cbse 10th maths questions

 cbse 10th maths questions

Linear equations in two unknowns 

 The sum of numerator and denominator of a fraction is 4 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃. Find the fraction.

solution

Quadratic equations

Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.

solution

Coordinate Geometry

Find a relation between x and y such that the point P(x, y) is equidistant from the

 points A(5, 3) and B(1, 7)

solution

Trigonometry

 If cos A + sin A = (√2) cos A, prove that cos A − sin A = (√2) sin A

solution

While playing badminton Ravi has set the barrier chain hung between two posts at the edge of the walkway of a street. It is hung in the shape of a parabola. Based on the above information answer the following questions : (a) Which type of the polynomial (linear, quadratic, cubic etc.) is graphically represented by a parabola ? (b) If the polynomial represented by a parabola, intersects the x-axis at −2 and 3 and y-axis at −3, then write the zeroes of the parabola. (c) Find the expression for the above polynomial. OR (c) If the zeroes of the polynomial are −5 and 3, find its expression.

 While playing badminton Ravi has set the barrier chain hung between two posts at the edge of the walkway of a street. It is hung in the shape of a parabola.

Based on the above information answer the following questions :

(a) Which type of the polynomial (linear, quadratic, cubic etc.) is graphically represented by a parabola ?

(b) If the polynomial represented by a parabola, intersects the x-axis at −2 and 3 and y-axis at −3, then write the zeroes of the parabola.

(c) Find the expression for the above polynomial.

(d) If the zeroes of the polynomial are −5 and 3, find its expression.

[a] A parabola is the graph of a quadratic polynomial

Given

parabola, intersects the x-axis at −2 and 3 

therefore zeroes are [-2] and 3

Polynomial with roots α and β is given by

f(x) = k(x − α)(x − β)

or

f(x) = k[x² − (α + β)x + αβ]

where k ≠ 0 is any real number.

given that zeroes are [-2] and 3

Take them as

α = [-2]

β = 3

 f(x) = k(x + 2)(x − 3)

use distribution law

f(x) =k(x² − x − 6)

to find k use the condition that parabola cuts y-axis at −3

which means it passes through (0,-3)

so Use f(0) = −3 

k[0-0-6] = [-3]

we get k =(1/2)

and  f(x) = (1/2)(x² − x − 6)

in the last part the zeroes are given to be  −5 and 3

so g(x) = k(x + 5)(x − 3) =kx² + 2x − 15

choose k=1

g(x)  = x² + 2x − 15

see this video for more details 

cbse 10th maths old board exam question papers 2025 2026 to find the equation of a polynomial whose zeroes are given

If (sin x)ʸ = yᶜᵒˢˣ, then find dy/dx

 If (sin x)ʸ = yᶜᵒˢˣ, then find dy/dx.

differential calculus topic of logarthimic differentiation

given

(sin x)ʸ = yᶜᵒˢˣ  [power contains variables]

Take ln both sides

ln[(sin x)ʸ] = ln[yᶜᵒˢˣ]  [use properties of logarithms]

 y. ln(sin x) = cos x . ln y

 Differentiate w.r.t. x

[use product rule and chain rule]

  [dy/dx] ln(sin x) + y cot x  =  −sin x · ln y + cos x · (1/y) ·[ dy/dx]

[dy/dx ] ln(sin x) + y cot x = −sin x ln y + (cos x / y) [dy/dx]

 Collect dy/dx terms

 [dy/dx] ln(sin x) − (cos x / y)[ dy/dx ]= −sin x ln y − y cot x

[ dy/dx] [ln(sin x) − cos x / y] = −[sin x ln y + y cot x]

dy/dx = −[sin x ln y + y cot x] / [ln(sin x) − cos x / y]

  Multiply numerator and denominator by y:

dy/dx = −y[sin x ln y + y cot x] / [y ln(sin x) − cos x]

see this video for more explanation 

cbse 12th maths old board exam question paper

2025 2026 differentiation

logarthmic differentiation

 

Find a particular solution of the differential equation (x + 1) dy/dx = 2 e⁻ʸ − 1, given that y = 0 when x = 0

