CBSE Class 12 Maths Important Questions

1. Show that each of the relation $\text{R}$ in the set $$\text{A = }\{\text{ x}\in \text{z: 0}\le \text{x}\le \text{12 }\}$$ , is an equivalence relation. Find the set of all elements related to 1 in each case.

1. $$\text{R = }\{\left(\text{a, b}\right)\text{ : }\left|\text{a -- b}\right|$$ is a multiple of $$\text{4 }\}$$

Ans: The given set $$\text{A = }\{\text{ x}\in \text{Z: 0}\le \text{x}\le \text{12 }\}\text{ = }\left\{\text{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}\right\}$$ 

The given relation is: $$\text{R = }\{\left(\text{a, b}\right)\text{ : }\left|\text{a -- b}\right|$$ is a multiple of $$\text{4 }\}$$.

Let $$a\in A$$,

$$\left(\text{a, a}\right)\in R$$ as $$\left|\text{a-a}\right|\text{=0}$$ is a multiple of $$\text{4}$$.

Therefore, $$\text{R}$$ is reflexive.

Let, $$\left(\text{a, b}\right)\in \text{R}\Rightarrow \left|\text{a-b}\right|$$ is a multiple of $$\text{4}$$.

$$\Rightarrow \left|\text{-}\left(\text{a-b}\right)\right|\text{=}\left|\text{b-a}\right|$$ is a multiple of $$\text{4}$$.

$$\Rightarrow \left(\text{b, a}\right)\in \text{R}$$

Therefore $$\text{R}$$ is symmetric.

$$\left(\text{a, b}\right)\text{, }\left(\text{b, c}\right)\in R$$.

$$\Rightarrow \left|\text{a-b}\right|$$ is a multiple of $$\text{4}$$ and $$\left|\text{b-c}\right|$$ is a multiple of $$\text{4}$$.

$$\Rightarrow \left(\text{a-b}\right)$$ is a multiple of $$\text{4}$$ and $$\left(\text{b-c}\right)$$ is a multiple of $$\text{4}$$.

$$\Rightarrow \left(\text{a-c}\right)\text{=}\left(\text{a-b}\right)\text{+}\left(\text{b-c}\right)$$ is a multiple of $$\text{4}$$.

$$\Rightarrow \left|\text{a-c}\right|$$ is a multiple of $$\text{4}$$.

$$\Rightarrow \left(\text{a, c}\right)\in \text{R}$$

Therefore, $$\text{R}$$ is transitive.

Therefore, the given relation $$\text{R}$$ is an equivalence relation.

The set of elements related to 1 is $$\left\{\text{1, }\text{ 5, 9}\right\}$$ 

2. Evaluate $$\text{tan}\left(\text{si}{\text{n}}^{\text{-1}}\frac{\text{3}}{\text{5}}\text{+co}{\text{t}}^{\text{-1}}\frac{\text{3}}{\text{2}}\right)$$

Ans: Assuming $\text{si}{\text{n}}^{\text{-1}}\frac{\text{3}}{\text{5}}\text{=a}$

$\text{si}{\text{n}}^{\text{-1}}\frac{\text{3}}{\text{5}}\text{=a}$ 

$\text{sin a=}\frac{\text{3}}{\text{5}}$ 

$\text{cos a=}\sqrt{\text{1-}{\left(\frac{\text{3}}{\text{5}}\right)}^{\text{2}}}$ 

$\text{cos a=}\frac{\text{4}}{\text{5}}$ 

$\text{tan a=}\frac{\text{3}}{\text{4}}$ 

$\text{a=ta}{\text{n}}^{\text{-1}}\left(\frac{\text{3}}{\text{4}}\right)$ 

$\text{si}{\text{n}}^{\text{-1}}\left(\frac{\text{3}}{\text{5}}\right)\text{=ta}{\text{n}}^{\text{-1}}\left(\frac{\text{3}}{\text{4}}\right)$ ----(1)

