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CBSE Important Questions for Class 12 Maths - 2025-26

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Class 12 Maths important questions with answers PDF download

CBSE Class 12 Maths Important Questions are designed to help students focus on key topics needed for their exams according to the CBSE Class 12 Maths Syllabus. These questions cover essential concepts from various chapters, making it easier to revise and practice. By working through these important questions, students can build their confidence and improve their problem-solving skills. This resource is valuable for both quick revisions and in-depth study, ensuring that students are well-prepared to tackle their exams successfully.

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CBSE Class 12 Maths Chapter-wise Important Questions

CBSE Class 12 Maths Chapter-wise Important Questions and Answers cover topics from all 13 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.


Competitive Exams after 12th Science
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10 Most Important Questions From Class 12 Maths Chapters

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

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

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

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-}\left( \frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{2}}{\text{3}} \right)} \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 \[A=\left[ \begin{matrix}  1 & 1 & 1  \\   1 & 1 & 1  \\   1 & 1 & 1  \\ \end{matrix} \right]\] , prove that \[{{A}^{n}}=\left[ \begin{matrix}   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\ \end{matrix} \right]\] , \[n\in N\].

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

By using the principles of mathematical induction.

For \[n=1\] , we have

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

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

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

\[\Rightarrow P\left( 1 \right)=A\]

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

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

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

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

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

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

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

Thus, the result is true for \[n=k+1\]

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

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


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

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

By using the principle of mathematical induction.

For \[n=1\] ,

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

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

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

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

\[{{A}^{k+1}}=A\cdot {{A}^{k}}\]

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

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

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

Thus, the result is true for \[n=k+1\]

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

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


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 \[\text{A=}\left[ \begin{matrix} \text{5} & \text{-1} & \text{4}  \\ \text{2} & \text{3} & \text{5}  \\ \text{3} & \text{-2} & \text{6}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{5}  \\ \text{2}  \\ \text{-1}  \\ \end{matrix} \right]\] 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}}} $ .

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

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

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

Therefore,  $ \underset{\text{x}\to \text{k}}{\mathop{\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 $x = \dfrac{2}{7}$ or $x = 2$

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

for $x$ close to $\dfrac{2}{7}$ and to right of $\dfrac{2}{7},{f^\prime }(x) > 0$. $x = \dfrac{2}{7}$ 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: $\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}$.

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

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

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

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

Integration of given expression is

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

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

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

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

$\therefore\int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\log \left| {{e}^{2x}}+{{e}^{-2x}} \right|+C$


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 $\text{ }\!\!\alpha\!\!\text{ }$ with coordinate axes

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

Now, we know that 

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

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

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

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

 

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!


Additional Study Materials for Class 12 Maths 

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FAQs on CBSE Important Questions for Class 12 Maths - 2025-26

1. What makes these questions "important" for the CBSE Class 12 Maths Board Exam 2025-26?

These questions are considered important because they are curated based on several factors: frequent appearance in past board papers, alignment with the high-weightage topics in the official CBSE syllabus for 2025-26, and their potential to test core conceptual understanding, including Higher-Order Thinking Skills (HOTS).

2. Which topics hold the most weightage in the Class 12 Maths board exam for 2025-26?

For the CBSE 2025-26 exam, the unit with the highest weightage is Calculus, covering topics like continuity, differentiability, applications of derivatives, integrals, and differential equations. Other crucial units include Vectors and 3D Geometry, and Probability. Focusing on important questions from these units is key to scoring well.

3. How should I approach solving Higher-Order Thinking Skills (HOTS) questions from this list?

When you encounter a HOTS question, first, identify the core concept being tested. Do not rush to a solution. Instead, break the problem down into smaller, manageable steps. Try to connect concepts from different chapters, as HOTS problems often require interdisciplinary application (e.g., using calculus in a geometry problem). Practising these specific questions builds the analytical skills needed for the exam.

4. What is the most effective strategy to score above 90% in Class 12 Maths using these important questions?

To score above 90%, use these important questions strategically. Follow this plan:

  • First, ensure your fundamental concepts from the NCERT textbook are clear.
  • Next, solve each important question chapter-wise without looking at the solution to test your knowledge.
  • Analyse your mistakes to identify weak areas.
  • Finally, practise these questions under timed conditions to improve your speed and accuracy for the board exam.

5. Should I focus only on these important questions or also include NCERT textbook exercises?

You should use both. The NCERT textbook builds your foundational understanding of every topic and formula. The important questions help you apply that knowledge in an exam-oriented way, focusing on question types and topics that are frequently tested by CBSE. A balanced approach is the best for a comprehensive preparation.

6. How do the important questions reflect the latest CBSE paper pattern and typology for 2025-26?

These questions are designed to align with the latest CBSE pattern, which includes a mix of question types:

  • Multiple Choice Questions (MCQs) to test quick recall.
  • Very Short Answer (VSA) questions of 2 marks.
  • Short Answer (SA) questions of 3 marks.
  • Long Answer (LA) and Case-Study based questions of 5 marks.
By practising them, you get accustomed to the format and marks distribution of the actual board paper.

7. Beyond just solving them, how can I use important questions to improve my time management?

To improve time management, set a timer while solving a set of important questions. For a 3-mark question, aim to finish it in 5-6 minutes, and for a 5-mark question, target 8-10 minutes. This practice helps you identify which types of problems slow you down, allowing you to work on your speed and develop a strategy for tackling the paper efficiently during the exam.

8. Is it better to memorise the solutions to important questions or to understand the underlying concepts?

It is always better to understand the underlying concepts. Memorising solutions is a flawed strategy because CBSE can, and often does, change the numerical values or tweak the language of the questions in the exam. A strong conceptual understanding allows you to solve any variation of a problem, not just the one you memorised.

9. How do the types of important questions for Calculus differ from those for Probability?

The question types differ significantly based on the skills being tested. Important questions in Calculus often involve lengthy, step-by-step procedures, proofs, and graphical interpretations (e.g., finding maxima/minima). In contrast, important questions in Probability are more analytical and logic-based, often requiring the application of specific formulas like Bayes' theorem or binomial distribution in word problems.

10. What is a common mistake students make when relying solely on important questions, and how can I avoid it?

The most common mistake is skipping the NCERT textbook entirely and practising only the important questions. This can leave gaps in your fundamental understanding, making it difficult to tackle unfamiliar or twisted questions in the exam. To avoid this, always complete the relevant NCERT chapter and its examples first before moving on to the important questions for that topic.