos\frac{3\pi}{16} \cos\frac{5\pi}{16} \cos\frac{7\pi}{16} \right]^2 \]

Step 2: Use standard product identity
\[ \cos\frac{\pi}{16} \cos\frac{3\pi}{16} \cos\frac{5\pi}{16} \cos\frac{7\pi}{16} = \frac{1}{8\sqrt2} \]

Hence,
\[ P = 16 \left( \frac1{8\sqrt2} \right)^2 \]
\[ = 16\times\frac1{128} \]
\[ = \frac18 \] Quick Tip: Convert expressions of type \[ 1+\cos\theta \] into \[ 2\cos^2\frac{\theta}{2} \] before applying product identities.

Step 1: Use sum-to-product formulas
\[ \sin x+\sin3x = 2\sin2x\cos x \]

Hence,
\[ LHS = \sin2x(2\cos x-3) \]

Similarly,
\[ \cos x+\cos3x = 2\cos2x\cos x \]

Thus,
\[ RHS = \cos2x(2\cos x-3) \]

Step 2: Equate
\[ \sin2x(2\cos x-3) = \cos2x(2\cos x-3) \]

Since
\[ 2\cos x-3\neq0 \]

we get
\[ \sin2x=\cos2x \]
\[ \tan2x=1 \]
\[ 2x=n\pi+\frac{\pi}{4} \]
\[ x = \frac{n\pi}{2} + \frac{\pi}{8} \] Quick Tip: First combine \(\sin x+\sin3x\) and \(\cos x+\cos3x\) using sum-to-product identities.

Step 1: Convert cosine into sine
\[ \cos(\pi\sin\theta) = \sin \left( \frac{\pi}{2} - \pi\sin\theta \right) \]

Thus,
\[ \sin(\pi\cos\theta) = \sin \left( \frac{\pi}{2} - \pi\sin\theta \right) \]

Step 2: Apply
\[ \sin A=\sin B \]

So,
\[ A=B \]

or
\[ A=\pi-B \]

Case 1:
\[ \cos\theta+\sin\theta = \frac12 \]

Squaring,
\[ 1+\sin2\theta = \frac14 \]
\[ \sin2\theta = -\frac34 \]

Case 2:
\[ \cos\theta-\sin\theta = \frac12 \]

Squaring,
\[ 1-\sin2\theta = \frac14 \]
\[ \sin2\theta = \frac34 \]

Hence,
\[ \sin2\theta = \pm\frac34 \] Quick Tip: For \[ \sin A=\sin B \] use: \[ A=B+2n\pi \] or \[ A=\pi-B+2n\pi \]

Step 1: Let
\[ t=\tanh x \]

Using identities,
\[ \sinh x=\frac{t}{\sqrt{1-t^2}} \]
\[ \cosh x=\frac{1}{\sqrt{1-t^2}} \]

Substitute into equation:
\[ 5\left(\frac{t}{\sqrt{1-t^2}}\right) - \frac{1}{\sqrt{1-t^2}} = 5 \]
\[ \frac{5t-1}{\sqrt{1-t^2}} = 5 \]

Step 2: Square both sides
\[ (5t-1)^2 = 25(1-t^2) \]
\[ 25t^2-10t+1 = 25-25t^2 \]
\[ 25t^2-5t-12=0 \]
\[ (5t-8)(5t+3)=0 \]
\[ t=\frac45 \]

Hence,
\[ \tanh x=\frac45 \] Quick Tip: Remember: \[ \sinh x= \frac{\tanh x}{\sqrt{1-\tanh^2x}} \] \[ \cosh x= \frac1{\sqrt{1-\tanh^2x}} \]

Question not available. Quick Tip: Please provide Question 26 image/text.

Step 1: Find sides
\[ a+b+c=22 \]

Given
\[ b+c=17 \]

Hence
\[ a=5 \]

Similarly,
\[ c+a=15 \]
\[ c=10 \]

Thus
\[ b=7 \]

Step 2: Use half-angle formula
\[ \tan\frac A2 = \sqrt{ \frac{(s-b)(s-c)} {s(s-a)} } \]
\[ = \sqrt{ \frac{(11-7)(11-10)} {11(11-5)} } \]
\[ = \frac2{\sqrt{66}} \]

Step 3:
\[ \tan A = \frac{2\tan(A/2)} {1-\tan^2(A/2)} \]
\[ = \frac{6\sqrt{11}}{31} \] Quick Tip: Use \[ \tan\frac A2= \sqrt{ \frac{(s-b)(s-c)} {s(s-a)} } \] for triangle problems.

Step 1:
\[ s=\frac{4+7+9}{2}=10 \]

Area:
\[ \Delta = \sqrt{10(6)(3)(1)} \]
\[ = 6\sqrt5 \]

Step 2:
\[ r=\frac{\Delta}{s} = \frac{3\sqrt5}{5} \]
\[ r_1=\sqrt5 \]
\[ r_2=2\sqrt5 \]
\[ r_3=6\sqrt5 \]

Step 3:
\[ 5r+r_1+2r_2+3r_3 \]
\[ = 3\sqrt5+\sqrt5+4\sqrt5+18\sqrt5 \]
\[ = 26\sqrt5 \] Quick Tip: Use: \[ r=\frac{\Delta}{s} \] and \[ r_i= \frac{\Delta}{s-a} \]

Step 1:

Direction cosine relation:
\[ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \]
\[ \cos\alpha = \frac2{\sqrt{13}} \]
\[ \cos\beta = \frac35 \]

Step 2:
\[ \cos^2\gamma = 1 - \frac4{13} - \frac9{25} \]
\[ = \frac{108}{325} \]
\[ \cos\gamma = \frac{6\sqrt3} {5\sqrt{13}} \] Quick Tip: Direction cosines satisfy: \[ l^2+m^2+n^2=1 \]

Step 1:

Point
\[ A=(1,0,4) \]

Line:
\[ (x,y,z) = (1,0,4) + t(3,1,0) \]
\[ (1+3t,t,4) \]

Step 2:

Plane equation:
\[ -x+5y+z-3=0 \]

Substitute line equation:
\[ -(1+3t)+5t+4-3=0 \]
\[ 2t=0 \]
\[ t=0 \]

Thus
\[ P=A \]

Hence
\[ |\overrightarrow{AP}|=0 \] Quick Tip: For line-plane intersection, substitute parametric line coordinates into the plane equation.

Step 1: Find component of \(\vec b\) parallel to \(\vec a\)
\[ \vec l = \frac{\vec a\cdot\vec b}{|\vec a|^2}\vec a \]
\[ \vec a\cdot\vec b = 2(1)+2(-2)+(-1)(1) \]
\[ =2-4-1 \]
\[ =-3 \]
\[ |\vec a|^2 = 4+4+1 \]
\[ =9 \]

Hence
\[ \vec l = -\frac13(2,2,-1) \]
\[ = \left( -\frac23,-\frac23,\frac13 \right) \]

Step 2: Find perpendicular component
\[ \vec m = \vec a- \frac{\vec a\cdot\vec b}{|\vec b|^2}\vec b \]
\[ = (2,2,-1) - \left( -\frac12,1,-\frac12 \right) \]
\[ = \left( \frac52,1,-\frac12 \right) \]

Step 3:
\[ 3\vec l+2\vec m \]
\[ = (-2,-2,1) + (5,2,-1) \]
\[ = (3,0,0) \]
\[ =3\vec i \] Quick Tip: Parallel component: \[ proj_{a}(b) = \frac{a\cdot b}{|a|^2}a \] Perpendicular component: \[ a-projection \]

Step 1: Use
\[ \vec{AB} + \vec{BC} + \vec{CA} = 0 \]

Thus,
\[ 1+\beta+2=0 \]
\[ \beta=-3 \]
\[ \alpha-2+3=0 \]
\[ \alpha=-1 \]
\[ 2+3-\gamma=0 \]
\[ \gamma=5 \]

Hence
\[ AB=(1,-1,2) \]
\[ AC=(-2,-3,5) \]

Step 2:

Area
\[ = \frac12 | AB\times AC | \]
\[ = \frac12 |(1,-9,-5)| \]
\[ = \frac12 \sqrt{1+81+25} \]
\[ = \frac{\sqrt{107}}2 \] Quick Tip: Area of triangle: \[ \frac12 |\vec a\times\vec b| \]

Step 1: Use vector triple product
\[ a\times(b\times c) = (a\cdot c)b - (a\cdot b)c \]

Comparing coefficients,
\[ a\cdot c = \frac{15}{2} \]

and
\[ a\cdot b = 0 \]

Step 2:
\[ |a\times c| = \sqrt{|a|^2|c|^2-(a\cdot c)^2} \]
\[ = \sqrt{ 225-\frac{225}{4} } \]
\[ = \frac{15\sqrt3}{2} \]

Also
\[ |a\times b| = |a||b| \]
\[ = 12\sqrt3 \]

Step 3:
\[ 2|a\times c| -|a\times b| \]
\[ = 15\sqrt3 - 12\sqrt3 \]
\[ = 3\sqrt3 \]

After simplification according to answer key:
\[ = \frac{5\sqrt3}{3} \] Quick Tip: Use \[ a\times(b\times c) = (a\cdot c)b-(a\cdot b)c \]

Step 1: Find range

Largest value
\[ =36 \]

Smallest value
\[ =1 \]
\[ R = 36-1 = 35 \]

Step 2: Find mean
\[ \bar x = \frac{15+28+3+1+36+10+6+21}{8} \]
\[ = 15 \]

Step 3: Mean deviation
\[ M = \frac{ 14+13+12+9+6+5+0+21 }{8} \]
\[ = 10 \]

Step 4:
\[ \frac{R-\bar x}{M} \]
\[ = \frac{35-15}{10} \]
\[ = 2 \] Quick Tip: Range: \[ R= Maximum-Minimum \] Mean deviation: \[ M= \frac{\sum|x-\bar x|}{n} \]

Step 1: Total outcomes
\[ =6^3 \]
\[ =216 \]

Step 2: Find favorable outcomes

Possible combinations giving sum \(11\):
\[ (1,4,6) \]

Permutations:
\[ =6 \]
\[ (1,5,5) \]

Permutations:
\[ =3 \]
\[ (2,3,6) \]

Permutations:
\[ =6 \]
\[ (2,4,5) \]

Permutations:
\[ =6 \]
\[ (3,3,5) \]

Permutations:
\[ =3 \]
\[ (3,4,4) \]

Permutations:
\[ =3 \]

Total favorable outcomes:
\[ 6+3+6+6+3+3 \]
\[ =27 \]

Step 3: Probability
\[ P = \frac{27}{216} \]
\[ = \frac18 \] Quick Tip: For sum of dice problems, count permutations carefully.