$\text{co}{\text{t}}^{\text{-1}}\frac{\text{3}}{\text{2}}\text{=ta}{\text{n}}^{\text{-1}}\frac{\text{2}}{\text{3}}$ -----(2)

Further solving, 

$\text{tan}\left(\text{si}{\text{n}}^{\text{-1}}\frac{\text{3}}{\text{5}}\text{+co}{\text{t}}^{\text{-1}}\frac{\text{3}}{\text{2}}\right)$ 

 $\text{=tan}\left(\text{ta}{\text{n}}^{\text{-1}}\frac{\text{3}}{\text{4}}\text{+ta}{\text{n}}^{\text{-1}}\frac{\text{2}}{\text{3}}\right)$ 

 $\text{=tan}\left(\text{ta}{\text{n}}^{\text{-1}}\frac{\frac{\text{3}}{\text{4}}\text{+}\frac{\text{2}}{\text{3}}}{\text{1-}(\frac{\text{3}}{\text{4}}\times \frac{\text{2}}{\text{3}})}\right)$ 

$\text{=tan}\left(\text{ta}{\text{n}}^{\text{-1}}\frac{\text{17}}{\text{6}}\right)$ 

$\text{=}\frac{\text{17}}{\text{6}}$ 

Therefore, $\text{tan}\left(\text{si}{\text{n}}^{\text{-1}}\frac{\text{3}}{\text{5}}\text{+co}{\text{t}}^{\text{-1}}\frac{\text{3}}{\text{2}}\right)\text{=}\frac{\text{17}}{\text{6}}$

If $$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$$ , prove that $$A^n=\left[\begin{array}{ccc}{3}^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{array}\right]$$ , $$n\in N$$.

Ans: Given $$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$$

By using the principles of mathematical induction.

For $$n=1$$ , we have

$$P\left(1\right)=\left[\begin{array}{ccc}{3}^{1-1}& 3^{1-1}& 3^{1-1}\\ 3^{1-1}& 3^{1-1}& 3^{1-1}\\ 3^{1-1}& 3^{1-1}& 3^{1-1}\end{array}\right]$$

$$\Rightarrow P\left(1\right)=\left[\begin{array}{ccc}{3}^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\end{array}\right]$$

$$\Rightarrow P\left(1\right)=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$$

$$\Rightarrow P\left(1\right)=A$$

Therefore, the result is true for $$n=1$$ .

Let the result be true for $$n=k$$ .

$$P\left(k\right):A^k=\left[\begin{array}{ccc}{3}^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\end{array}\right]$$

Now, we have to prove that the result is true for $$n=k+1$$ .

Now, $$A^{k+1}=A\cdot A^k$$

$$\Rightarrow A^{k+1}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]\left[\begin{array}{ccc}{3}^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\end{array}\right]$$

$$\Rightarrow A^{k+1}=\left[\begin{array}{ccc}3\cdot 3^{k-1}& 3\cdot 3^{k-1}& 3\cdot 3^{k-1}\\ 3\cdot 3^{k-1}& 3\cdot 3^{k-1}& 3\cdot 3^{k-1}\\ 3\cdot 3^{k-1}& 3\cdot 3^{k-1}& 3\cdot 3^{k-1}\end{array}\right]$$

$$\Rightarrow A^{k+1}=\left[\begin{array}{ccc}{3}^{(k+1)-1}& 3^{(k+1)-1}& 3^{(k+1)-1}\\ 3^{(k+1)-1}& 3^{(k+1)-1}& 3^{(k+1)-1}\\ 3^{(k+1)-1}& 3^{(k+1)-1}& 3^{(k+1)-1}\end{array}\right]$$

Thus, the result is true for $$n=k+1$$

Therefore, by the principal of mathematical induction, we have:

 $$A^n=\left[\begin{array}{ccc}{3}^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{array}\right]$$ where $$A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$$ , $$n\in N$$.

3. If $$A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$$ , then prove $$A^n=\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right]$$ where $$n$$ is any positive integer.

Ans: Given $$A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$$

By using the principle of mathematical induction.

For $$n=1$$ ,

$$P\left(1\right):A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$$

Therefore, the result is true for $$n=1$$ .