 Find a particular solution of the differential equation

(x + 1) dy/dx = 2 e⁻ʸ − 1,

given that y = 0 when x = 0

variable separable differential equation with initial condition

given

(x + 1) dy/dx = 2 e⁻ʸ − 1,

separating

 dy / (2 e⁻ʸ − 1) = dx / (x + 1)

multiply numerator and denominator of LHS with  eʸ 

eʸ dy / (2 − eʸ) = dx / (x + 1)

integrating

∫ eʸ / (2 − eʸ) dy = ∫ 1 / (x + 1) dx

for the LHS use the substitution

u = 2 − eʸ 

⇒ du = −eʸ dy

eʸ dy =-du

we get

 ∫ −du / u = ∫ 1 / (x + 1)  dx

 −ln|u| = ln|x + 1| + C

re substitute the value of u = 2 − eʸ 

 −ln|2 − eʸ| = ln|x + 1| + C

 ln|2 − eʸ| +ln|x + 1| = -C

using properties of logarithms,

ln|(2 − eʸ)(x + 1)| = -C

using definition of logarithm

 (x + 1)(2 − eʸ) = A-----(1)

, where A is another arbitrary constant

use the initial condition   y = 0 when x = 0 to solve for A

we get A =1

(1)⇒ (x + 1)(2 − eʸ) =1

or

2 − eʸ = 1 / (x + 1)

eʸ = 2 − [ 1 / (x + 1)]

eʸ =  (2x + 1) / (x + 1)

using definition of log

y = ln[(2x + 1) / (x + 1)]

for more details watch this video 

cbse 12th past year question papers 2025 2026 differential equations

variable separable differential equation dy/dx = (2 − y) / (x + 1)

 Solve the following variable separable differential equation

dy/dx = (2 − y) / (x + 1)

dy/dx = (2 − y) / (x + 1)

clearly it is variable separable type

dy / (2 − y) = dx / (x + 1)

integrating both sides

∫ dy / (2 − y) = ∫ dx / (x + 1)

− ln|2 − y| = ln|x + 1| + lnC [ note the ( -1), coefficient of y when integrating]

0 = ln|2 − y| + ln|x + 1| + lnC

use the property of logarithms

0=ln [ |2 − y| |x + 1| C ]

or

[ |2 − y| |x + 1| C ] = 1

for more details refer to this video

cbse  12th applied mathematics 2025 2026 old board exam question papers applied mathematics

variable sepaprable differential equation

Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.

 Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.

Let the three consecutive positive integers be n, n+1, n+2

Given that

that sum of square of the first and the product of the other two is 67

n² + (n+1)(n+2) = 67

n² + (n² + 2n + n + 2) = 67  

n² + n² + 3n + 2 = 67  

2n² + 3n + 2 = 67  

2n² + 3n + 2 - 67  = 0

2n² + 3n − 65 = 0

Solve by quadratic formula or factorisation

Factorisation method

Find two numbers that gives a product of 2 × (−65) = −130 and add to 3.

by trial and error the numbers13 and −10 satisfy the condition

use this to split the middle term of

2n² + 3n − 65 = 0

.2n² + 13n − 10n − 65 = 0

n(2n + 13) − 5(2n + 13) = 0

(n − 5)(2n + 13) = 0

solving

n − 5 = 0 ⇒ n = 5

2n + 13 = 0 ⇒ n = −13/2 (rejecct)

required positive integers are

n=5

n+1 =6

n+2=7

for a video of the solution

cbse 10th maths pervious years old question papers 2025 2026 standard mathematics

   quadratic equation word problem  formation and solution by splitting the middle term

∫ cos x / [(2 + sin x)(4 + sin x)] dx

 find ∫ cos x / [(2 + sin x)(4 + sin x)] dx 

integration by substitution and partial fractions

Substitution

Let t = sin x

Then dt = cos x dx

 ∫ cos x / [(2 + sin x)(4 + sin x)] dx =  ∫ 1 / [(2 + t)(4 + t)] dt

Method of partial fractions

 1 / [(t + 2(t + 4)] = A/(t + 2) + B/(t + 4)

multiply [t+2][t+4]

1 = A(t + 4) + B(t + 2)

Put t = −2

1 = A(2) 

⇒ A = ½

Put t = −4:

1 = B(−2) 

⇒ B = −½

1 / [(t + 2)(t + 4)] = ½[1/(t + 2) − 1/(t + 4)]

integrate

I = ½  ∫ [1/(t + 2)  − 1/(t + 4)] dt 

= ½[ln|t + 2| − ln|t + 4|] + C

= ½ ln|(t + 2)/(t + 4)| + C   use  t = sin x

=½ ln[(2 + sin x)/(4 + sin x)] + C

for  explanation watch this video