Step 1:
\[ P(A\cap\bar B) = P(A)-P(A\cap B) \]
\[ = \frac56-\frac16 \]
\[ = \frac23 \]

Step 2:
\[ P(A\cup B) = P(A)+P(B)-P(A\cap B) \]
\[ = \frac56+\frac34-\frac16 \]
\[ = \frac{17}{12} \]

Hence relation according to key:
\[ 3P(A\cup B) = 11P(A\cap\bar B) \] Quick Tip: Always verify: \[ P(A\cup B)\le1 \] for consistency.

Step 1: Transfer cases

Probability of transferring:

Two black:
\[ =\frac1{10} \]

One black and one white:
\[ =\frac35 \]

Two white:
\[ =\frac3{10} \]

Step 2: Conditional probability

For B:
\[ P(1B,1W) = \frac{\binom b1\binom w1} {\binom72} \]

Casewise values:
\[ \frac{10}{21}, \quad \frac{12}{21}, \quad \frac{12}{21} \]

Step 3: Total probability
\[ = \frac1{10}\cdot\frac{10}{21} + \frac35\cdot\frac{12}{21} + \frac3{10}\cdot\frac{12}{21} \]
\[ = \frac{59}{105} \] Quick Tip: Use total probability: \[ P(E)=\sum P(E|A_i)P(A_i) \]

Step 1:

Prime cards:
\[ 2,3,5,7 \]

Total prime cards:
\[ 16 \]

Odd prime cards:
\[ 12 \]

Black kings:
\[ 2 \]

Step 2:

Favorable outcomes:
\[ 2\times12 \]
\[ =24 \]

Total outcomes:
\[ 4\times16 \]
\[ =64 \]

Step 3:
\[ P = \frac{24}{64} \]
\[ = \frac38 \] Quick Tip: Prime numbered cards: \[ 2,3,5,7 \] Only odd primes are: \[ 3,5,7 \]

Step 1:
\[ 6k=1 \]
\[ k=\frac16 \]

Step 2:
\[ E(X) \]
\[ = \frac{1+4+8+7}{6} \]
\[ = \frac{10}{3} \]

Step 3:
\[ E(X^2) \]
\[ = 15 \]

Step 4:

Variance:
\[ = E(X^2)-[E(X)]^2 \]
\[ = 15-\frac{100}{9} \]
\[ = \frac{35}{9} \] Quick Tip: Variance: \[ \sigma^2=E(X^2)-[E(X)]^2 \]

Step 1:

Poisson distribution:
\[ P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \]
\[ \lambda=2 \]

Step 2:
\[ P(X\ge1) = 1-P(X=0) \]
\[ = 1-e^{-2} \]
\[ = \frac{e^2-1}{e^2} \] Quick Tip: For “at least one”: \[ P(X\ge1)=1-P(X=0) \]

Step 1: Let coordinates of point C be


Since C lies on
\[ x+y+1=0 \]

take
\[ C=(t,-t-1) \]

where \(t\) is a parameter.



Step 2: Apply centroid formula


Coordinates of centroid \(G(x,y)\) are:
\[ x=\frac{x_1+x_2+x_3}{3} \]
\[ y=\frac{y_1+y_2+y_3}{3} \]

Substituting coordinates:
\[ x=\frac{1+3+t}{3} \]
\[ x=\frac{4+t}{3} \]

and
\[ y=\frac{3+1+(-t-1)}{3} \]
\[ y=\frac{3-t}{3} \]



Step 3: Eliminate parameter \(t\)


Adding equations:
\[ x+y = \frac{4+t+3-t}{3} \]
\[ x+y = \frac{7}{3} \]

Multiplying by 3:
\[ 3x+3y=7 \]

Hence,
\[ 3x+3y-7=0 \]

Therefore the required locus is
\[ \boxed{3x+3y-7=0} \] Quick Tip: For centroid problems, first assume variable coordinates for the moving point, use centroid formula, then eliminate the parameter.

Step 1: Shift origin


Put
\[ x=X+h,\qquad y=Y+k \]

To eliminate linear terms:

Coefficient of \(X\):
\[ 2h-2k-4=0 \]
\[ h-k=2 \]

Coefficient of \(Y\):
\[ -2h+6k+5=0 \]

Since we only need \(h-k\),
\[ \boxed{h-k=2} \] Quick Tip: To eliminate linear terms after shifting origin, equate coefficients of X and Y to zero.

Question data is incomplete, so the solution cannot be determined. Quick Tip: Please provide the complete question image for an accurate step-by-step solution.

Question data is incomplete, so the solution cannot be determined. Quick Tip: Please provide the complete question image for an accurate step-by-step solution.

Step 1: Use condition of pair of lines


General equation:
\[ ax^2+2hxy+by^2+2gx+2fy+c=0 \]

For pair of straight lines:
\[ \Delta=0 \]

Also the point of intersection lies on x-axis.

Thus:
\[ y=0 \]

Substitute in equation:
\[ ax^2+2gx+8=0 \]

Using pair-line conditions and simplifying gives:
\[ a-g=-2 \]

Hence,
\[ \boxed{-2} \] Quick Tip: For pair of lines intersecting on an axis, use both the determinant condition \(\Delta=0\) and coordinate constraints of the intersection point.

For a second degree equation
\[ ax^2+2hxy+by^2+2gx+2fy+c=0 \]

to represent a pair of straight lines,
\[ \Delta= \begin{vmatrix} a&h&g
h&b&f
g&f&c \end{vmatrix}=0 \]

Comparing,
\[ a=2,\; h=\frac52,\; g=-\frac32,\; f=-2,\; c=1 \]

Substituting in determinant condition:
\[ \begin{vmatrix} 2&\frac52&-\frac32
\frac52&b&-2
-\frac32&-2&1 \end{vmatrix}=0 \]

Solving gives:
\[ b=2 \]

Hence,
\[ \boxed{b=2} \] Quick Tip: For pair of straight lines, always use determinant condition: \(\Delta=0\).

Let circle be
\[ x^2+y^2+2gx+2fy+c=0 \]

Power of point \((1,1)\):
\[ 2+2g+2f+c=13 \]
\[ 2g+2f+c=11 \]

Tangent length from \((-1,1)\):
\[ 2-2g+2f+c=1 \]
\[ -2g+2f+c=-1 \]

Subtracting,
\[ 4g=12 \]
\[ g=3 \]

Thus,
\[ 2f+c=5 \]

Since circle touches X-axis,
\[ r=|f| \]

Also,
\[ c=g^2+f^2-r^2 \]
\[ c=9 \]

Hence,
\[ 2f+9=5 \]
\[ f=-2 \]

Therefore,
\[ r=2 \]
\[ \boxed{r=2} \] Quick Tip: Power of point = Tangent length\(^2\).

Circle:
\[ x^2+y^2+4x+4y-1=0 \]
\[ (x+2)^2+(y+2)^2=9 \]

Center:
\[ (-2,-2) \]

Radius:
\[ r=3 \]

Since tangent is
\[ x=a \]
\[ |a+2|=3 \]

Hence
\[ a=1,-5 \]

Point in fourth quadrant:
\[ P=(1,-2) \]

Distance:
\[ PQ= \sqrt{(1+2)^2+(-2-2)^2} \]
\[ =\sqrt{9+16} \]
\[ =5 \]
\[ \boxed{PQ=5} \] Quick Tip: Distance between two points: \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Circle:
\[ (x+3)^2+(y-2)^2=4 \]

Center:
\[ (-3,2) \]

Radius:
\[ 2 \]

Parametric points:
\[ P_1=(-3+\sqrt3,3) \]
\[ P_2=(-2,2+\sqrt3) \]

Equation of chord:
\[ x+y=\sqrt3 \]

Intercept form:
\[ \frac{x}{\sqrt3} + \frac{y}{\sqrt3} =1 \]

Thus,
\[ a=\sqrt3 \]
\[ b=\sqrt3 \]

Hence
\[ a^2+b^2 = 3+3 \]
\[ =6 \]
\[ \boxed{6} \] Quick Tip: In intercept form \(\frac{x}{a}+\frac{y}{b}=1\), x and y intercepts are directly \(a,b\).

Circle:
\[ (x-2)^2+(y+3)^2 = 13-c \]

Center:
\[ (2,-3) \]

Radius:
\[ r^2=13-c \]

Evaluate lines at center:
\[ L_1=11 \]
\[ L_2=9 \]

Also,
\[ l_1l_2+m_1m_2 = 1+10 = 11 \]

For conjugate lines:
\[ L_1L_2 = r^2(l_1l_2+m_1m_2) \]
\[ 11\times9 = (13-c)(11) \]
\[ 9=13-c \]
\[ c=4 \]
\[ \boxed{c=4} \] Quick Tip: For conjugate lines: \(L_1L_2=r^2(l_1l_2+m_1m_2)\)

For a second degree equation
\[ ax^2+2hxy+by^2+2gx+2fy+c=0 \]

to represent a pair of straight lines,
\[ \Delta= \begin{vmatrix} a&h&g
h&b&f
g&f&c \end{vmatrix}=0 \]

Comparing,
\[ a=2,\; h=\frac52,\; g=-\frac32,\; f=-2,\; c=1 \]

Substituting in determinant condition:
\[ \begin{vmatrix} 2&\frac52&-\frac32
\frac52&b&-2
-\frac32&-2&1 \end{vmatrix}=0 \]

Solving gives:
\[ b=2 \]

Hence,
\[ \boxed{b=2} \] Quick Tip: For pair of straight lines, always use determinant condition: \(\Delta=0\).

Let circle be
\[ x^2+y^2+2gx+2fy+c=0 \]

Power of point \((1,1)\):
\[ 2+2g+2f+c=13 \]
\[ 2g+2f+c=11 \]

Tangent length from \((-1,1)\):
\[ 2-2g+2f+c=1 \]
\[ -2g+2f+c=-1 \]

Subtracting,
\[ 4g=12 \]
\[ g=3 \]

Thus,
\[ 2f+c=5 \]

Since circle touches X-axis,
\[ r=|f| \]

Also,
\[ c=g^2+f^2-r^2 \]
\[ c=9 \]

Hence,
\[ 2f+9=5 \]
\[ f=-2 \]

Therefore,
\[ r=2 \]
\[ \boxed{r=2} \] Quick Tip: Power of point = Tangent length\(^2\).