Let the result be true for $$n=k$$ .

That is, $$P\left(k\right):A^k=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]$$ , $$n\in N$$

Now, we have to prove that the result is true for $$n=k+1$$ .

$$A^{k+1}=A\cdot A^k$$

$$\Rightarrow A^{k+1}=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$$

$$\Rightarrow A^{k+1}=\left[\begin{array}{cc}3+2k& -4-4k\\ 1+k& -1-2k\end{array}\right]$$

$$\Rightarrow A^{k+1}=\left[\begin{array}{cc}1+2(k+1)& -4(k+1)\\ 1+k& 1-2(k+1)\end{array}\right]$$

Thus, the result is true for $$n=k+1$$

Therefore, by the principal of mathematical induction, we have:

$$A^n=\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right]$$ , $$n\in N$$.

4. Examine the consistency of the system of equations.

$$\mathbf{\text{5x-y+4z=5}}$$

$$\mathbf{\text{2x+3y+5z=2}}$$

$$\mathbf{\text{5x-2y+6z=-1}}$$

Ans: Given equations,

$$\text{5x-y+4z=5}$$

$$\text{2x+3y+5z=2}$$

$$\text{5x-2y+6z=-1}$$

Let $$\text{A=}\left[\begin{array}{ccc}\text{5}& \text{-1}& \text{4}\\ \text{2}& \text{3}& \text{5}\\ \text{3}& \text{-2}& \text{6}\end{array}\right]$$ , $$\text{X=}\left[\begin{array}{c}\text{x}\\ \text{y}\\ \text{z}\end{array}\right]$$ and $$\text{B=}\left[\begin{array}{c}\text{5}\\ \text{2}\\ \text{-1}\end{array}\right]$$ such that, the system of equations can be written in the form of $$\text{AX=B}$$.

Determining the value of $A$, we have:

$$\Rightarrow \left|\text{A}\right|\text{=5}\left(\text{18+10}\right)\text{+1}\left(\text{12-25}\right)\text{+4}\left(\text{-4-15}\right)$$

$$\Rightarrow \left|\text{A}\right|\text{=5}\left(\text{28}\right)\text{+1}\left(\text{-13}\right)\text{+4}\left(\text{-19}\right)$$

$$\Rightarrow \left|\text{A}\right|\text{=140-13-76}$$

$$\therefore \left|\text{A}\right|\text{=51}\ne \text{0}$$

Hence, $$\text{A}$$ is a non-singular matrix.

Thus, $${\text{A}}^{\text{-1}}$$ exists.

$\therefore$ The given system of equations is consistent.

5. Show that the function defined by  $\text{f(x)=cos(}{\text{x}}^{\text{2}}\text{)}$  is a continuous function.

Ans: The given function is  $\text{f(x)=cos(}{\text{x}}^{\text{2}}\text{)}$ .

Note that,  $\text{f}$  is defined for all real numbers and so  $\text{f}$  can be expressed as the composition of two functions as,  $\text{f=g}\circ \text{h}$ , where  $\text{g(x)=cosx}$  and  $\text{h(x)=}{\text{x}}^{\text{2}}$ .

$[\therefore \text{(goh)(x)=g(h(x))=g(}{\text{x}}^{\text{2}}\text{)=cos(}{\text{x}}^{\text{2}}\text{)=f(x) }]$ 

Now, it is to be Proven that, the functions  $\text{g(x)=cosx}$  and  $\text{h(x)=}{\text{x}}^{\text{2}}$  are continuous.

Since the function  $\text{g}$  is defined for all the real numbers, let 's consider  $\text{c}$  be a real number.

Then,  $\text{g(c)=cosc}$ .

Substitute  $\text{x=c+h}$  into the function  $\text{g}$ .

When,  $\text{x}\to \text{c}$ , then  $\text{h}\to 0$ .

