Solved Examples Bank | Ishu (My Wife)

Chapter 1: Relations & Functions ~5 Marks Board

Example 15 MarksEquivalence Relation

Question

Show that R = {(a,b) : 2 divides (a-b)} defined on Z is an equivalence relation.

1

Reflexive: For any aZ, a-a=0=20. So 2|(a-a) (a,a)R for all a. R is Reflexive.

2

Symmetric: Let (a,b)R a-b=2k. Then b-a = 2(-k) 2|(b-a) (b,a)R. R is Symmetric.

3

Transitive: (a,b)R and (b,c)R a-b=2k, b-c=2k. So a-c = 2(k+k) (a,c)R. R is Transitive.

Conclusion

Since R is Reflexive, Symmetric and Transitive R is an Equivalence Relation.

CBSE Tip: Prove all 3 properties separately. Each = 1 mark. End with "Hence R is an Equivalence Relation". Never skip any step!

Example 23 MarksOne-One & Onto

Question

Show that f:RR defined by f(x) = 2x+3 is bijective.

1

One-One: Let f(x)=f(x) 2x+3=2x+3 x=x. f is One-One.

2

Onto: For any yR, let x=(y-3)/2 R. Then f(x)=2(y-3)/2+3=y. f is Onto.

Conclusion

f is One-One and Onto f is Bijective.

CBSE Tip: One-One: start with f(x)=f(x), conclude x=x. Onto: take arbitrary y in codomain, find x in domain.

Chapter 2: Inverse Trigonometry ~5 Marks Board

Example 12 MarksPrincipal Value

Question

Find the principal value of $\cos^{-1}\!\left(-\dfrac{\sqrt{3}}{2}\right)$

1

Use identity: $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$

$\cos^{-1}\!\left(-\frac{\sqrt{3}}{2}\right) = \pi - \cos^{-1}\!\left(\frac{\sqrt{3}}{2}\right)$

2

$\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ so $\cos^{-1}\!\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$

$= \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Final Answer

$\boxed{\dfrac{5\pi}{6}}$

CBSE Tip: cos(-x) = π - cos(x). NOT = -cos(x)! Most common mistake. sin(-x) = -sin(x) (odd function). cos is NOT odd!

Example 23 MarksSimplify ITF

Question

Prove: $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{2}{11} = \tan^{-1}\frac{3}{4}$

1

Check: $xy = \frac{1}{2}\times\frac{2}{11}=\frac{1}{11} < 1$ Use direct formula.

2

$\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{2}{11} = \tan^{-1}\!\left(\frac{\frac{1}{2}+\frac{2}{11}}{1-\frac{1}{2}\cdot\frac{2}{11}}\right) = \tan^{-1}\!\left(\frac{\frac{15}{22}}{\frac{20}{22}}\right) = \tan^{-1}\frac{15}{20} = \tan^{-1}\frac{3}{4}$

Hence Proved

LHS = $\tan^{-1}\dfrac{3}{4}$ = RHS

CBSE Tip: Always verify xy < 1 before applying addition formula. If xy > 1, add π. Write "LHS = ... = RHS, Hence Proved".

Chapter 3: Matrices ~5 Marks Board

Example 13 MarksSym + Skew Decomposition

Question

Express $A=\begin{pmatrix}3&5\\1&-1\end{pmatrix}$ as sum of a Symmetric and a Skew-Symmetric matrix.

1

Find $A^T = \begin{pmatrix}3&1\\5&-1\end{pmatrix}$
Formula: $A = \underbrace{\frac{A+A^T}{2}}_{P} + \underbrace{\frac{A-A^T}{2}}_{Q}$

2

Symmetric $P = \frac{A+A^T}{2} = \frac{1}{2}\begin{pmatrix}6&6\\6&-2\end{pmatrix} = \begin{pmatrix}3&3\\3&-1\end{pmatrix}$. Verify: $P^T=P$

3

Skew-Sym $Q = \frac{A-A^T}{2} = \frac{1}{2}\begin{pmatrix}0&4\\-4&0\end{pmatrix} = \begin{pmatrix}0&2\\-2&0\end{pmatrix}$. Verify: $Q^T=-Q$

Final Answer

$A = \begin{pmatrix}3&3\\3&-1\end{pmatrix} + \begin{pmatrix}0&2\\-2&0\end{pmatrix}$

CBSE Tip: Verify P=Pᵀ and Q=-Qᵀ explicitly. Diagonal of skew-symmetric matrix is ALWAYS zero. This is a common 3-mark question!

