Class 12 Mathematics Chapter-Wise Subjective Question Bank PDF — 15 Questions Per Chapter (Calculus & Vectors)

To prove that R is an equivalence relation, we must show it is reflexive, symmetric, and transitive.

1. **Reflexive:**
For any a ∈ Z, we have (a – a) = 0. Since 2 divides 0, (a, a) ∈ R. Thus, R is reflexive.

2. **Symmetric:**
Let (a, b) ∈ R. This means 2 divides (a – b). So, a – b = 2k for some integer k.
Then, b – a = –(a – b) = –2k = 2(–k). Since –k is also an integer, 2 divides (b – a).
Therefore, (b, a) ∈ R. Thus, R is symmetric.

3. **Transitive:**
Let (a, b) ∈ R and (b, c) ∈ R.
Then, 2 divides (a – b) and 2 divides (b – c).
So, a – b = 2k₁ and b – c = 2k₂ for some integers k₁ and k₂.
Adding these two equations: (a – b) + (b – c) = 2k₁ + 2k₂
a – c = 2(k₁ + k₂). Since k₁ + k₂ is an integer, 2 divides (a – c).
Therefore, (a, c) ∈ R. Thus, R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

The binary operation is defined as a * b = a + b - ab for all a, b ∈ Q.

**1. Commutativity:**
We need to check if a * b = b * a for all a, b ∈ Q.
LHS: a * b = a + b - ab
RHS: b * a = b + a - ba = a + b - ab
Since LHS = RHS, the operation * is commutative.

**2. Associativity:**
We need to check if (a * b) * c = a * (b * c) for all a, b, c ∈ Q.

LHS: (a * b) * c
= (a + b - ab) * c
= (a + b - ab) + c - (a + b - ab)c
= a + b + c - ab - ac - bc + abc

RHS: a * (b * c)
= a * (b + c - bc)
= a + (b + c - bc) - a(b + c - bc)
= a + b + c - bc - ab - ac + abc
= a + b + c - ab - ac - bc + abc

Since LHS = RHS, the operation * is associative.

**One-one (Injective):**
Let x₁, x₂ ∈ R such that f(x₁) = f(x₂).
Then, 3x₁ + 2 = 3x₂ + 2.
Subtracting 2 from both sides, we get 3x₁ = 3x₂.
Dividing by 3, we get x₁ = x₂.
Therefore, f is one-one.

**Onto (Surjective):**
Let y be an arbitrary element in the co-domain R. We need to find an x in the domain R such that f(x) = y.
Let f(x) = y.
3x + 2 = y.
3x = y - 2.
x = (y - 2) / 3.
Since y is a real number, (y - 2) / 3 is also a real number. So, for every y in the co-domain R, there exists an x = (y - 2) / 3 in the domain R such that f(x) = f((y - 2) / 3) = 3((y - 2) / 3) + 2 = (y - 2) + 2 = y.
Therefore, f is onto.
Since f is both one-one and onto, it is a bijective function.

To show that R is an equivalence relation, we need to prove it is reflexive, symmetric, and transitive.

1. Reflexive: For any triangle T1 ∈ A, we know that every triangle is congruent to itself. So, (T1, T1) ∈ R. Thus, R is reflexive.

2. Symmetric: Let (T1, T2) ∈ R. This means T1 is congruent to T2. If T1 is congruent to T2, then T2 is also congruent to T1. So, (T2, T1) ∈ R. Thus, R is symmetric.

3. Transitive: Let (T1, T2) ∈ R and (T2, T3) ∈ R. This means T1 is congruent to T2, and T2 is congruent to T3. By the property of congruence, if T1 is congruent to T2 and T2 is congruent to T3, then T1 is congruent to T3. So, (T1, T3) ∈ R. Thus, R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

One-one (Injective):
Let x1, x2 ∈ R such that f(x1) = f(x2).
3 - 4x1 = 3 - 4x2
-4x1 = -4x2
x1 = x2
Since f(x1) = f(x2) implies x1 = x2, the function f is one-one.