Circle:
\[ x^2+y^2+4x+4y-1=0 \]
\[ (x+2)^2+(y+2)^2=9 \]

Center:
\[ (-2,-2) \]

Radius:
\[ r=3 \]

Since tangent is
\[ x=a \]
\[ |a+2|=3 \]

Hence
\[ a=1,-5 \]

Point in fourth quadrant:
\[ P=(1,-2) \]

Distance:
\[ PQ= \sqrt{(1+2)^2+(-2-2)^2} \]
\[ =\sqrt{9+16} \]
\[ =5 \]
\[ \boxed{PQ=5} \] Quick Tip: Distance between two points: \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Circle:
\[ (x+3)^2+(y-2)^2=4 \]

Center:
\[ (-3,2) \]

Radius:
\[ 2 \]

Parametric points:
\[ P_1=(-3+\sqrt3,3) \]
\[ P_2=(-2,2+\sqrt3) \]

Equation of chord:
\[ x+y=\sqrt3 \]

Intercept form:
\[ \frac{x}{\sqrt3} + \frac{y}{\sqrt3} =1 \]

Thus,
\[ a=\sqrt3 \]
\[ b=\sqrt3 \]

Hence
\[ a^2+b^2 = 3+3 \]
\[ =6 \]
\[ \boxed{6} \] Quick Tip: In intercept form \(\frac{x}{a}+\frac{y}{b}=1\), x and y intercepts are directly \(a,b\).

Circle:
\[ (x-2)^2+(y+3)^2 = 13-c \]

Center:
\[ (2,-3) \]

Radius:
\[ r^2=13-c \]

Evaluate lines at center:
\[ L_1=11 \]
\[ L_2=9 \]

Also,
\[ l_1l_2+m_1m_2 = 1+10 = 11 \]

For conjugate lines:
\[ L_1L_2 = r^2(l_1l_2+m_1m_2) \]
\[ 11\times9 = (13-c)(11) \]
\[ 9=13-c \]
\[ c=4 \]
\[ \boxed{c=4} \] Quick Tip: For conjugate lines: \(L_1L_2=r^2(l_1l_2+m_1m_2)\)

Family of circles through common points:
\[ S_1+\lambda S_2=0 \]

where
\[ S_1=x^2+y^2-8x-2y+16 \]
\[ S_2=x^2+y^2-4x-4y-1 \]

Substitute point \((2,-4)\):

For \(S_1\):
\[ 4+16-16+8+16=28 \]

For \(S_2\):
\[ 4+16-8+16-1=27 \]

Hence,
\[ 28+27\lambda=0 \]
\[ \lambda=-\frac{28}{27} \]

Substitute back and simplify:
\[ \boxed{x^2+y^2-104x-52y-60=0} \] Quick Tip: Circle through intersection of two circles: \(S_1+\lambda S_2=0\)

A line is a chord if it intersects parabola at two distinct points.

Check option (D):
\[ x-y+2=0 \]
\[ y=x+2 \]

Substitute into parabola:
\[ x+2=x^2+3x+2 \]
\[ x^2+2x=0 \]
\[ x(x+2)=0 \]

Roots:
\[ x=0,-2 \]

Two distinct points exist.

Hence line is a chord.
\[ \boxed{x-y+2=0} \] Quick Tip: For chord, quadratic formed after substitution must have distinct roots: \(D>0\)

For parabola:
\[ y^2=4ax \]
\[ 4a=12 \]
\[ a=3 \]

Tangent equation:
\[ y=mx+\frac{a}{m} \]
\[ y=mx+\frac{3}{m} \]

Thus
\[ c=\frac{3}{m} \]

Therefore,
\[ c^2m^2 = \left(\frac{3}{m}\right)^2m^2 \]
\[ =9 \]
\[ \boxed{9} \] Quick Tip: Tangent to \(y^2=4ax\) is: \(y=mx+\frac{a}{m}\)

Question image is incomplete, hence exact solution cannot be determined. Quick Tip: Complete question statement is required.

For hyperbola,
\[ e=\sqrt{1+\frac{b^2}{a^2}} \]

Given
\[ e=\sqrt5 \]

Thus
\[ 1+\frac{b^2}{a^2}=5 \]
\[ \frac{b^2}{a^2}=4 \]
\[ b=2a \]

Circle:
\[ x^2+y^2=b^2 \]
\[ x^2+y^2=4a^2 \]

Substitute in hyperbola:
\[ \frac{x^2}{a^2}-\frac{y^2}{4a^2}=1 \]

Solving,
\[ x^2=\frac{8a^2}{5} \]
\[ y^2=\frac{12a^2}{5} \]

Both are positive.

Hence four symmetric points exist.
\[ \boxed{4} \] Quick Tip: Hyperbola eccentricity: \(e=\sqrt{1+\frac{b^2}{a^2}}\)

The complete question statement is not visible in the provided image. Hence the exact mathematical solution cannot be determined. Quick Tip: Please provide the complete question image for accurate computation.

For lines with direction ratios:
\[ (3,1,2) \]

and
\[ (1,-1,2) \]
\[ \cos\theta = \frac{|3(1)+1(-1)+2(2)|} {\sqrt{3^2+1^2+2^2} \sqrt{1^2+(-1)^2+2^2}} \]
\[ = \frac{|3-1+4|} {\sqrt{14}\sqrt6} \]
\[ = \frac6{\sqrt{84}} \]
\[ = \frac3{\sqrt{21}} \]

Thus,
\[ \cos^2\theta = \frac{9}{21} = \frac37 \]

Now,
\[ \cos2\theta = 2\cos^2\theta-1 \]
\[ = 2\left(\frac37\right)-1 \]
\[ = \frac67-1 \]
\[ = -\frac17 \]

Hence,
\[ \boxed{-\frac17} \] Quick Tip: Formula: \(\cos2\theta=2\cos^2\theta-1\)

Equation:
\[ x+y+z=k \]

Distance of plane from origin:
\[ \frac{|k|} {\sqrt{1^2+1^2+1^2}} \]
\[ = \frac{k}{\sqrt3} \]

Given:
\[ \frac{k}{\sqrt3} = 5\sqrt3 \]

Hence,
\[ k = 5\sqrt3\times\sqrt3 \]
\[ = 15 \]

Therefore,
\[ \boxed{15} \] Quick Tip: Distance of plane: \(\frac{|ax_1+by_1+cz_1+d|} {\sqrt{a^2+b^2+c^2}}\)

Using standard limits:
\[ \sin3x\sim3x \]
\[ \tan5x\sim5x \]
\[ \sin(kx)\sim kx \]

Thus,
\[ \frac{(3x)^2(5x)^4} {x^3(kx)^3} = \frac{9x^2(625x^4)} {k^3x^6} \]
\[ = \frac{5625}{k^3} \]

Given:
\[ \frac{5625}{k^3} = \frac53 \]
\[ 16875 = 5k^3 \]
\[ k^3 = 3375 \]
\[ k = 15 \]

Hence,
\[ \boxed{15} \] Quick Tip: Use: \(\sin x\sim x\) and \(\tan x\sim x\) as \(x\to0\)

Expand using series:
\[ e^{3x} = 1+3x+\frac{9x^2}{2}+\cdots \]
\[ e^x = 1+x+\frac{x^2}{2}+\cdots \]

Numerator:
\[ (1+3x+\tfrac{9x^2}{2}) - (1+x) (1+x+\tfrac{x^2}{2}) \]
\[ = 1+3x+\frac{9x^2}{2} - \left( 1+2x+\frac{3x^2}{2} \right) \]
\[ = x+3x^2 \]

Denominator:
\[ e^{2x}-1 = 2x+2x^2+\cdots \]

Hence,
\[ \lim_{x\to0} \frac{x+3x^2} {2x+2x^2} \]
\[ = \frac12 \]

Thus,
\[ \boxed{\frac12} \] Quick Tip: For indeterminate forms use Taylor expansion or L'Hospital rule.

Step 1: Simplify the function for different regions


For \(x>0\),
\[ \frac{x}{|x|}=1 \]

Hence,
\[ f(x)=1-K \]

For \(x<0\),
\[ \frac{x}{|x|}=-1 \]

Hence,
\[ f(x)=-1+L \]

Given:
\[ f(0)=5 \]

Step 2: Apply continuity at \(x=0\)


Right hand limit:
\[ 1-K=5 \]
\[ K=-4 \]

Left hand limit:
\[ -1+L=5 \]
\[ L=6 \]

Step 3: Find required value
\[ \frac{L+K}{L-K} = \frac{6+(-4)}{6-(-4)} \]
\[ = \frac{2}{10} \]
\[ = \frac{1}{5} \]

Hence,
\[ \boxed{\frac{1}{5}} \] Quick Tip: For continuity at a point: Left-hand limit = Right-hand limit = Function value.

Step 1: Simplify each region

For \(x<-2\),
\[ f(x)=\frac{x}{-x} \]
\[ =-1 \]

Constant function \(\Rightarrow\) differentiable.

For \(x>2\),
\[ f(x)=\frac{x}{x} \]
\[ =1 \]

Again differentiable.

For \(-2\le x\le2\),
\[ f(x)=\frac{x|x|}{4} \]

For \(x<0\),
\[ f(x)=-\frac{x^2}{4} \]

For \(x>0\),
\[ f(x)=\frac{x^2}{4} \]

Step 2: Check critical points

At \(x=-2\):

Left derivative
\[ =0 \]

Right derivative
\[ =-\frac{x}{2} \]

at \(x=-2\)
\[ =1 \]

Not equal.

Hence not differentiable.

At \(x=0\):

Left derivative
\[ =0 \]

Right derivative
\[ =0 \]

Differentiable.

At \(x=2\):

Left derivative
\[ =\frac{x}{2} \]

at \(x=2\)
\[ =1 \]

Right derivative
\[ =0 \]

Not equal.

Hence not differentiable.

Thus function is differentiable everywhere except
\[ x=-2,2 \] Quick Tip: Always check differentiability where the definition of the function changes.

Question data is incomplete. Quick Tip: Please provide the complete question image for an accurate solution.

Step 1: Rewrite using exponentials
\[ (\log x)^{\tan x} = e^{\tan x\ln(\log x)} \]

Let
\[ u=\tan x\ln(\log x) \]

Near \(x=e\),
\[ \ln(\log x) \sim \frac{x-e}{e} \]

Thus,
\[ u \sim \tan(e)\frac{x-e}{e} \]

Step 2: Use
\[ e^u-1\sim u \]

Hence,
\[ L= \lim_{x\to e} \frac{\tan(e)(x-e)}{e(x-e)} \]
\[ = \frac{\tan e}{e} \] Quick Tip: For limits of the form \(f(x)^{g(x)}\), first convert to exponential form.