Chapter 4: Determinants ~8 Marks Board

Example 15 MarksSolve System via Matrix

Question

Solve using matrices: x+y+z=6, 2y+5z=4, 2x+5yz=27

1

Write AX=B: $A=\begin{pmatrix}1&1&1\\0&2&5\\2&5&-1\end{pmatrix}$, $X=\begin{pmatrix}x\\y\\z\end{pmatrix}$, $B=\begin{pmatrix}6\\-4\\27\end{pmatrix}$

2

Find $|A| = 1(2\cdot(-1)-5\cdot5)-1(0-10)+1(0-4) = (-27)+10-4 = -21 \neq 0$

3

Find adj(A) by computing all 9 cofactors with signs $(-1)^{i+j}$, then $A^{-1}=\frac{1}{|A|}$adj(A). Then $X=A^{-1}B$.

Final Answer

x = 1, y = 2, z = 3

CBSE Tip: Always check |A|0 first. Cofactor sign = (-1)^(i+j). Show each cofactor calculation each cofactor carries marks!

Chapter 5: Continuity & Differentiability ~8 Marks Board

Example 13 MarksFind k for Continuity

Question

Find k so that $f(x)=\begin{cases}kx+1,& x\leq5\\ 3x-5,& x>5\end{cases}$ is continuous at x=5.

1

For continuity at x=5: LHL = RHL = f(5)

2

LHL: $\lim_{x\to5^-}(kx+1) = 5k+1$

3

RHL: $\lim_{x\to5^+}(3x-5) = 15-5 = 10$

4

f(5)=5k+1. Equating LHL=RHL: $5k+1=10 \Rightarrow k=\frac{9}{5}$

Final Answer

$k = \dfrac{9}{5}$

CBSE Tip: Write "For continuity, LHL=RHL=f(c)" before solving this line alone gives 1 mark. Show LHL, RHL, f(c) in separate steps!

Chapter 6: Application of Derivatives ~7 Marks Board

Example 15 MarksOptimization

Question

Find dimensions of rectangle of maximum area inscribed in a circle of radius r.

1

Let sides = 2x, 2y. Diagonal = 2r so $x^2+y^2=r^2 \Rightarrow y=\sqrt{r^2-x^2}$

2

Area A = 4xy. Maximize $S=A^2=16x^2(r^2-x^2)$

3

$\frac{dS}{dx}=16(2xr^2-4x^3)=0 \Rightarrow x^2=\frac{r^2}{2} \Rightarrow x=\frac{r}{\sqrt{2}}$

4

$y=\frac{r}{\sqrt{2}}$. Verify: $\frac{d^2S}{dx^2}<0$ (Maximum)

Final Answer

Dimensions: $r\sqrt{2} \times r\sqrt{2}$ (a square gives max area!)

CBSE Tip: Always verify max/min using 2nd derivative test. Write conclusion: "Maximum area = __ sq units at x = __". State units!

Chapter 7: Integrals ~8 Marks Board

Example 15 MarksIntegration by Parts

Question

Evaluate: $\int x^2 e^x\,dx$

1

ILATE: u=x (Algebraic), dv=eˣdx

$\int u\,dv = uv - \int v\,du$

2

$= x^2e^x - \int e^x\cdot2x\,dx = x^2e^x - 2\int xe^x\,dx$

3

Apply by parts again: $\int xe^x dx = xe^x - e^x$

4

$= x^2e^x - 2(xe^x-e^x) = e^x(x^2-2x+2)$

Final Answer

$\boxed{e^x(x^2-2x+2)+C}$

CBSE Tip: Always write +C! Apply ILATE (Algebraic before Exponential). Show every step of By Parts explicitly.

Example 25 MarksDefinite Integral Property

Question

Evaluate: $\int_0^{\pi/2}\dfrac{\sin x}{\sin x+\cos x}\,dx$

1

Let $I=\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx$ ...(i)

2

Apply $\int_0^a f(x)dx=\int_0^a f(a-x)dx$:

$I=\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}dx$ ...(ii)

3

Add (i)+(ii): $2I=\int_0^{\pi/2}1\,dx=\frac{\pi}{2}$

Final Answer

$\boxed{I = \dfrac{\pi}{4}}$

CBSE Tip: "King Property" comes EVERY year! Form I, apply property, add both I's, divide. This exact method must be shown.

Chapter 8: Application of Integrals ~5 Marks Board

Example 15 MarksArea Bounded by Curves

Question

Find the area bounded by the parabola $y^2=4x$ and the line $x=3$.

1

Sketch: parabola $y^2=4x$ opens rightward, symmetric about x-axis. Line x=3 is vertical.

2

By symmetry, Total Area = 2 Area above x-axis.

$A = 2\int_0^3\sqrt{4x}\,dx = 4\int_0^3 x^{1/2}\,dx$

3

$= 4\left[\frac{x^{3/2}}{3/2}\right]_0^3 = \frac{8}{3}\cdot3\sqrt{3} = 8\sqrt{3}$

Final Answer

$A = 8\sqrt{3}$ sq. units

CBSE Tip: Draw a rough sketch! Always write "sq. units" compulsory for full marks. Use symmetry when possible to simplify limits.