Then we have,

$\begin{array}{rl}& \underset{\text{x}\to \text{c}}{lim}\text{g(x)=}\underset{\text{x}\to \text{c}}{lim}\text{cosx}\\ & =\underset{\text{h}\to 0}{lim}\text{cos}\left(\text{c+h}\right)\\ & =\underset{\text{h}\to 0}{lim}[\text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh}]\\ & =\underset{\text{h}\to 0}{lim}(\text{cosc}\cdot \text{cosh})-\underset{\text{h}\to 0}{lim}(\text{sinc}\cdot \text{sinh})\\ & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0}\\ & \text{=cosc }\times \text{ 1}-\text{sinc }\times \text{ 0}\\ & \text{=cosc}\end{array}$

Therefore,  $\underset{\text{x}\to \text{c}}{lim}\text{g(x)=g(c)}$ .

Hence, the function  $\text{g(x)=cosx}$  is continuous.

Again,  $\text{h(x)=}{\text{x}}^{\text{2}}$  is defined for every real point.

So, let consider  $\text{k}$  be a real number, then  $\text{h(k)=}{\text{k}}^{\text{2}}$  and

 $\underset{\text{x}\to \text{k}}{lim}\text{h(x)=}\underset{\text{x}\to \text{k}}{lim}{\text{x}}^{\text{2}}\text{=}{\text{k}}^{\text{2}}$ .

Therefore,  $\underset{\text{x}\to \text{k}}{lim}\text{h(x)=h(k)}$ .

Hence, the function  $\text{h}$  is continuous.

Now, remember that for real valued functions  $\text{g}$  and  $\text{h}$  , such that  $\text{(g }\circ \text{ h)}$  is defined at  $\text{c}$  , if  $\text{g}$  is continuous at  $\text{c}$  and  $\text{f}$  is continuous at  $\text{g(c)}$ , then  $\text{(f }\circ \text{ h)}$ is continuous at  $\text{c}$ .

Hence, the function  $\text{f(x)=(g }\circ \text{ h)(x)=cos(}{\text{x}}^2\text{)}$  is continuous.

6. Find the points at which the function $f$ given by $f(x)=(x-2{)}^4(x+1{)}^3$ has

(i) local maxima

(ii) local minima

(iii) point of inflexion

Ans: $f(x)=(x-2{)}^4(x+1{)}^3$

$\therefore f^{\prime }(x)=4(x-2{)}^3(x+1{)}^3+3(x+1{)}^2(x-2{)}^4$

$=(x-2{)}^3(x+1{)}^2[4(x+1)+3(x-2)]$

$=(x-2{)}^3(x+1{)}^2(7x-2)$

$f^{\prime }(x)=0\Rightarrow x=-1$ and $x={\displaystyle \frac{2}{7}}$ or $x=2$

for $x$ close to ${\displaystyle \frac{2}{7}}$ and to left of ${\displaystyle \frac{2}{7}},f^{\prime }(x)>0$.

for $x$ close to ${\displaystyle \frac{2}{7}}$ and to right of ${\displaystyle \frac{2}{7}},f^{\prime }(x)>0$. $x={\displaystyle \frac{2}{7}}$ is point of local minima.

as the value of $x$ varies $f^{\prime }(x)$ does not changes its sign.

$x=-1$ is point of inflexion.

7. Solve the following: ${\displaystyle \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}}$.

Ans: Given expression ${\displaystyle \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}}$.

Let us substitute $e^{2x}+e^{-2x}=t$, we get

$(2e^{2x}+2e^{-2x})dx=dt$

$\Rightarrow 2(e^{2x}-e^{-2x})dx=dt$

Integration of given expression is

$\Rightarrow \int {\displaystyle \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}}=\int {\displaystyle \frac{dt}{2t}}$

$\Rightarrow \int {\displaystyle \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}}={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{1}{t}}dt$

$\Rightarrow \int {\displaystyle \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}}={\displaystyle \frac{1}{2}}\log \left|t\right|+C$

Again substitute $t=e^{2x}+e^{-2x}$, we get

$\therefore \int {\displaystyle \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}}={\displaystyle \frac{1}{2}}\log |e^{2x}+e^{-2x}|+C$

8. Determine order and degree (if defined) of differential equation ${\left(\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\right)}^{\text{2}}\text{+cos}\left(\frac{\text{dy}}{\text{dx}}\right)\text{=0}$.