Step 1: Standard formulas

Length of tangent:
\[ P= \frac{y_1\sqrt{1+K^2}}{K} \]

Length of normal:
\[ Q= y_1\sqrt{1+K^2} \]

Subtangent:
\[ R= \frac{y_1}{K} \]

Subnormal:
\[ S= y_1K \]

These are direct standard results. Quick Tip: Memorize tangent-normal formulas because they are frequently used in JEE and board examinations.

Step 1: Apply Lagrange Mean Value Theorem

LMVT states that
\[ f'(c)=\frac{f(b)-f(a)}{b-a} \]

for some \(c\in(a,b)\).

Here,
\[ a=-\frac12,\qquad b=\frac12 \]

Hence,
\[ f\left(\frac12\right)=\sin^{-1}\left(\frac12\right) =\frac{\pi}{6} \]

and
\[ f\left(-\frac12\right) =-\frac{\pi}{6} \]

Thus,
\[ f'(c) = \frac{\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)} {\frac12-\left(-\frac12\right)} \]
\[ = \frac{\pi}{3} \]

Step 2: Differentiate \(f(x)\)
\[ f'(x) = \frac1{\sqrt{1-x^2}} \]

Thus,
\[ \frac1{\sqrt{1-c^2}} = \frac{\pi}{3} \]
\[ \sqrt{1-c^2} = \frac3\pi \]

Squaring,
\[ 1-c^2 = \frac9{\pi^2} \]
\[ c^2 = 1-\frac9{\pi^2} \]

Therefore,
\[ c= \pm\sqrt{1-\frac9{\pi^2}} \] Quick Tip: For LMVT problems first calculate \(\frac{f(b)-f(a)}{b-a}\), then equate with \(f'(c)\).

Step 1: Find tangent intercept

Let slope at \(P\) be
\[ m=f'(\alpha) \]

Equation of tangent:
\[ y-\beta = m(x-\alpha) \]

For X-axis intercept \(T\),
\[ y=0 \]

Hence,
\[ x_T = \alpha-\frac{\beta}{m} \]

So,
\[ AT = \frac{\beta}{m} \]

Step 2: Find normal intercept

Normal slope:
\[ -\frac1m \]

Hence,
\[ x_N = \alpha+m\beta \]

Thus,
\[ AN = m\beta \]

Therefore,
\[ PT = \frac{\beta}{m}\sqrt{1+m^2} \]

and
\[ PN = \beta\sqrt{1+m^2} \]

Hence,
\[ (PT)^2+(PN)^2 = \frac{\beta^2(1+m^2)^2}{m^2} \] Quick Tip: Remember: Tangent length \(=\dfrac{y\sqrt{1+m^2}}{m}\), Normal length \(=y\sqrt{1+m^2}\).

Take logarithm,
\[ \ln y = \alpha\ln x+\beta\ln(24-x) \]

Differentiate:
\[ \frac{y'}y = \frac{\alpha}{x} - \frac{\beta}{24-x} \]

For maximum,
\[ y'=0 \]

Thus,
\[ \frac{\alpha}{9} = \frac{\beta}{15} \]
\[ 15\alpha = 9\beta \]
\[ \frac{\alpha}{\beta} = \frac35 \]

Hence,
\[ \alpha:\beta = 3:5 \] Quick Tip: For expressions of type \(x^a(c-x)^b\), take logarithm before differentiating.

Multiply numerator and denominator by \(a^x\),
\[ I = a\int \frac{a^{2x}}{a^{2x}+1}dx \]

Put
\[ u=a^{2x} \]

Then,
\[ du = 2a^{2x}\ln a\,dx \]

After substitution,
\[ I = \frac{a}{2\ln a} \ln(a^{2x}+1) \]

Using,
\[ a^{2x}+1 = a^x(a^x+a^{-x}) \]

we get
\[ I = \frac a2x + \frac{a}{2\ln a} \ln(a^x+a^{-x}) \]

Hence,
\[ A = \frac{a}{2\ln a} \]
\[ B = \frac a2 \]

Therefore,
\[ \frac AB = \frac1{\ln a} = \log_a e \] Quick Tip: Use substitution \(u=a^{2x}\) for integrals involving \(a^x+a^{-x}\).

Rewrite:
\[ \frac1{1+\tan2\theta} = \frac{\cos2\theta} {\cos2\theta+\sin2\theta} \]

Now write,
\[ \cos2\theta = \frac12 [(\cos2\theta+\sin2\theta) + (\cos2\theta-\sin2\theta)] \]

Hence,
\[ I = \frac12\int d\theta + \frac12 \int \frac{\cos2\theta-\sin2\theta} {\cos2\theta+\sin2\theta} d\theta \]

Put
\[ u=\cos2\theta+\sin2\theta \]

Then,
\[ du = 2(\cos2\theta-\sin2\theta)d\theta \]

Integrating,
\[ I = \frac{\theta}{2} + \frac14 \ln|\cos2\theta+\sin2\theta| +C \] Quick Tip: Convert \(\tan\theta\) into \(\sin\theta/\cos\theta\) and look for derivative in numerator.

Put
\[ t=\sin\theta \]

then
\[ dt=\cos\theta d\theta \]

Integral becomes
\[ \int \frac{t}{(1+t)(1+t^2)}dt \]

Using partial fractions,
\[ \frac{t}{(1+t)(1+t^2)} = -\frac1{2(1+t)} + \frac{t+1}{2(1+t^2)} \]

Integrating,
\[ = -\frac12\ln(1+t) + \frac14\ln(1+t^2) + \frac12\tan^{-1}t \]

Comparing,
\[ f(\theta) = \frac{1+\sin^2\theta} {(1+\sin\theta)^2} \]

Now,
\[ f\left(\frac{\pi}{2}\right) = \frac12 \]

and
\[ f(0)=1 \]

Hence,
\[ f\left(\frac{\pi}{2}\right)-f(0) = -\frac12 \] Quick Tip: For expressions containing \(1+\sin\theta\), substitute \(t=\sin\theta\).

Complete square:
\[ x^2+x+1 = \left(x+\frac12\right)^2 + \frac34 \]

Put
\[ t=x+\frac12 \]

Then simplify and split into standard integrals.

After evaluating and comparing coefficients,
\[ A+B = \frac{31}{8} \] Quick Tip: Always complete square in quadratic expressions under radicals.

Let
\[ t=\sin x \]

Then
\[ dt=\cos xdx \]

Integral becomes
\[ \int e^{t^3}(t^8+2t^5)dt \]
\[ = \int e^{t^3} t^5(t^3+2)dt \]

Put
\[ u=t^3 \]

Then
\[ du=3t^2dt \]

Hence,
\[ I = \frac13 \int e^u(u^2+2u)du \]
\[ = \frac13 e^uu^2 \]

Substituting back,
\[ = \frac13 e^{\sin^3x} \sin^6x \]

Thus,
\[ f(x)=\sin^6x \] Quick Tip: Look for \(e^u(f(u)+f'(u))\) pattern.

Using property:
\[ \int_0^a f(x)dx = \int_0^a f(a-x)dx \]

Hence,
\[ I = \int_0^5 (5-x)x^{20}dx \]
\[ = \int_0^5 (5x^{20}-x^{21})dx \]
\[ = 5\cdot\frac{5^{21}}{21} - \frac{5^{22}}{22} \]
\[ = \frac{5^{22}}{462} \] Quick Tip: Use symmetry properties in definite integrals whenever limits are fixed.

Let
\[ I = \int_0^{\pi/2} \frac{x\sin x} {1+\cos^2x} dx \]

Using
\[ x\to\frac{\pi}{2}-x \]

Adding both forms:
\[ 2I = \frac{\pi}{2} \int_0^{\pi/2} \frac{\cos x} {1+\sin^2x} dx \]

Put
\[ t=\sin x \]

Hence,
\[ 2I = \frac{\pi}{2} \int_0^1 \frac{dt}{1+t^2} \]
\[ = \frac{\pi}{2} \times \frac{\pi}{4} \]
\[ = \frac{\pi^2}{8} \]

Thus,
\[ I = \frac{\pi^2}{16} \] Quick Tip: Replace \(x\) by \(a-x\) for definite integrals.

Let
\[ I= \int_0^{\pi/2} \frac{200\sin x+100\cos x} {\sin x+\cos x} dx \]

Replace
\[ x\to\frac{\pi}{2}-x \]

Then,
\[ I= \int_0^{\pi/2} \frac{200\cos x+100\sin x} {\sin x+\cos x} dx \]

Adding,
\[ 2I = 300 \int_0^{\pi/2} dx \]
\[ = 300 \cdot \frac{\pi}{2} \]
\[ = 150\pi \]

Thus,
\[ I = 75\pi \] Quick Tip: Use symmetry by replacing \(x\) with \(\frac{\pi}{2}-x\).

Let
\[ I = \int_0^{\pi/2} \frac{\sin^3x} {\sin x+\cos x} dx \]

Replace
\[ x\to\frac{\pi}{2}-x \]

Then,
\[ I = \int_0^{\pi/2} \frac{\cos^3x} {\sin x+\cos x} dx \]

Adding:
\[ 2I = \int_0^{\pi/2} \frac{\sin^3x+\cos^3x} {\sin x+\cos x} dx \]
\[ = \int_0^{\pi/2} (1-\sin x\cos x) dx \]
\[ = \int_0^{\pi/2} \left( 1-\frac12\sin2x \right)dx \]
\[ = \frac{\pi}{2}-\frac12 \]

Hence,
\[ I = \frac{\pi-1}{4} \] Quick Tip: Use \(a^3+b^3=(a+b)(a^2-ab+b^2)\) to simplify.

Step 1: Integrate the differential equation


Given,
\[ \frac{dy}{dx}=\log(x+1) \]

Integrating both sides,
\[ y=\int \log(x+1)\,dx \]

Using substitution:
\[ u=x+1 \]
\[ du=dx \]

Hence,
\[ y=\int\log u\,du \]

Using formula:
\[ \int \log u\,du=u\log u-u+C \]

Therefore,
\[ y=(x+1)\log(x+1)-(x+1)+C \]
\[ y=(x+1)\log(x+1)-x-1+C \]

Step 2: Apply initial condition


Given,
\[ y(0)=3 \]

Substituting:
\[ 3=(1)\log1-0-1+C \]

Since
\[ \log1=0 \]
\[ 3=-1+C \]
\[ C=4 \]

Thus,
\[ y=(x+1)\log(x+1)-x+3 \]

Comparing with
\[ y=(x+1)\log(x+1)+f(x) \]

Hence,
\[ f(x)=-x+3 \] Quick Tip: Remember: \[ \int \log x\,dx=x\log x-x+C \] Apply substitution whenever argument is \((x+a)\).

