Bihar Board Class 11 Half-Yearly Mathematics Question Paper 2025-26 (Dec 16) with Answer Key & Solution

Bihar Board Class 11 Half-Yearly Mathematics Exam 2025-26 – Overview

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Bihar Board Class 11 Half-Yearly Mathematics Question Paper 2025-26

Bihar board class 11 half-yearly maths exam was conducted successfully on December 16, 2025. Students can now downloadthe complete Bihar board class 11 half-yearly question paper 2025-26 by clicking on the link given in the table below.

Bihar Board Class 11 Half-Yearly Mathematics Answer Key 2025-26

Bihar board class 11 half-yearly maths answer key with solution is provided on this page of careers360. Students match the correct answer and solution with the given BSEB 11th half-yearly mathematics solution 2025-26.

Q1. Value of $\sin 75^{\circ}$

$
\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}=\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2} \cdot \frac{1}{2}
$
So, $\sin 75^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$.
Q2. Solve: $\sin x=-\frac{\sqrt{3}}{2}$
Reference angle is $60^{\circ}$ (or $\pi / 3$ ).
Sine is negative in III and IV quadrants.
General solution:

$
x=240^{\circ}+360^{\circ} k \text { or } x=300^{\circ}+360^{\circ} k
$

or in radians,

$
x=\frac{4 \pi}{3}+2 \pi k \quad \text { or } \quad x=\frac{5 \pi}{3}+2 \pi k
$

where $k \in \mathbb{Z}$.

Q3. Principal solution of $\tan x=\sqrt{3}$
Reference angle is $60^{\circ}$ (since $\tan 60^{\circ}=\sqrt{3}$ ).
Principal value is taken in $\left(-90^{\circ}, 90^{\circ}\right)$, so

$
x=60^{\circ} \text { or } x=\frac{\pi}{3}
$

Q4. Express $3(7+i 7)+i(7+i 7)$ in $a+i b$ form
First simplify inside brackets:

$
7+i 7=7(1+i)
$

$
\begin{gathered}
3(7+i 7)=3 \cdot 7(1+i)=21+21 i, \\
i(7+i 7)=i \cdot 7(1+i)=7 i+7 i^2=7 i-7 .
\end{gathered}
$

$
(21+21 i)+(-7+7 i)=(21-7)+(21 i+7 i)=14+28 i
$

So, $a+i b=14+28 i$.

Q5. Multiplicative inverse of $\sqrt{5}-3 i$
For $z=a+i b$, inverse is $\frac{1}{z}=\frac{\bar{z}}{|z|^2}$.
Here $z=\sqrt{5}-3 i$, so $\bar{z}=\sqrt{5}+3 i$,

$
|z|^2=(\sqrt{5})^2+(-3)^2=5+9=14
$

Thus,

$
\frac{1}{\sqrt{5}-3 i}=\frac{\sqrt{5}+3 i}{14} .
$
So the multiplicative inverse is $\frac{\sqrt{5}+3 i}{14}$.

Q6. Modulus of $z_1+z_2$
Given $z_1=2-i, z_2=2+i$.
$z_1+z_2=(2-i)+(2+i)=4$.
Modulus $|4|=4$.

Q7. Prove $1+i^2+i^4+i^6=0$
Use powers of $i$ :
$i^2=-1$,
$i^4=\left(i^2\right)^2=(-1)^2=1$,
$i^{-6}=i^4 \cdot i^2=1 \cdot(-1)=-1$.

So,

$
1+i^2+i^4+i^6=1+(-1)+1+(-1)=0
$

Q8. Solve $3 x^2+15=0$

$
3 x^2+15=0 \Rightarrow 3 x^2=-15 \Rightarrow x^2=-5 \Rightarrow x= \pm \sqrt{-5}= \pm i \sqrt{5}
$
Answer: $x=i \sqrt{5}$ or $x=-i \sqrt{5}$.

Q9. Conjugate of $(2-5 i)^2$
First square:

$
(2-5 i)^2=4-20 i+25 i^2=4-20 i-25=-21-20 i
$
Conjugate of $-21-20 i$ is $-21+20 i$.

