Important Questions For Class 12 Maths - 2025-26

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=$\frac{44}{91}$

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