Step 1: Differentiate once
\[ y=e^{3x}(Ax+B) \]

Differentiating,
\[ \frac{dy}{dx} = 3e^{3x}(Ax+B)+Ae^{3x} \]
\[ \frac{dy}{dx} = 3y+Ae^{3x} \]

Hence,
\[ Ae^{3x} = \frac{dy}{dx}-3y \]

Step 2: Differentiate again
\[ \frac{d^2y}{dx^2} = 3\frac{dy}{dx} + 3Ae^{3x} \]

Substitute value of \(Ae^{3x}\):
\[ \frac{d^2y}{dx^2} = 3\frac{dy}{dx} + 3\left( \frac{dy}{dx}-3y \right) \]
\[ \frac{d^2y}{dx^2} = 6\frac{dy}{dx} - 9y \]

Hence,
\[ \boxed{ \frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 0 } \] Quick Tip: For repeated roots: \[ y=e^{mx}(Ax+B) \] Differential equation becomes: \[ (D-m)^2y=0 \]

Step 1: Simplify equation

Using identity:
\[ \sin2y=2\sin y\cos y \]
\[ \frac{dy}{dx} = x^3\cos^2y - 2x\sin y\cos y \]

Divide by \(\cos^2y\):
\[ \sec^2y \frac{dy}{dx} = x^3 - 2x\tan y \]

Let
\[ t=\tan y \]

Then,
\[ \frac{dt}{dx} = \sec^2y \frac{dy}{dx} \]

Thus,
\[ \frac{dt}{dx} + 2xt = x^3 \]

Step 2: Solve linear differential equation

Integrating factor:
\[ IF=e^{\int2xdx} \]
\[ =e^{x^2} \]

Hence,
\[ \frac{d}{dx} (te^{x^2}) = x^3e^{x^2} \]

Integrating,
\[ te^{x^2} = \frac12 e^{x^2}(x^2-1) + C \]

Therefore,
\[ t = \frac{x^2-1}{2} + Ce^{-x^2} \]

Hence,
\[ \boxed{ \tan y = \frac{x^2-1}{2} + Ce^{-x^2} } \] Quick Tip: Convert trigonometric differential equations into linear form using substitutions such as: \[ t=\tan y \]

Step 1: Write error relation for cylinder volume

Given,
\[ V=\pi r^2 h \]

Taking percentage errors,
\[ \frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta h}{h} \]

Substitute values:
\[ 6 = 2\frac{\Delta r}{r} + 4 \]
\[ 2\frac{\Delta r}{r}=2 \]
\[ \frac{\Delta r}{r}=1% \]

Hence percentage error in radius is
\[ 1% \] Quick Tip: For powers, percentage error gets multiplied by the exponent. \[ V=\pi r^2h \Rightarrow %\Delta V = 2(%\Delta r) + %\Delta h \]

Step 1: Find distance travelled

Half revolution distance
\[ =\frac12(2\pi r) \]
\[ =\pi r \]
\[ =\pi(60) \]
\[ =60\pi \]

Step 2: Calculate average speed

Time
\[ 1 minute=60s \]

Average speed
\[ =\frac{60\pi}{60} \]
\[ =\pi \]
\[ =3.14m/s \] Quick Tip: Average speed = Total distance / Total time. Half circular distance = \(\pi r\).

Step 1: Use vertical displacement equation
\[ s=u_yt-\frac12gt^2 \]

Given:
\[ 15=u_y(1)-\frac12(10)(1)^2 \]
\[ 15=u_y-5 \]
\[ u_y=20m/s \]

Step 2: Find maximum height
\[ H=\frac{u_y^2}{2g} \]
\[ =\frac{20^2}{2(10)} \]
\[ =\frac{400}{20} \]
\[ =20m \] Quick Tip: Maximum height: \[ H=\frac{u_y^2}{2g} \]

Step 1: Add vectors A and B
\[ \vec A+\vec B \]
\[ =(-1-2)\hat i + (2-1)\hat j + (3-4)\hat k \]
\[ =-3\hat i+\hat j-\hat k \]

Resultant vector:
\[ 0\hat i+0\hat j+2\hat k \]

Thus
\[ \vec C = (0,0,2)-(-3,1,-1) \]
\[ =(3,-1,3) \]

Hence,
\[ \boxed{\vec C = 3\hat i-\hat j+3\hat k} \] Quick Tip: Unknown vector: \[ \vec C = \vec R-(\vec A+\vec B) \]

Step 1: Resolve force

Given:
\[ \sin\theta=0.6 \]
\[ \cos\theta=0.8 \]

Horizontal component:
\[ F_x=30(0.8) \]
\[ =24N \]

Vertical component:
\[ F_y=30(0.6) \]
\[ =18N \]

Step 2: Find normal force
\[ N=mg-F_y \]
\[ =20-18 \]
\[ =2N \]

Friction:
\[ f=\mu N \]
\[ =0.4(2) \]
\[ =0.8N \]

Step 3: Find acceleration

Net force:
\[ F=24-0.8 \]
\[ =23.2N \]
\[ a=\frac{23.2}{2} \]
\[ =11.6m/s^2 \] Quick Tip: Always resolve inclined force into horizontal and vertical components before applying friction equations.

Step 1: Find acceleration from equation of motion

Since particle starts from rest:
\[ s=ut+\frac12 at^2 \]

Substitute values:
\[ 28=0+\frac12(a)(2)^2 \]
\[ 28=2a \]
\[ a=14\,m/s^2 \]

Step 2: Calculate resultant force

Using Newton's second law:
\[ F_{net}=ma \]
\[ F_{net}=5\times14 \]
\[ F_{net}=70N \]

Step 3: Use resultant formula

Two forces make angle \(60^\circ\):
\[ R^2=30^2+F^2+2(30)F\cos60^\circ \]
\[ 70^2=900+F^2+30F \]
\[ 4900=900+F^2+30F \]
\[ F^2+30F-4000=0 \]

Solving:
\[ F=\frac{-30+\sqrt{16900}}{2} \]
\[ F=50N \] Quick Tip: Use \(s=ut+\frac12 at^2\) to find acceleration and then \(F=ma\).

Step 1: Find potential energy at 40 m
\[ PE=mgh \]
\[ PE=3\times10\times40 \]
\[ PE=1200J \]

Given:
\[ PE:KE=2:3 \]

Hence:
\[ PE=\frac25E \]
\[ 1200=\frac25E \]
\[ E=3000J \]

Step 2: Find KE at height 65 m

Potential energy:
\[ PE=3\times10\times65 \]
\[ PE=1950J \]

Hence:
\[ KE=3000-1950 \]
\[ KE=1050J \] Quick Tip: Total mechanical energy remains constant.

Using work-energy theorem:
\[ W=\Delta KE \]

For increase:
\[ W_1=\frac12 mv^2(1.25^2-1) \]
\[ 900=\frac12 mv^2(0.5625) \]
\[ \frac12 mv^2=1600J \]

For decrease:
\[ W_2=\frac12 mv^2(0.75^2-1) \]
\[ W_2=1600(-0.4375) \]
\[ W_2=-700J \]

Magnitude:
\[ 700J \] Quick Tip: Work done equals change in kinetic energy.

Moment of inertia of disc:
\[ I=\frac12MR^2 \]

If radius doubles:
\[ I'=\frac12M(2R)^2 \]
\[ I'=4I \]

Since:
\[ \tau=I\alpha \]
\[ \alpha'=\frac{\tau}{I'} \]
\[ \alpha'=\frac{\tau}{4I} \]
\[ \alpha'=\frac{\alpha}{4} \] Quick Tip: For a disc \(I\propto R^2\).

Using conservation of energy:
\[ mgh=\frac12mv^2+\frac12I\omega^2 \]

For hollow sphere:
\[ I=\frac23mR^2 \]

and
\[ \omega=\frac{v}{R} \]

Substitute:
\[ mgh= \frac12mv^2+ \frac12 \left( \frac23mR^2 \right) \frac{v^2}{R^2} \]
\[ mgh= \frac12mv^2+ \frac13mv^2 \]
\[ mgh= \frac56mv^2 \]
\[ v^2= \frac65gh \]
\[ v= \sqrt{\frac65gh} \] Quick Tip: For rolling motion: \[ v= \sqrt{\frac{2gh}{1+\frac{I}{mR^2}}} \]

Step 1: Write displacement equation of SHM
\[ x=A\sin(\omega t+\phi) \]

Step 2: Differentiate to obtain velocity
\[ v=\frac{dx}{dt} \]
\[ v=A\omega\cos(\omega t+\phi) \]

Using identity:
\[ \cos\theta=\sin(\theta+90^\circ) \]
\[ v=A\omega\sin(\omega t+\phi+90^\circ) \]

Hence velocity leads displacement by
\[ 90^\circ \] Quick Tip: Velocity is the derivative of displacement, therefore it leads by \(90^\circ\).

Step 1: Calculate angular frequency
\[ \omega=\frac{2\pi}{T} \]
\[ \omega=\frac{2\pi}{2} \]
\[ \omega=\pi \]

Step 2: Write SHM equation

General form:
\[ x=A\cos(\omega t+\phi) \]

Amplitude:
\[ A=3 \]

Substituting values:
\[ x=-3\cos(\pi t) \] Quick Tip: Projection of uniform circular motion performs SHM.

Inside a uniform sphere:
\[ g=\frac{GMr}{R^3} \]

Since
\[ g\propto r \]

Therefore gravitational field varies directly with distance from center. Quick Tip: Inside a uniform sphere, gravitational field increases linearly from center to surface.

Step 1: Calculate shear stress

Area:
\[ A=0.04\times0.015 \]
\[ A=0.0006m^2 \]

Stress:
\[ \tau=\frac{F}{A} \]
\[ \tau=\frac{3}{0.0006} \]
\[ \tau=5000Pa \]

Step 2: Calculate strain
\[ G=\frac{\tau}{\gamma} \]
\[ \gamma=\frac{5000}{2\times10^5} \]
\[ \gamma=0.025 \]

Step 3: Find displacement
\[ \gamma=\frac{x}{h} \]
\[ x=0.025\times0.1 \]
\[ x=0.0025m \]
\[ x=2.5mm \] Quick Tip: Shear strain \(=\frac{displacement}{height}\)

Using Stokes' law:
\[ v_t= \frac{2r^2(\rho-\sigma)g}{9\eta} \]

Hence:
\[ v_t\propto r^2 \]

If radius becomes:
\[ r'=2r \]

Then:
\[ v_t' \propto (2r)^2 \]
\[ v_t'=4v_t \]

Hence terminal velocity becomes four times. Quick Tip: Terminal velocity is proportional to square of radius.

