CBSE Board 2020


Question Paper Solution

Section B

2 Marks Questions

Question 21: A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper. The following were the answers given by the students.

`2x+3`, `3x^2+7x+2`, `4x^3+3x^2+2`, `x^3sqrt(3x)+7`, `7x+sqrt7`, `5x^3-7x+2`, `2x^2+3-5/x`, `5x-1/2`, `ax^3+bax^2+cx+d`, `x+1/x`

Answer the following questions:

  1. How many of the above ten, are not polynomials?
  2. How many of the above ten are quadratic polynomials?

Answer: (i) 3, (ii) 1

Question 22: A child has a die whose six faces show the letters as shown here: A, B, C, D, E, A

The die is thrown once. What is the probability of getting (i) A, (ii) D?

Answer: (i) `1/3` (ii) `1/6`

Question 23: In following figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that



Answer: Let us draw altitudes AM and DN on BC; respectively from A and D

similar triangles exercise solution

`text(ar ABC)/text(ar DBC)=(1/2xxBCxxAM)/(1/2xxBCxxDN)`



∠ AMO = ∠ DNO (Right angle)

∠ AOM = ∠ DON (Opposite angles)

Hence; ΔAMO ∼ ΔDNO



Or, `text(ar ABC)/text(ar DBC)=(AO)/(DO)`


In following figure, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2

Similar triangles

Answer: In Δ ADB

AD2 = AB2 - BD2


AD2 = AC2 - CD2

So, AB2 - BD2 = AC2 - CD2

Or, AB2 - BD2 + CD2 = AC2

Or, AB2 + CD2 = BD2 + AC2 proved

Question 24: Prove that `1+(text(cot)^2α)/(1+text(cosecα))=text(cosecα)`

Answer: LHS =`1+(text(cot)^2α)/(1+text(cosecα))`


`=1+text(cosecα)-1=text(cosecα)` proved


Show that tan4θ + tan2θ = sec4θ - sec2θ

Answer: RHS = sec4θ - sec2θ

= (1 + tan2θ)2 - (1 + tan2θ)

= 1 + tan4θ + 2 tan2θ - 1 – tan2θ

= tan4θ + tan2θ = LHS proved

Question 25: Find the mode of the following frequency distribution:


Answer: Here, modal class is 30 – 35, l = 30, h = 5, f1 = 10, f0 = 9, f2 = 3

Mode can be calculated as follows:

`=l+(f_1-f_0)/(2f_1-f_0-f_2)× x`

`=30+(10-9)/(2×10-9-3)× 5`



Question 26: From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of the same height and same base radius is removed. Find the volume of the remaining solid.

Answer: Volume of cylinder `=πr^2h`

`=(22)/7×6^2×14=1584` cm3

Volume of cone `=1/3×πr^2h`

`=(1584)/3=528` cm3

So, volume of remaining solid `=1584-528=1056` cm3

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