CBSE Board 2020

Mathematics

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

`text(arΔABC)/text(areΔDBC)-(AO)/(DO)`

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

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

`=(AM)/(DN)`

In ΔAMO and ΔDNO;

∠ AMO = ∠ DNO (Right angle)

∠ AOM = ∠ DON (Opposite angles)

Hence; ΔAMO ∼ ΔDNO

Hence;

`(AM)/(DN)=(AO)/(DO)`

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

OR

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

Answer: In Δ ADB

AD² = AB² - BD²

In Δ ADC

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-1))/(1+text(cosecα))`

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

OR

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`

`=30+1/8×5`

`=(240+5)/8=(245)/8=30.625`

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