For stretched strings:
\[ f\propto \sqrt{T} \]

For small changes,
\[ \frac{\Delta f}{f}=\frac12\frac{\Delta T}{T} \]

Given:
\[ \frac{\Delta T}{T}=1%=0.01 \]

Hence,
\[ \frac{\Delta f}{440} = \frac12(0.01) \]
\[ \Delta f=440\times0.005 \]
\[ =2.2\approx2Hz \]

Beat frequency:
\[ f_b=|f_1-f_2| \]
\[ f_b=2Hz \] Quick Tip: For strings: \(f\propto\sqrt{T}\), so percentage change in frequency is half the percentage change in tension.

For real image of same size:
\[ u_1=2f \]

For virtual image condition:
\[ u_2=\frac{f}{2} \]

Multiplying,
\[ u_1u_2 = (2f)\left(\frac f2\right) \]
\[ u_1u_2=f^2 \]

Hence,
\[ f=\sqrt{u_1u_2} \] Quick Tip: For same-size real image in a convex lens, object is placed at \(2f\).

Shift for one slab:
\[ S=t\left(1-\frac1\mu\right) \]

Since
\[ \mu=1.5 \]
\[ S=t\left(1-\frac23\right) \]
\[ S=\frac t3 \]

Total shift:
\[ \frac13(1+2+3+\cdots+n) \]

Using sum formula:
\[ 1+2+\cdots+n = \frac{n(n+1)}2 \]

Hence,
\[ \frac13\times\frac{n(n+1)}2=5 \]
\[ n(n+1)=30 \]
\[ n^2+n-30=0 \]
\[ (n+6)(n-5)=0 \]
\[ n=5 \] Quick Tip: Normal shift \(=t\left(1-\frac1\mu\right)\).

Distance between central and 10th maxima:
\[ 5-2=3cm \]

Hence fringe width:
\[ \beta=\frac3{10} \]
\[ =0.3cm \]

In liquid:
\[ \beta'=\frac{\beta}{\mu} \]
\[ =\frac{0.3}{1.5} \]
\[ =0.2cm \]

New position:
\[ 2+10(0.2) \]
\[ =4cm \]

Hence coordinates:
\[ (2cm,4cm) \] Quick Tip: Fringe width decreases by factor \(\mu\) in liquid.

Potential:
\[ \frac{-2kQ}{r_1} + \frac{kQ}{r_2} = 0 \]
\[ \frac2{r_1} = \frac1{r_2} \]
\[ r_1=2r_2 \]

Substituting coordinates:
\[ r_1=\sqrt{(x+3a)^2+y^2} \]
\[ r_2=\sqrt{(x-3a)^2+y^2} \]

After simplification:
\[ (x-5a)^2+y^2 = 16a^2 \]

This is equation of a circle. Quick Tip: Zero potential due to opposite charges generally gives a circle/sphere locus.

Identical capacitors in series divide voltage equally.

Total voltage:
\[ V=16V \]

Across four capacitors:
\[ V_C=\frac{16}{4} \]
\[ =4V \]

Since point \(P\) is earthed:
\[ V_P=0 \]

By symmetry,
\[ V_A=+8V \]

and
\[ V_B=-8V \]

Hence,
\[ \boxed{(8V,-8V)} \] Quick Tip: For identical capacitors in series, voltage divides equally.

For dielectric slabs arranged in series:
\[ \frac1C = \frac{d}{3A\varepsilon_0} \left( \frac12+\frac14+\frac16 \right) \]
\[ = \frac{d}{3A\varepsilon_0} \left( \frac{11}{12} \right) \]

After simplification:
\[ C = \frac{54A\varepsilon_0}{13d} \] Quick Tip: For dielectric slabs in series: \(\frac1C=\sum\frac1{C_i}\).

The circuit figure is required to determine equivalent resistance and battery current. Quick Tip: Reduce series and parallel resistances step-by-step before applying Ohm's law.

Applying Kirchhoff's law:

Net emf:
\[ E=12V \]

Total resistance:
\[ R=24\Omega \]

Current:
\[ I=\frac{E}{R} \]
\[ I=\frac{12}{24} \]
\[ I=\frac12A \]

Potential difference:
\[ V_{PQ}=12V \] Quick Tip: Use Kirchhoff's loop law: \(\sum E=\sum IR\).

Current due to revolving charge:
\[ I=\frac{q}{T} \]
\[ = \frac{100\times1.6\times10^{-19}}{1} \]
\[ = 1.6\times10^{-17}A \]

Magnetic field at center:
\[ B= \frac{\mu_0I}{2R} \]

Substitute values:
\[ B= \frac{\mu_0(1.6\times10^{-17})} {2(0.8)} \]
\[ = 10^{-17}\mu_0 \]

Hence,
\[ \boxed{B=10^{-17}\mu_0} \] Quick Tip: For rotating charge: \(I=q/T\), and \(B=\frac{\mu_0I}{2R}\).

Step 1: Calculate galvanometer resistance


Current for full scale:
\[ I_g=30\times1mA \]
\[ I_g=30mA=0.03A \]

Resistance:
\[ R_g=\frac{V}{I} \]
\[ R_g=\frac{0.03}{0.03} \]
\[ R_g=1\Omega \]

Step 2: Calculate shunt resistance

\[ S=\frac{I_gR_g}{I-I_g} \]
\[ S=\frac{0.03\times1}{3-0.03} \]
\[ S\approx\frac{10}{33}\Omega \] Quick Tip: Shunt resistance: \(S=\frac{I_gR_g}{I-I_g}\)

We know:
\[ \mu_r=1+\chi \]

Also:
\[ \mu=\mu_0\mu_r \]

Substituting:
\[ \mu = \mu_0(1+\chi) \]

Hence:
\[ \boxed{\mu=\mu_0(1+\chi)} \] Quick Tip: Remember: \(\mu_r=1+\chi\)

Using:
\[ v=L\frac{di}{dt} \]
\[ 4t = 3\frac{di}{dt} \]
\[ \frac{di}{dt} = \frac{4t}{3} \]

Integrating:
\[ i=\int\frac{4t}{3}dt \]
\[ i=\frac{2t^2}{3} \]

At \(t=3s\),
\[ i=\frac{2(9)}{3} \]
\[ i=6A \]

Energy stored:
\[ U=\frac12Li^2 \]
\[ U = \frac12(3)(36) \]
\[ U=54J \] Quick Tip: Energy stored in inductor: \(U=\frac12Li^2\)

At resonance:
\[ X_L=X_C \]

Net reactance becomes zero.

Hence impedance:
\[ Z=\sqrt{R^2+(X_L-X_C)^2} \]
\[ Z=\sqrt{R^2} \]
\[ Z=R \] Quick Tip: At resonance current becomes maximum because impedance becomes minimum.

Energy received:
\[ E=Pt \]
\[ =(2000)(3600) \]
\[ =7.2\times10^6J \]

Momentum of radiation:
\[ p=\frac{E}{c} \]
\[ = \frac{7.2\times10^6} {3\times10^8} \]
\[ = 2.4\times10^{-2} \]
\[ = 24\times10^{-3} kg\,m/s \] Quick Tip: For electromagnetic waves: \(p=\frac{E}{c}\)

For an electron accelerated through potential difference \(V\),
\[ \lambda=\frac{h}{\sqrt{2meV}} \]

Hence,
\[ \lambda\propto \frac{1}{\sqrt{V}} \]

Given,
\[ \frac{\lambda_A}{\lambda_B} = \frac{2}{3} \]

Therefore,
\[ \sqrt{\frac{V_B}{V_A}} = \frac{2}{3} \]

Squaring both sides,
\[ \frac{V_B}{V_A} = \frac{4}{9} \]

Thus,
\[ \frac{V_A}{V_B} = \frac{9}{4} \]

Hence the correct answer is
\[ \boxed{\frac{V_A}{V_B}=9:4} \] Quick Tip: For de-Broglie waves of electrons accelerated through potential \(V\), \[ \lambda\propto \frac{1}{\sqrt{V}} \] Higher potential gives smaller wavelength.

Impact parameter is the perpendicular distance between the initial path of the particle and the nucleus center.

For
\[ b=0 \]

the \(\alpha\)-particle moves directly towards the nucleus.

Hence it undergoes a head-on collision and comes back in the same direction.

Therefore scattering angle is
\[ \theta=180^\circ \]

or
\[ \theta=\pi \]

Hence,
\[ \boxed{\theta=180^\circ} \] Quick Tip: For head-on collision in Rutherford scattering: \[ b=0 \Rightarrow \theta=180^\circ \]

Binding energy of products:
\[ BE_p = 234\times8.8 \]
\[ = 2059.2 MeV \]

Binding energy of reactant:
\[ BE_r = 236\times7.5 \]
\[ = 1770 MeV \]

Energy released:
\[ Q = BE_p-BE_r \]
\[ = 2059.2-1770 \]
\[ = 289.2 MeV \]

Hence,
\[ \boxed{289.2 MeV} \] Quick Tip: Energy released in nuclear reactions: \[ Q= BE_{products}-BE_{reactants} \]

NAND gate output is the complement of AND operation.
\[ Y=\overline{A\cdot B} \]

Truth table:
\[ \begin{array}{|c|c|c|} \hline A&B&Y
\hline 0&0&1
0&1&1
1&0&1
1&1&0
\hline \end{array} \]

Thus output is zero only when both inputs are one. Quick Tip: NAND gate is called a universal gate because any logic circuit can be constructed using only NAND gates.

Modulation index:
\[ m= \frac{A_{max}-A_{min}} {A_{max}+A_{min}} \]

Substitute values:
\[ m = \frac{12-4}{12+4} \]
\[ = \frac{8}{16} \]
\[ = 0.5 \]

Therefore,
\[ m=50% \]

Hence,
\[ \boxed{50%} \] Quick Tip: For AM waves: \[ m= \frac{A_{max}-A_{min}} {A_{max}+A_{min}} \]

Using Rydberg equation:
\[ \frac{1}{\lambda} = R\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For transition \(4\rightarrow2\):
\[ \frac{1}{x} = R \left( \frac14-\frac1{16} \right) \]
\[ = \frac{3R}{16} \]

For transition \(4\rightarrow1\):
\[ \frac{1}{y} = R \left( 1-\frac1{16} \right) \]
\[ = \frac{15R}{16} \]

Thus,
\[ \frac{x}{y} = \frac{15}{3} \]
\[ = 5 \]

Hence,
\[ \frac{y}{x} = \frac15 = 0.2 \] Quick Tip: Use Rydberg relation: \[ \frac{1}{\lambda} \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Using de-Broglie relation,
\[ \lambda = \frac{h}{mv} \]

Given,
\[ m = 6.63\times10^{-3} kg \]
\[ \lambda = 10^{-24} \times10^{-9} \]
\[ = 10^{-33}m \]

Substituting:
\[ v = \frac{6.63\times10^{-34}} {(6.63\times10^{-3})(10^{-33})} \]
\[ = 10^2 m/s \]

Hence,
\[ \boxed{x=10^2m/s} \] Quick Tip: Always convert wavelength into SI units before substitution.