Q10. Solve $\frac{5-2 x}{3} \leq \frac{x}{6}-5$
Multiply bath sides by 6:

$
6 \cdot \frac{5-2 x}{3} \leq 6\left(\frac{x}{6}-5\right) \Rightarrow 2(5-2 x) \leq x-30 \Rightarrow 10-4 x \leq x-30 .
$

Bring like terms together:

$
10+30 \leq x+4 x \Rightarrow 40 \leq 5 x \Rightarrow x \geq 8 .
$

Answer: $x \geq 8$.
Q11. Solve $5 x-7<3 x+1$ for integer $x$

$
5 x-7<3 x+1 \Rightarrow 5 x-3 x<1+7 \Rightarrow 2 x<8 \Rightarrow x<4 .
$

For integer $x$, all integers $x<4$ satisfy it ( $,-2,-1,0,1,2,3$ ).
Answer: All integers $x$ such that $x<4$.
Q12. If $\frac{{ }^n P_4}{n-1 P_4}=\frac{5}{3}, n>4$, find $n$
Use permutation formula: " $P_4=\frac{n!}{(n-4)!}$.
Then

$
\frac{{ }^n P_4}{{ }^{n-1} P_4}=\frac{\frac{n!}{(n-4)!}}{\frac{(n-1)!}{(n-5)!}}=\frac{n!}{(n-4)!} \cdot \frac{(n-5)!}{(n-1)!}=\frac{n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)(n-4)}=\frac{n}{n-4} .
$

Given $\frac{n}{n-4}=\frac{5}{3}$

$
3 n=5(n-4) \Rightarrow 3 n=5 n-20 \Rightarrow 2 n=20 \Rightarrow n=10 .
$

Since $n>4, n=10$ is valid.
Answer: $n=10$.

Q13. If ${ }^n C_{10}={ }^n C_4$, find $n$
For combinations, ${ }^n C_{\mathrm{r}}={ }^n C_{n-r}$
So 10 and 4 must be complementary: $n-10=4 \Rightarrow n=14$.

$
{ }^{14} C_{10}={ }^{14} C_{4-}
$

Answer: $n=14$ and ${ }^n C_4={ }^{14} C_4$.

Q14. If $\frac{1}{6}+\frac{1}{\binom{7}{x}}=\frac{1}{8}$, find $x$

$
\frac{1}{\binom{7}{x}}=\frac{1}{8}-\frac{1}{6}=\frac{3-4}{24}=-\frac{1}{24} .
$

So $\binom{7}{x}=-24$, but a binomial coefficient is always non-negative.
Hence no real value of $x$ satisfies this equation.
Answer: No solution (no such $x$ ).

Q15. How many chords through 21 points on a circle?
Any chord is determined by choosing 2 distinct points.
Number of chords $=\binom{21}{2}=\frac{21-20}{2}=210$.
Answer: 210 chords.

Q16. Find the value of $\binom{8}{6}+{ }^{10} P_8$

$
\begin{gathered}
\binom{8}{6}=\binom{8}{2}=\frac{8 \cdot 7}{2}=28 . \\
{ }^{20} P_8=\frac{10!}{(10-8)!}=\frac{10!}{2!}=\frac{3628800}{2}=1814400 .
\end{gathered}
$

Sum:

$
28+1814400=1814428
$

Answer: 1814428.

Q17. Number of permutations of ALLAHABAD
Letters: A, L, L, A, H, A, B, A, D.
Total letters $=9$; repeated letters A (4 times), L (2 times).
Number of distinct permutations:

$
\frac{9!}{4!2!}=\frac{362880}{24 \cdot 2}=\frac{362880}{48}=7560 .
$

Answer: 7560 permutations.
Q18. Expansion of $(2 x-3)^5$
Use binomial theoremc $(a+b)^5=\sum_{k-0}^5\binom{5}{k} a^{5-k} b^k$.
with $a=2 x, b=-3$.

$
(2 x-3)^5=32 x^5-240 x^4+720 x^3-1080 x^2+810 x-243 .
$

Answer: $(2 x-3)^5=32 x^5-240 x^4+720 x^3-1080 x^2+810 x-243$.

Q19. 6th term in expansion of $(x-2 y)^{12}$
General term:

$
T_{k+1}=\binom{12}{k} x^{12-k}(-2 y)^k .
$

For 6 th term: $k+1=6 \Rightarrow k=5$.

$
T_6=\binom{12}{5} x^7(-2 y)^5=\binom{12}{5} x^7\left(-32 y^5\right) .
$

$\binom{12}{5}=792$.

$
T_6=792 \cdot(-32) x^7 y^5=-25344 x^7 y^5 .
$

Answer: 6th term is $-25344 x^7 y^5$.