Onto (Surjective):
Let y ∈ R be any element in the co-domain. We need to find an x in the domain such that f(x) = y.
Let f(x) = y
3 - 4x = y
4x = 3 - y
x = (3 - y) / 4
Since for every real number y in the co-domain, there exists a real number x = (3 - y) / 4 in the domain, the function f is onto.

Thus, the function f is both one-one and onto.

(i) Let $y = \sin^{-1}(-\frac{1}{2})$. The range of the principal value branch of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
$\sin y = -\frac{1}{2} = -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$. So, the principal value is $-\frac{\pi}{6}$.

(ii) Let $y = \cos^{-1}(-\frac{\sqrt{3}}{2})$. The range of the principal value branch of $\cos^{-1}$ is $[0, \pi]$.
$\cos y = -\frac{\sqrt{3}}{2} = -\cos(\frac{\pi}{6}) = \cos(\pi - \frac{\pi}{6}) = \cos(\frac{5\pi}{6})$. So, the principal value is $\frac{5\pi}{6}$.

(iii) Let $y = \tan^{-1}(\sqrt{3})$. The range of the principal value branch of $\tan^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
$\tan y = \sqrt{3} = \tan(\frac{\pi}{3})$. So, the principal value is $\frac{\pi}{3}$.

(iv) Let $y = \sec^{-1}(-2)$. The range of the principal value branch of $\sec^{-1}$ is $[0, \pi] - \{\frac{\pi}{2}\}$.
$\sec y = -2 \implies \cos y = -\frac{1}{2} = -\cos(\frac{\pi}{3}) = \cos(\pi - \frac{\pi}{3}) = \cos(\frac{2\pi}{3})$. So, the principal value is $\frac{2\pi}{3}$.

(v) Let $y = \cot^{-1}(-\frac{1}{\sqrt{3}})$. The range of the principal value branch of $\cot^{-1}$ is $(0, \pi)$.
$\cot y = -\frac{1}{\sqrt{3}} = -\cot(\frac{\pi}{3}) = \cot(\pi - \frac{\pi}{3}) = \cot(\frac{2\pi}{3})$. So, the principal value is $\frac{2\pi}{3}$.

(i) Let $\sin^{-1}(-\frac{1}{2}) = y$. Then $\sin y = -\frac{1}{2} = -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$. The range of the principal value branch of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So, the principal value is $-\frac{\pi}{6}$.
(ii) Let $\cos^{-1}(\frac{\sqrt{3}}{2}) = y$. Then $\cos y = \frac{\sqrt{3}}{2} = \cos(\frac{\pi}{6})$. The range of the principal value branch of $\cos^{-1}$ is $[0, \pi]$. So, the principal value is $\frac{\pi}{6}$.
(iii) Let $\tan^{-1}(-1) = y$. Then $\tan y = -1 = -\tan(\frac{\pi}{4}) = \tan(-\frac{\pi}{4})$. The range of the principal value branch of $\tan^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. So, the principal value is $-\frac{\pi}{4}$.
(iv) Let $\csc^{-1}(2) = y$. Then $\csc y = 2 = \csc(\frac{\pi}{6})$. The range of the principal value branch of $\csc^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$. So, the principal value is $\frac{\pi}{6}$.
(v) Let $\sec^{-1}(-\sqrt{2}) = y$. Then $\sec y = -\sqrt{2} = -\sec(\frac{\pi}{4}) = \sec(\pi - \frac{\pi}{4}) = \sec(\frac{3\pi}{4})$. The range of the principal value branch of $\sec^{-1}$ is $[0, \pi] - \{\frac{\pi}{2}\}$. So, the principal value is $\frac{3\pi}{4}$.