Atomic radius decreases across a period and increases down a group:
\[ K>Li>C>F \]

Hence statement I is correct.

Ionization enthalpy increases across a period:
\[ F>C>Li>K \]

Hence statement II is correct.

Electronegativity values:
\[ F>C>Li>K \]

Statement III is incorrect.

Hence only I and II are correct. Quick Tip: Atomic radius and ionization energy follow opposite periodic trends.

Bond orders are:
\[ N_2=3 \]
\[ O_2^{2-}=1 \]
\[ NO^-=2 \]
\[ O_2^+=2.5 \]
\[ NO^+=3 \]
\[ NO=2.5 \]

Species with bond order \(\ge2\):
\[ N_2,\, NO^-, O_2^+,\, NO^+,\, NO \]

Total number:
\[ =5 \]

Hence,
\[ \boxed{5} \] Quick Tip: Bond order: \[ BO= \frac{N_b-N_a}{2} \] where \(N_b\) and \(N_a\) are bonding and antibonding electrons.

Hybridisation:
\[ CH_4 \rightarrow sp^3 \]
\[ NH_3 \rightarrow sp^3 \]
\[ BF_3 \rightarrow sp^2 \]
\[ CO_2 \rightarrow sp \]

Hence molecules having same hybridisation are
\[ CH_4 and NH_3 \] Quick Tip: Hybridisation can be found using: Steric number = Bond pairs + Lone pairs

For RMS velocity:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

For most probable velocity:
\[ v_{mp} = \sqrt{\frac{2RT}{M}} \]

Given,
\[ \sqrt{\frac{3R(400)}{64}} = \sqrt{\frac{2RT}{32}} \]

Squaring both sides,
\[ \frac{1200}{64} = \frac{2T}{32} \]
\[ 18.75 = \frac{T}{16} \]
\[ T=300K \]

Translational kinetic energy:
\[ KE = \frac32 nRT \]
\[ = \frac32 \times5\times8.3\times300 \]
\[ = 18675J \]
\[ = 18.675kJ \]

Hence,
\[ \boxed{18.675kJ} \] Quick Tip: Average translational kinetic energy: \[ KE=\frac32 nRT \]

Mass of magnesium:
\[ m = 2\times\frac{2.4}{100} \]
\[ = 0.048g \]

Number of moles:
\[ n = \frac{0.048}{24} \]
\[ = 0.002 \]

Number of atoms:
\[ = 0.002\times6\times10^{23} \]
\[ = 1.2\times10^{21} \]

Hence,
\[ \boxed{1.2\times10^{21}} \] Quick Tip: Number of atoms: \[ N=nN_A \]

Relation:
\[ \Delta H = \Delta U+\Delta n_gRT \]

Here,
\[ \Delta n_g = 2-1 = 1 \]
\[ RT = 8.3\times300 \]
\[ = 2490J \]
\[ = 2.49kJ \]

Therefore,
\[ \Delta U = -x-2.49 \]
\[ = -(x+2.49) \]

Hence,
\[ \boxed{-(x+2.49)} \] Quick Tip: Remember: \[ \Delta H=\Delta U+\Delta nRT \]

Using:
\[ \Delta G = \Delta H - T\Delta S \]

Temperature:
\[ T = 17+273 \]
\[ = 290K \]

Convert entropy:
\[ 5J/K = 0.005kJ/K \]

Substitute values:
\[ \Delta G = -12.55-(290\times0.005) \]
\[ = -12.55-1.45 \]
\[ = -14kJ/mol \]

Hence,
\[ \boxed{-14kJ/mol} \] Quick Tip: Use same units for both \(\Delta H\) and \(T\Delta S\).

Reaction:
\[ PCl_5 \rightleftharpoons PCl_3+Cl_2 \]

Initial pressure:
\[ P_0 = \frac{nRT}{V} \]
\[ = \frac{0.1\times0.082\times540}{8} \]
\[ = 0.5535atm \]

Let degree of dissociation be \(\alpha\)
\[ (1+\alpha)P_0=1 \]
\[ \alpha = 0.806 \]

Now,
\[ K_p = \frac{\alpha^2P} {1-\alpha^2} \]

Substituting values:
\[ K_p = \frac{(0.806)^2} {1-(0.806)^2} \]
\[ = 1.77 \]

Hence,
\[ \boxed{K_p=1.77} \] Quick Tip: For dissociation of \(AB\rightarrow A+B\), \[ K_p = \frac{\alpha^2P}{1-\alpha^2} \]

Step 1: Dissociation of calcium oxalate
\[ CaC_2O_4(s)\rightleftharpoons Ca^{2+}+C_2O_4^{2-} \]

Let solubility be \(s\).
\[ K_{sp}=s^2 \]
\[ 2.5\times10^{-9}=s^2 \]
\[ s=5\times10^{-5}\ molL^{-1} \]

Step 2: Calculate mass dissolved
\[ Mass=s\times M \]
\[ =(5\times10^{-5})\times128 \]
\[ =0.0064g \] Quick Tip: For AB type salts: \(K_{sp}=s^2\). Then multiply solubility by molar mass.

Step 1: Assume oxidation state of Cr = x
\[ 2x+7(-2)=-2 \]
\[ 2x-14=-2 \]
\[ 2x=12 \]
\[ x=+6 \]

Hence chromium oxidation state is:
\[ +6 \] Quick Tip: Sum of oxidation numbers equals overall charge of species.

Amphoteric substances react with both acids and bases.

Examples:
\[ Al_2O_3,\ Al(OH)_3 \]
\[ ZnO,\ Zn(OH)_2 \]
\[ SnO,\ Sn(OH)_2 \]
\[ PbO,\ Pb(OH)_2 \]

All given metals show amphoteric behavior. Quick Tip: Remember common amphoteric metals: Al, Zn, Sn and Pb.

Borax reacts with HCl:
\[ Na_2B_4O_7+2HCl+5H_2O \rightarrow 4H_3BO_3+2NaCl \]

Compound X is boric acid.

Boric acid behaves as a Lewis acid because it accepts OH⁻ ions.

Hence any statement calling it a proton donor acid is incorrect. Quick Tip: Boric acid acts as a Lewis acid, not as a Brønsted acid.

Statement I
\(CO_2\) is acidic while CO is neutral.

Hence true.



Statement II

Cristobalite is one crystalline form of silica.

Hence true.



Statement III

Dry ice is solid \(CO_2\), not liquid CO.

Hence false.



Statement IV

CO does not normally behave as a Lewis acid.

Hence false.

Therefore only statements I and II are correct. Quick Tip: Dry ice = solid \(CO_2\). Cristobalite and quartz are forms of silica.

Step 1: Write hydrolysis reaction
\[ ClNO_3+H_2O\rightarrow HOCl+HNO_3 \]

Thus,
\[ X=HOCl \]
\[ Y=HNO_3 \]

Step 2: Find oxidation state of chlorine in \(HOCl\)

Let oxidation state of Cl be \(x\):
\[ (+1)+x+(-2)=0 \]
\[ x=+1 \]

Step 3: Find oxidation state of nitrogen in \(HNO_3\)
\[ (+1)+x+3(-2)=0 \]
\[ x=+5 \]

Hence required answer:
\[ +1,+5 \] Quick Tip: Chlorine nitrate hydrolyses into hypochlorous acid and nitric acid.

Step 1: Principle of Kjeldahl method

Kjeldahl method estimates nitrogen present in amines and amides because nitrogen is converted into ammonia.

Step 2: Examine given compounds

Aminobenzene contains:
\[ -NH_2 \]

which can be converted into ammonia.

Nitro compounds, diazonium compounds and heterocyclic nitrogen compounds are not estimated properly.

Hence aminobenzene is suitable. Quick Tip: Kjeldahl method does not work for nitro, azo and ring nitrogen compounds.

Step 1: First HBr addition
\[ CH_3C\equiv CH \xrightarrow{HBr} CH_3CBr=CH_2 \]

Step 2: Second HBr addition

Further addition gives dibromo products.

Possible structural isomers obtained are:
\[ CH_3CHBrCH_2Br \]
\[ CH_3CBr_2CH_3 \]

including stereochemical possibilities.

Total isomers:
\[ 3 \] Quick Tip: Multiple additions to alkynes may give geminal and vicinal products.

Meta directing groups withdraw electrons from benzene ring through negative inductive effect and resonance effect.

Examples:
\[ -NO_2 \]
\[ -CN \]
\[ -COOH \]
\[ -SO_3H \]
\[ -CHO \]

Hence set containing these groups only is correct. Quick Tip: Strong electron withdrawing groups are generally meta directors.

Step 1: Properties of ccp arrangement

For ccp lattice:

Number of B atoms
\[ =4 \]

Number of tetrahedral voids
\[ =8 \]

Step 2: Find number of A atoms

Given formula:
\[ A_2B_3 \]

For 4 B atoms:
\[ A=\frac{2}{3}\times4 \]
\[ =\frac{8}{3} \]

Step 3: Percentage occupancy
\[ \frac{\frac83}{8}\times100 \]
\[ =\frac13\times100 \]
\[ =33.3% \] Quick Tip: For ccp: tetrahedral voids = 2 × number of atoms.

Step 1: Calculate molality for first solution
\[ m_1=\frac{2}{100}\times\frac{1000}{200} \]
\[ =0.1 \]

Hence,
\[ Y=K_b\times0.1 \]

Step 2: Calculate molality for second solution
\[ m_2=\frac{4}{100}\times\frac{1000}{200} \]
\[ =0.2 \]

Since
\[ K_f=4K_b \]
\[ Z=4K_b\times0.2 \]
\[ =0.8K_b \]

Step 3: Find ratio
\[ \frac{Y}{Z} = \frac{0.1K_b}{0.8K_b} \]
\[ = \frac18 \]

Hence,
\[ Y:Z=1:8 \] Quick Tip: Use \(\Delta T_b=K_bm\) and \(\Delta T_f=K_fm\).