Q20
Expression: $\left(\frac{2}{3} x-\frac{3}{2} y\right)^{20}$.
General term:

$
T_{r+1}=\binom{20}{r}\left(\frac{2}{3} x\right)^{20-r}\left(-\frac{3}{2} y\right)^r
$

For power 20 , the middle term is $\left(\frac{20}{2}+1\right)=11$ th, so put $r=10$ :

$
T_{11}=\binom{20}{10}\left(\frac{2}{3} x\right)^{10}\left(-\frac{3}{2} y\right)^{10}=\binom{20}{10} x^{10} y^{10}
$

Middle term $=\binom{20}{10} x^{10} y^{10}$.

Q21

$
a_n=\frac{n}{n+1}
$

Take $n=1,2,3,4$ :
- $a_1=\frac{1}{2}$
- $a_2=\frac{2}{3}$
- $a_3=\frac{3}{4}$
- $a_4=\frac{4}{5}$

Q22
7, 2, $\mathrm{x}, \mathrm{y}$ are in A.P.
Common difference $d=2-7=-5$.
Then
- $x=2+d=2-5=-3$
- $y=x+d=-3-5=-8$.

So $x=-3, y=-8$.

Q23
A.P: 50, 46, 42, -

First term $a=50$, common difference $d=46-50=-4$, number of terms $n=10$. $n$th term:

$
a_{10}=a+(10-1) d=50+9(-4)=14
$

Sum:

$
S_{10}=\frac{n}{2}\left(a+a_{10}\right)=\frac{10}{2}(50+14)=5 \cdot 64=320
$

Q24
G.P. with first term $a$, ratio $r$.

Second term $a r=24$.
Fifth term $a r^4=81$.
Divide: $r^3=\frac{81}{24}=\frac{27}{8} \Rightarrow r=\frac{3}{2}$.
Then $a=\frac{24}{r}=\frac{24}{3 / 2}=16$.
So first term $=16$.

Q25
G.P.: $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$

First term $a=\sqrt{7}$.
Common ratio $r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$.
Sum of $n$ terms of G.P.:

$
S_n=\frac{a\left(r^n-1\right)}{r-1}=\frac{\sqrt{7}\left((\sqrt{3})^n-1\right)}{\sqrt{3}-1}
$

Q26: Prove the trigonometric identity
Prove that

$
\cos ^2 x+\cos ^2\left(x+\frac{\pi}{3}\right)+\cos ^2\left(x-\frac{\pi}{3}\right)=\frac{3}{2}
$

Use $\cos ^2 \theta=\frac{1+\cos 2 \theta}{2}$.

$
\begin{aligned}
\cos ^2 x & +\cos ^2\left(x+\frac{\pi}{3}\right)+\cos ^2\left(x-\frac{\pi}{3}\right) \\
& =\frac{1+\cos 2 x}{2}+\frac{1+\cos \left(2 x+\frac{2 \pi}{3}\right)}{2}+\frac{1+\cos \left(2 x-\frac{2 \pi}{3}\right)}{2} \\
& =\frac{3}{2}+\frac{1}{2}\left[\cos 2 x+\cos \left(2 x+\frac{2 \pi}{3}\right)+\cos \left(2 x-\frac{2 \pi}{3}\right)\right]
\end{aligned}
$

Now use $\cos (A+B)+\cos (A-B)=2 \cos A \cos B$ :

$
\cos \left(2 x+\frac{2 \pi}{3}\right)+\cos \left(2 x-\frac{2 \pi}{3}\right)=2 \cos 2 x \cos \frac{2 \pi}{3}=2 \cos 2 x\left(-\frac{1}{2}\right)=-\cos 2 x .
$

So the bracket becomes

$
\cos 2 x-\cos 2 x=0
$

Therefore

$
\cos ^2 x+\cos ^2\left(x+\frac{\pi}{3}\right)+\cos ^2\left(x-\frac{\pi}{3}\right)=\frac{3}{2}+\frac{1}{2} \cdot 0=\frac{3}{2}
$

proved.