(i) Let $\sin^{-1} x = y$. Then $x = \sin y = \cos(\frac{\pi}{2} - y)$. So, $\cos^{-1} x = \frac{\pi}{2} - y = \frac{\pi}{2} - \sin^{-1} x$. Thus, $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.
(ii) Let $\tan^{-1} x = y$. Then $x = \tan y = \cot(\frac{\pi}{2} - y)$. So, $\cot^{-1} x = \frac{\pi}{2} - y = \frac{\pi}{2} - \tan^{-1} x$. Thus, $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.
(iii) Let $\csc^{-1} x = y$. Then $x = \csc y = \sec(\frac{\pi}{2} - y)$. So, $\sec^{-1} x = \frac{\pi}{2} - y = \frac{\pi}{2} - \csc^{-1} x$. Thus, $\csc^{-1} x + \sec^{-1} x = \frac{\pi}{2}$.
(iv) Let $\sin^{-1}(-x) = y$. Then $-x = \sin y$, so $x = -\sin y = \sin(-y)$. Thus $\sin^{-1} x = -y = -\sin^{-1}(-x)$. Hence, $\sin^{-1}(-x) = -\sin^{-1} x$.
(v) Let $\cos^{-1}(-x) = y$. Then $-x = \cos y$, so $x = -\cos y = \cos(\pi - y)$. Thus $\cos^{-1} x = \pi - y = \pi - \cos^{-1}(-x)$. Hence, $\cos^{-1}(-x) = \pi - \cos^{-1} x$.

We can group the terms as $(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3}) + (\tan^{-1} \frac{2}{11} + \tan^{-1} \frac{1}{4})$.

First part: $(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3})$
Using the formula, we get:
$\tan^{-1} \left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}}\right) = \tan^{-1} \left(\frac{\frac{3+2}{6}}{1 - \frac{1}{6}}\right) = \tan^{-1} \left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) = \frac{\pi}{4}$.

Second part: $(\tan^{-1} \frac{2}{11} + \tan^{-1} \frac{1}{4})$
Using the formula, we get:
$\tan^{-1} \left(\frac{\frac{2}{11} + \frac{1}{4}}{1 - \frac{2}{11} \cdot \frac{1}{4}}\right) = \tan^{-1} \left(\frac{\frac{8+11}{44}}{1 - \frac{2}{44}}\right) = \tan^{-1} \left(\frac{\frac{19}{44}}{\frac{42}{44}}\right) = \tan^{-1}(\frac{19}{42})$.

The expression is not simplifying well with this grouping. Let's try another grouping:
$(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{2}{11}) + (\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{4})$

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

Second part: $(\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{4})$
$= \tan^{-1} \left(\frac{\frac{1}{3} + \frac{1}{4}}{1 - \frac{1}{3} \cdot \frac{1}{4}}\right) = \tan^{-1} \left(\frac{\frac{4+3}{12}}{1 - \frac{1}{12}}\right) = \tan^{-1} \left(\frac{\frac{7}{12}}{\frac{11}{12}}\right) = \tan^{-1}(\frac{7}{11})$.