Chapter 9: Differential Equations ~6 Marks Board

Example 15 MarksLinear DE (I.F. Method)

Question

Solve: $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$

1

Form: $\frac{dy}{dx}+Py=Q$ where $P=\frac{1}{x}$, $Q=x^2$

2

I.F. $= e^{\int P\,dx}=e^{\int\frac{dx}{x}}=e^{\ln x}=x$

3

Multiply both sides by I.F.=x: $\frac{d}{dx}(xy)=x^3$

4

Integrate: $xy=\int x^3dx=\frac{x^4}{4}+C$

General Solution

$xy = \dfrac{x^4}{4} + C$

CBSE Tip: Steps (mandatory): (1) Identify P,Q (2) Find I.F. (3) Multiply (4) Integrate RHS. ALWAYS write +C. Formula: yIF = QIF dx + C

Chapter 10: Vector Algebra ~6 Marks Board

Example 14 MarksDot Product & Angle

Question

If $\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b}=3\hat{i}-\hat{j}+2\hat{k}$, find $\vec{a}\cdot\vec{b}$ and angle between them.

1

$\vec{a}\cdot\vec{b}=(1)(3)+(2)(-1)+(-3)(2)=3-2-6=-5$

2

$|\vec{a}|=\sqrt{1+4+9}=\sqrt{14}$, $|\vec{b}|=\sqrt{9+1+4}=\sqrt{14}$

3

$\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}=\frac{-5}{\sqrt{14}\cdot\sqrt{14}}=\frac{-5}{14}$

$\theta=\cos^{-1}\!\left(\frac{-5}{14}\right)$

Final Answer

$\vec{a}\cdot\vec{b}=-5$, $\quad\theta=\cos^{-1}\!\left(\dfrac{-5}{14}\right)$

CBSE Tip: dot=0. cross=0. Compute magnitudes separately. Area of triangle = |ab|. Always show formula then substitution.

Chapter 11: 3D Geometry ~7 Marks Board

Example 13 MarksDistance: Point to Plane

Question

Find the distance from P(2,3,1) to the plane 3x4y+12z+13=0.

1

Formula: $d=\dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$

2

Numerator: $|3(2)+(-4)(3)+12(-1)+13|=|6-12-12+13|=|-5|=5$

3

Denominator: $\sqrt{9+16+144}=\sqrt{169}=13$

Final Answer

$d = \dfrac{5}{13}$ units

CBSE Tip: Distance is NEVER negative always use |absolute value|. Write "units" at the end. Memorize this formula perfectly!

Chapter 12: Linear Programming ~5 Marks Board

Example 15 MarksMaximize Z

Question

Maximize Z=5x+3y subject to: 3x+5y15, 5x+2y10, x0, y0.

1

Plot constraints and shade feasible region. Find all corner points.

2

Corner points: (0,0), (2,0), (0,3). Intersection of 3x+5y=15 & 5x+2y=10:

Solving: x=20/19, y=45/19

3

Evaluate Z at all corners:

Point	Z=5x+3y
(0,0)	0
(2,0)	10
(20/19, 45/19)	235/19 MAX
(0,3)	9

Final Answer

Maximum Z = 235/19 at (20/19, 45/19)

CBSE Tip: Corner-point table is COMPULSORY. Shade feasible region. State conclusion: "Maximum value of Z is __ at point (__, __)". Never skip the table!

Chapter 13: Probability ~8 Marks Board

Example 15 MarksBayes' Theorem

Question

Bag I: 4 red + 4 black. Bag II: 2 red + 6 black. One bag chosen at random, one ball drawn is red. Find P(Bag I was chosen).

1

Let E=Bag I, E=Bag II, A=Red ball.

$P(E_1)=\frac{1}{2},\quad P(E_2)=\frac{1}{2}$

2

$P(A|E_1)=\frac{4}{8}=\frac{1}{2},\quad P(A|E_2)=\frac{2}{8}=\frac{1}{4}$

3

Total probability P(A):

$=\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$

4

Bayes' Theorem:

$P(E_1|A)=\frac{P(E_1)\cdot P(A|E_1)}{P(A)}=\frac{\frac{1}{4}}{\frac{3}{8}}=\frac{2}{3}$

Final Answer

$P(E_1|A) = \dfrac{2}{3}$

CBSE Tip: 4 steps mandatory: (1) Define events (2) Write P(Eᵢ) (3) Write P(A|Eᵢ) (4) Compute P(A) THEN Bayes. Missing any step = marks cut!