Q27: Convert $-4+i 4 \sqrt{3}$ to polar form
Given complex number:

$
z=-4+i 4 \sqrt{3}
$

1. Modulus $r$

$
r=|z|=\sqrt{(-4)^2+(4 \sqrt{3})^2}=\sqrt{16+48}=\sqrt{64}=8
$

2. Argument $\theta$

$
\tan \theta=\frac{\mathrm{Im}}{\mathrm{Re}}=\frac{4 \sqrt{3}}{-4}=-\sqrt{3}
$

The point $(-4,4 \sqrt{3})$ lies in quadrant II (negative x , positive y ). In quadrant II, $\tan \theta=-\sqrt{3} \Rightarrow \theta=\frac{2 \pi}{3}$.

So polar (trigonometric) form:

$
z=8\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)
$

(You may also write $8 \mathrm{cis} \frac{2 \pi}{3}$.)

Q28: Solve $x^2+x+\frac{1}{\sqrt{2}}=0$
Quadratic:

$
x^2+x+\frac{1}{\sqrt{2}}=0 .
$

Use quadratic formula $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$ with $a=1, b=1, c=1 / \sqrt{2}$

$
\begin{aligned}
\Delta & =b^2-4 a c=1-4 \cdot 1 \cdot \frac{1}{\sqrt{2}}=1-\frac{4}{\sqrt{2}} \\
& =1-2 \sqrt{2}
\end{aligned}
$

Note that $1-2 \sqrt{2}=(\sqrt{2}-1)^2-4 \sqrt{2}+2$ is negative; more directly.

$
(\sqrt{2})^2=2>1 \Rightarrow 2 \sqrt{2}>1 \Rightarrow 1-2 \sqrt{2}<0
$

So roots are complex.
Write $\Delta=-(2 \sqrt{2}-1)$. Then

$
\sqrt{\Delta}=i \sqrt{2 \sqrt{2}-1}
$

Hence

$
x=\frac{-1 \pm i \sqrt{2 \sqrt{2}-1}}{2} .
$

So the solutions are

$
x_1=\frac{-1+i \sqrt{2 \sqrt{2}-1}}{2}, \quad x_2=\frac{-1-i \sqrt{2 \sqrt{2}-1}}{2} .
$

Q29: Solve system of inequations graphically
System:

$
\left\{\begin{array}{l}
5 x+4 y \leq 40, \\
x \geq 2, \\
y \geq 3 .
\end{array}\right.
$

1. Line $5 x+4 y=40 \mathrm{E}$
- If $x=0 \Rightarrow y=10$.
- If $y=0 \Rightarrow x=8$.

Draw line through $(0,10)$ and $(8,0)$.
Region $5 x+4 y \leq 40$ is below this line.
2. Line $x=2$ : vertical line through $x=2$.

Region $x \geq 2$ is to the right of this line.
3. Line $y=3$ : horizontal line through $y=3$.

Region $y \geq 3$ is above this line.
The solution set is the common shaded region:
- bounded on the left by $x=2$,
- below by line $5 x+4 y=40$,
- below also by that line and above by $y=3$.

Find intersection of $5 x+4 y=40$ with $y=3$ :

$
5 x+4(3)=40 \Rightarrow 5 x+12=40 \Rightarrow 5 x=28 \Rightarrow x=\frac{28}{5}=5.6 .
$

So the feasible region is the quadrilateral with vertices:

$
(2,3),(2,8),(8,3),\left(\frac{28}{5}, 3\right)
$

restricted by $5 x+4 y \leq 40, x \geq 2, y \geq 3$.
Any point in that region (for example $(3,4)$ ) satisfies all three inequalities.

Bihar Board Class 11 Half-Yearly Mathematics Paper Analysis 2025-26

BSEB class 11 half-yearly maths exam is conducted successfully in all the schools of BSEB. The overall paper was easy to moderate. Check the detailed analysis, which is given below:

• Overall the difficulty level was easy to moderate

• MCQ section was the easiest section among all the sections.

• No out-of-syllabus questions were asked in the paper.

• Students who have solved the questions from NCERT are able to solve the complete paper easily.

Preparation Tips for Bihar Board Class 11 Half-Yearly Mathematics

• Go through the National Council for Educational Research and Training(NCERT), Bihar Board textbooks-solved examples and exercises.

• Precise focus on high-weightage chapters/topics.

• Solved previous year and sample papers, for understanding weightage & pattern.

• Make a formula sheet or short notes and revise every day for faster recall.

• Set a timer to solve question papers-this will improve overall writing speed and temperament in the exam.

• Avoid silly mistakes by double-checking: algebraic signs, steps, and final answers.

• Daily revision: 30-45 minutes on complicated topics, along with 10-15 problems per day.

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