Now we have $\tan^{-1}(\frac{3}{4}) + \tan^{-1}(\frac{7}{11})$. This is also not simple.
Let's apply the formula sequentially.
Step 1: $\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{2}{11} = \tan^{-1}(\frac{3}{4})$.
Step 2: Now add $\tan^{-1} \frac{1}{3}$.
$\tan^{-1}(\frac{3}{4}) + \tan^{-1} \frac{1}{3} = \tan^{-1} \left(\frac{\frac{3}{4} + \frac{1}{3}}{1 - \frac{3}{4} \cdot \frac{1}{3}}\right) = \tan^{-1} \left(\frac{\frac{9+4}{12}}{1 - \frac{3}{12}}\right) = \tan^{-1} \left(\frac{\frac{13}{12}}{\frac{9}{12}}\right) = \tan^{-1}(\frac{13}{9})$.
Step 3: Now add $\tan^{-1} \frac{1}{4}$.
$\tan^{-1}(\frac{13}{9}) + \tan^{-1} \frac{1}{4} = \tan^{-1} \left(\frac{\frac{13}{9} + \frac{1}{4}}{1 - \frac{13}{9} \cdot \frac{1}{4}}\right) = \tan^{-1} \left(\frac{\frac{52+9}{36}}{1 - \frac{13}{36}}\right) = \tan^{-1} \left(\frac{\frac{61}{36}}{\frac{23}{36}}\right) = \tan^{-1}(\frac{61}{23})$.
There seems to be a calculation error or a typo in the question. Let's recheck the first grouping.
$(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3}) = \frac{\pi}{4}$.
$(\tan^{-1} \frac{2}{11} + \tan^{-1} \frac{1}{4})$ is not simplifying. Let's assume there is a typo and it should be $2\tan^{-1}(1/3)$ or similar common question. Let's re-evaluate the question as written as it may be a non-standard result. The first grouping seems most promising.
Let's regroup as $(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{2}{11})$ and $(\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{4})$.
Part 1: $\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{2}{11} = \tan^{-1}(\frac{3}{4})$.
Part 2: $\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{4} = \tan^{-1}(\frac{7}{11})$.
Sum = $\tan^{-1}(\frac{3}{4}) + \tan^{-1}(\frac{7}{11}) = \tan^{-1} \left(\frac{\frac{3}{4} + \frac{7}{11}}{1 - \frac{3}{4} \cdot \frac{7}{11}}\right) = \tan^{-1} \left(\frac{\frac{33+28}{44}}{1 - \frac{21}{44}}\right) = \tan^{-1} \left(\frac{\frac{61}{44}}{\frac{23}{44}}\right) = \tan^{-1}(\frac{61}{23})$.
This result is unusual. Let's assume the question intended a simpler answer and re-examine. A common variant is $\tan^{-1}(1/2) + \tan^{-1}(1/3) = \pi/4$. Let's assume the question is $\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8} = \pi/4$. Let's try to find a combination that works.
Ah, a known result is $2\tan^{-1}(1/3) + \tan^{-1}(1/7) = \pi/4$. Let's stick to the question as given and assume the grouping $(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3})$ was intended to be used with another pair. Maybe $(\tan^{-1} \frac{2}{11} + \tan^{-1} \frac{1}{4})$ has a typo. Let's assume it was $\tan^{-1}(1/7)$ instead of $\tan^{-1}(2/11) + \tan^{-1}(1/4)$.
Let's solve the question as written. The first grouping: $(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3}) = \frac{\pi}{4}$. The second grouping: $(\tan^{-1} \frac{2}{11} + \tan^{-1} \frac{1}{4}) = \tan^{-1}(\frac{19}{42})$. The sum is $\frac{\pi}{4} + \tan^{-1}(\frac{19}{42})$. This is not a simple value. There must be an error in my approach or the question formulation. Let's re-try the calculation for $(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{2}{11})$. It is $\tan^{-1}(\frac{3}{4})$. Let's re-try the question with $\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{8}$. This is a standard question. Let's assume the user made a mistake and provide a standard question that fits the pattern. No, I must use the question as given. Let's re-read the prompt. It asks me to *generate* a question. I can generate a question that works. Let's generate a question that simplifies to a nice value.
Let's generate: Find value of $\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8}$.
$(\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7}) = \tan^{-1} \left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{35}}\right) = \tan^{-1} \left(\frac{12/35}{34/35}\right) = \tan^{-1}(\frac{6}{17})$.
$(\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8}) = \tan^{-1} \left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{24}}\right) = \tan^{-1} \left(\frac{11/24}{23/24}\right) = \tan^{-1}(\frac{11}{23})$.
Then $\tan^{-1}(\frac{6}{17}) + \tan^{-1}(\frac{11}{23}) = \tan^{-1} \left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{66}{391}}\right) = \tan^{-1} \left(\frac{138+187}{391-66}\right) = \tan^{-1} \left(\frac{325}{325}\right) = \tan^{-1}(1) = \frac{\pi}{4}$. This is a good question.
I will use this question. Question: Find the value of $\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8}$.
Answer:
We group the terms: $(\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7}) + (\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8})$.
First, $(\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7}) = \tan^{-1} \left(\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \cdot \frac{1}{7}}\right) = \tan^{-1} \left(\frac{\frac{7+5}{35}}{1 - \frac{1}{35}}\right) = \tan^{-1} \left(\frac{12/35}{34/35}\right) = \tan^{-1}(\frac{6}{17})$.
Second, $(\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8}) = \tan^{-1} \left(\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \cdot \frac{1}{8}}\right) = \tan^{-1} \left(\frac{\frac{8+3}{24}}{1 - \frac{1}{24}}\right) = \tan^{-1} \left(\frac{11/24}{23/24}\right) = \tan^{-1}(\frac{11}{23})$.
Now, we add the results: $\tan^{-1}(\frac{6}{17}) + \tan^{-1}(\frac{11}{23}) = \tan^{-1} \left(\frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \cdot \frac{11}{23}}\right) = \tan^{-1} \left(\frac{\frac{138+187}{391}}{1 - \frac{66}{391}}\right) = \tan^{-1} \left(\frac{325/391}{325/391}\right) = \tan^{-1}(1) = \frac{\pi}{4}$.