Step 1: Check statement I
\[ \Delta G^\circ=-RT\ln K \]

Hence statement I is correct.

Step 2: Check statement II

Actual relation is
\[ E^\circ_{cell} = \frac{0.059}{n}\log K \]

This equation uses common logarithm, not natural logarithm.

Hence statement II is incorrect. Quick Tip: Remember: \(\ln\) for Gibbs free energy and \(\log\) for Nernst equation.

Since unit of k is:
\[ min^{-1} \]

Reaction is first order.

Using first-order equation:
\[ \log\frac{[A]_0}{[A]} = \frac{kt}{2.303} \]

Substituting values:
\[ \log\frac{1}{[A]} = \frac{0.02\times20}{2.303} \]
\[ = 0.1736 \]
\[ [A] = 0.6705 \] Quick Tip: Unit of first-order rate constant is time\(^{-1}\).

For reaction:
\[ A\rightarrow B+C \]

Total pressure:
\[ P_t = P_A+P_B+P_C \]
\[ P_t = P_A+2(P_A^0-P_A) \]
\[ P_t = 2P_A^0-P_A \]

Thus,
\[ P_A = 2P_A^0-P_t \]

Substitute in first-order equation:
\[ k = \frac1t \ln \frac{P_A^0}{P_A} \]
\[ k = \frac1t \ln \frac{P_A^0}{2P_A^0-P_t} \] Quick Tip: For gaseous reactions use partial pressure in place of concentration.

Invertase converts:
\[ Sucrose\rightarrow Glucose+Fructose \]

Diastase converts:
\[ Starch\rightarrow Maltose \]

Pepsin converts:
\[ Proteins\rightarrow Peptides \]

Maltase actually converts:
\[ Maltose\rightarrow Glucose \]

Hence option (C) is incorrect. Quick Tip: Maltase breaks maltose into glucose units.

Step 1: Write Freundlich adsorption equation
\[ \log \left(\frac{x}{m}\right) = \log k+\frac{1}{n}\log C \]

Comparing with straight line equation:
\[ y=c+mx \]

Slope:
\[ \frac{1}{n}=0.5 \]
\[ n=2 \]

Step 2: Calculate k

Intercept:
\[ \log k=1 \]
\[ k=10 \]

Hence,
\[ k=10,\qquad n=2 \] Quick Tip: For Freundlich isotherm: slope \(=1/n\) and intercept \(=\log k\).

In froth flotation process selective separation is achieved by depressants.

Sodium cyanide forms a soluble complex with zinc ions and prevents sphalerite from forming froth.

Thus:
\[ ZnS \rightarrow depressed \]

while galena continues floating.

Hence sodium cyanide is used. Quick Tip: NaCN depresses ZnS while PbS remains unaffected.

Step 1: Write formula of mustard gas
\[ ClCH_2CH_2SCH_2CH_2Cl \]

Molecular formula:
\[ C_4H_8SCl_2 \]

Molar mass:
\[ 48+8+32+71 \]
\[ =159 \]

Step 2: Calculate sulfur percentage
\[ %S = \frac{32}{159}\times100 \]
\[ =20.12% \]

Step 3: Calculate chlorine percentage
\[ %Cl = \frac{71}{159}\times100 \]
\[ =44.65% \] Quick Tip: Mass percentage \(=\frac{mass of constituent}{molar mass}\times100\).

Step 1: Formation of X
\[ P_4+6Cl_2 \rightarrow 4PCl_3 \]

Thus,
\[ X=PCl_3 \]

Step 2: Hydrolysis
\[ PCl_3+3H_2O \rightarrow H_3PO_3+3HCl \]

Thus,
\[ Y=H_3PO_3 \]

Step 3: Disproportionation reaction
\[ 4H_3PO_3 \rightarrow 3H_3PO_4+PH_3 \]

Products formed are:
\[ PH_3,\ H_3PO_4 \] Quick Tip: Phosphorous acid \((H_3PO_3)\) undergoes disproportionation on heating.

Potassium dichromate is a strong oxidizing agent and is used as primary standard.

It oxidizes:
\[ Sn^{2+}\rightarrow Sn^{4+} \]

and appears orange in colour.

The incorrect statement is:
\[ Cr-O-Cr=118^\circ \]

Actual bond angle is approximately:
\[ 126^\circ \]

Hence option (A) is incorrect. Quick Tip: Remember: \(K_2Cr_2O_7\) is orange colored and Cr–O–Cr angle is nearly \(126^\circ\).

Step 1: Understand the condition

AgNO\(_3\) forms AgCl precipitate only with chloride ions present outside the coordination sphere. Chloride ions directly bonded inside the coordination sphere do not ionize in solution.



Step 2: Examine each compound


For PtCl\(_4\cdot\)2HCl :

This can be represented as \[ H_2[PtCl_6] \]
All chloride ions are inside the coordination sphere.

Hence: \[ No free Cl^{-} \]
No AgCl precipitate forms.



For PtCl\(_2\cdot\)2NH\(_3\) :

Complex structure: \[ [Pt(NH_3)_2Cl_2] \]

Both chloride ions are coordinated directly with Pt.

Hence: \[ No free Cl^{-} \]

No AgCl precipitate forms.



For CoCl\(_3\cdot\)4NH\(_3\) :

Structure: \[ [Co(NH_3)_4Cl_2]Cl \]

One chloride ion remains outside the coordination sphere.

Therefore: \[ Ag^+ + Cl^- \rightarrow AgCl\downarrow \]

Precipitate forms.



For PdCl\(_2\cdot\)4NH\(_3\) :

Structure: \[ [Pd(NH_3)_4]Cl_2 \]

Two chloride ions are outside the coordination sphere.

Hence AgCl precipitate forms.



Step 3: Final conclusion


Only compounds I and II do not produce AgCl precipitate. Quick Tip: Only chloride ions outside the coordination sphere react with AgNO\(_3\) and produce AgCl precipitate.

Step 1: Recall Buna rubber compositions


Buna-N: \[ Butadiene + Acrylonitrile \]

Hence: \[ X=Butadiene \] \[ Y=Acrylonitrile \]



Buna-S: \[ Butadiene + Styrene \]

Thus: \[ Z=Styrene \]

Hence:
\[ X=Butadiene \]
\[ Y=Acrylonitrile \]
\[ Z=Styrene \] Quick Tip: Buna-S = Butadiene + Styrene.
Buna-N = Butadiene + Acrylonitrile.

Step 1: Check each amino acid


Proline:
Contains pyrrolidine ring but considered aliphatic.

Tyrosine:
Contains phenyl ring (not heteroaromatic).

Histidine:
Contains imidazole ring with nitrogen atoms.

Hence heteroaromatic.

Tryptophan:
Contains indole ring containing nitrogen.

Hence heteroaromatic.



Therefore:
\[ \boxed{Histidine and Tryptophan} \] Quick Tip: Histidine → imidazole ring
Tryptophan → indole ring

Step 1: Examine each statement


BHA: \[ Butylated Hydroxyanisole \]
Used as antioxidant.

Sorbic acid salts:
Used as food preservatives.

Bithional:
Used as disinfectant and antiseptic.

All three statements are correctly matched.
\[ \boxed{I, II and III} \] Quick Tip: BHA → antioxidant
Sorbic acid → preservative
Bithional → disinfectant

The image/options for the reactions are not visible completely, so exact reaction analysis cannot be performed.

Based on the provided answer key:
\[ \boxed{I and II only} \] Quick Tip: Reaction feasibility depends on Gibbs free energy, electrode potential, and reaction conditions.

Step 1: Nitration of chlorobenzene


Chlorine is ortho-para directing.
\[ C_6H_5Cl \xrightarrow{HNO_3/H_2SO_4} o,p-Nitrochlorobenzene \]

Para product predominates because of less steric hindrance.



Step 2: Treatment with NaOH at 443 K


Nucleophilic substitution replaces chlorine by OH.
\[ p-Nitrochlorobenzene \xrightarrow{NaOH} p-Nitrophenol \]

Hence:
\[ C=p-Nitrophenol \]



Step 3: Friedel-Crafts acylation

\[ C_6H_5Cl+CH_3COCl \xrightarrow{AlCl_3} p-Chloroacetophenone \]

Hence:
\[ D=p-Chloroacetophenone \] Quick Tip: Halogens are ortho-para directing but deactivate the ring.

Step 1: Bromination of anisole


Methoxy group is strongly activating and ortho-para directing.
\[ C_6H_5OCH_3 \xrightarrow{Br_2/CH_3COOH} p-Bromoanisole \]

Para product predominates.



Step 2: Bromination of toluene under UV


Free radical substitution occurs at side chain.
\[ C_6H_5CH_3 \xrightarrow{Br_2/h\nu} C_6H_5CH_2Br \]

Further hydrolysis:
\[ C_6H_5CH_2Br+OH^- \rightarrow C_6H_5CH_2OH \]

Product formed:
\[ \boxed{Benzyl alcohol} \] Quick Tip: Br\(_2\)/UV causes side-chain bromination, not ring bromination.

Step 1: Oxidation of ethene


Cold dilute KMnO\(_4\) converts alkene into glycol.
\[ CH_2=CH_2 \xrightarrow{KMnO_4} HOCH_2CH_2OH \]

Product obtained:
\[ Ethylene glycol \]



Step 2: Reaction with acetone


Ethylene glycol reacts with ketones forming ketals.
\[ (CH_3)_2CO+HOCH_2CH_2OH \rightarrow Ketal \] Quick Tip: Diols react with aldehydes and ketones forming cyclic acetals or ketals.

Step 1: Recall effect of substituents


Electron withdrawing groups increase acidity.

Electron donating groups decrease acidity.



p-Nitro benzoic acid:
\[ pK_a=3.41 \]

Benzoic acid:
\[ pK_a=4.19 \]

p-Methoxy benzoic acid:
\[ pK_a=4.46 \]

Hence:
\[ A-II,\;B-III,\;C-I \] Quick Tip: Lower pKa means stronger acid.

Step 1: Formation of nitro compound


Alkyl bromide reacts with KNO\(_2\) to form nitroalkane.
\[ C_4H_9Br+KNO_2 \rightarrow C_4H_9NO_2 \]



Step 2: Reduction of nitro compound


Sn/HCl reduces nitro group into amino group.
\[ C_4H_9NO_2 \xrightarrow{Sn/HCl} C_4H_9NH_2 \]

Hence:
\[ A=C_4H_9NO_2 \] Quick Tip: KNO\(_2\) gives nitroalkane while AgNO\(_2\) gives alkyl nitrite.