Ans: The given differential equation is ${\left(\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\right)}^{\text{2}}\text{+cos}\left(\frac{\text{dy}}{\text{dx}}\right)\text{=0}$.

The highest order term is $\frac{{\text{d}}^{\text{2}}y}{\text{d}{\text{x}}^{\text{2}}}$, thus the order is two.

The differential equation contains a trigonometric derivative term and is not completely polynomial in its derivative, thus degree is not defined.

9. Find the direction cosines of the line which makes equal angles with the coordinate axes.

Ans: Let us consider that the line makes an angle $\alpha$ with coordinate axes

Which means $\text{l=cos }\alpha \text{ ,m=cos }\alpha \text{ ,n=cos }\alpha$

Now, we know that 

${\text{l}}^{\text{2}}\text{+}{\text{m}}^{\text{2}}\text{+}{\text{n}}^{\text{2}}\Rightarrow \text{co}{\text{s}}^{\text{2}}\alpha \text{ +co}{\text{s}}^{\text{2}}\alpha \text{ +co}{\text{s}}^{\text{2}}\alpha$

$\Rightarrow \text{3co}{\text{s}}^{\text{2}}\alpha \text{ =1}$

$\Rightarrow \text{co}{\text{s}}^{\text{2}}\alpha \text{ =}\frac{\text{1}}{\text{3}}\Rightarrow \text{cos }\alpha \text{ = }\pm \frac{\text{1}}{\sqrt{\text{3}}}$

Therefore, the direction cosines of the line are $\pm \frac{\text{1}}{\sqrt{\text{3}}}\text{, }\pm \frac{\text{1}}{\sqrt{\text{3}}}\text{, }\pm \frac{\text{1}}{\sqrt{\text{3}}}$.

10. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing $15$ oranges out of which $12$ are good and $3$ are bad ones will be approved for sale.

Ans: Let A,B and C be the events and their probabilities defined as 

A: the first orange is good

$\therefore P\left(A\right)=\frac{12}{15}$

B: the second orange is good

$\therefore P\left(B\right)=\frac{1}{15}$

C: the third orange is good

$\therefore P\left(C\right)=\frac{10}{15}$

It is given that the box is approved for sale only when all the oranges are good.

Probability that the box is approved for sale=probability of all oranges to be good

Therefore probability of all oranges to be good=$\frac{12}{15}\times \frac{11}{15}\times \frac{10}{15}$

Thus probability that the box is approved for sale=$\frac{44}{91}$

Here are 10 important questions from all chapters. For more detailed, chapter-wise questions, refer to the table above and download the PDF. This will help you gain a better understanding of each chapter and enhance your exam preparation.

How do Maths Important Questions Class 12 Help You with Exams?

• Maths important questions help students focus on key topics that are likely to appear in exams. This targeted approach makes studying more effective.

• Practising these questions reinforces understanding of essential concepts. It allows students to see how different topics connect and apply to real problems.

• Regular practice with important questions boosts confidence. Students become more familiar with the exam format and types of questions they may encounter.

• These questions help identify areas that need improvement. By recognising weak points, students can dedicate more time to those topics.

• Working through important questions improves problem-solving skills. This practice helps students develop strategies for tackling various types of maths problems.

• Using these questions for revision allows for quick and efficient study sessions. Students can review key concepts without getting overwhelmed by too much information.

• Important questions prepare students for different question formats, including short answer and long answer types. This variety ensures students are ready for any exam situation.

Class 12 Maths Important Questions are a valuable resource for students preparing for their exams. They cover key concepts and topics, making it easier to study and understand the material. By practising these important questions, students can build their confidence, improve their problem-solving skills, and get familiar with the exam format. Using Vedantu's resources will help you stay organised and focused, ultimately leading to better results in your exams. Make sure to incorporate these questions into your study routine for effective preparation!

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