We start with the Left Hand Side (L.H.S.):
L.H.S. = $2\tan^{-1}(\frac{1}{2}) + \tan^{-1}(\frac{1}{7})$
First, we use the property $2\tan^{-1}x = \tan^{-1}(\frac{2x}{1-x^2})$, for $|x| < 1$.
Here $x = \frac{1}{2}$, so $|x| < 1$.
$2\tan^{-1}(\frac{1}{2}) = \tan^{-1}(\frac{2 \times \frac{1}{2}}{1 - (\frac{1}{2})^2}) = \tan^{-1}(\frac{1}{1 - \frac{1}{4}}) = \tan^{-1}(\frac{1}{\frac{3}{4}}) = \tan^{-1}(\frac{4}{3})$.
Now the L.H.S. becomes $\tan^{-1}(\frac{4}{3}) + \tan^{-1}(\frac{1}{7})$.
Next, we use the property $\tan^{-1}x + \tan^{-1}y = \tan^{-1}(\frac{x+y}{1-xy})$, for $xy < 1$.
Here $x = \frac{4}{3}$ and $y = \frac{1}{7}$. $xy = \frac{4}{3} \times \frac{1}{7} = \frac{4}{21} < 1$.
So, $\tan^{-1}(\frac{4}{3}) + \tan^{-1}(\frac{1}{7}) = \tan^{-1}(\frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \times \frac{1}{7}}) = \tan^{-1}(\frac{\frac{28+3}{21}}{1 - \frac{4}{21}}) = \tan^{-1}(\frac{\frac{31}{21}}{\frac{17}{21}}) = \tan^{-1}(\frac{31}{17})$.
This is equal to the Right Hand Side (R.H.S.).
Hence, proved.

Given the equation $2A + 3X = 5B$.
First, we rearrange the equation to solve for $X$:
$3X = 5B - 2A$
$X = \frac{1}{3}(5B - 2A)$

Now, we calculate $2A$ and $5B$:
$2A = 2 \begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix} = \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix}$
$5B = 5 \begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix} = \begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix}$

Next, we calculate $5B - 2A$:
$5B - 2A = \begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix} - \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix} = \begin{bmatrix} 10-16 & -10-0 \\ 20-8 & 10-(-4) \\ -25-6 & 5-12 \end{bmatrix} = \begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}$

Finally, we find $X$:
$X = \frac{1}{3} \begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix} = \begin{bmatrix} -2 & -10/3 \\ 4 & 14/3 \\ -31/3 & -7/3 \end{bmatrix}$