CBSE Class 10 Mathematics Solved Question Paper 2017

S = {−3, −2, −1, 0, 1, 2, 3}
Let E be the event of getting a number whose square is less than or equal to 1.
So, E = {−1, 1, 0}
P(E)=3/7.
Hence, the probability of getting a number whose square is less than or equal to is 3/7.

The angle of elevation of the sun from the ground is θ.
We have, x:y =√3:1
Now, In ∆ABC

1) Draw a line segment AB = 8 cm.
2) Draw a ray AX making an acute angle ∠BAX=60° with AB.
3) Draw a ray BY parallel to AX by making an acute angle ∠ABY=∠BAX.
4) Mark four points A₁, A₂, A₃, A₄ on AX and five points B₁,B₂,B₃,B₄,B₅ on BY in such a way that AA₁=A₁A₂=A₂A₃=A₃A₄.
5) Join A₄B₅
6) Let this line intersect AB at a point P.
Thus, P is the point dividing the line segment AB internally in the ratio of 4:5.

Let the points X (4,k) and Y(1,0)
It is given that the distance XY is 5 units.
By using the distance formula,

Let the radius of the hemisphere be x units.
Volume of a hemisphere = Surface area of the hemisphere

Hence, the diameter of the hemisphere is equal to 9 units.

Let AB be the diameter of a circle, with centre O. The tangents PQ and RS are drawn at points A and B, respectively.

We know that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OA ⊥ PQ and OB ⊥ RS
⇒ ∠OBR = 90°
∠OBS = 90°
∠OAP = 90°
∠OAQ = 90°
We can observe the following:
∠OBR = ∠OAQ and ∠OBS = ∠OAP
Also, these are the pair of alternate interior angles.
Since alternate interior angles are equal, the lines PQ and RS are parallel to each other.
Hence, proved.

The first term between 200 and 500 divisible by 8 is 208, and the last term is 496.
So, first term (a) = 208
Common difference (d) = 8
a_n=a+(n−1)d=496
⇒208+(n−1)8=496
⇒(n−1)8=288
⇒n−1=36⇒n=37
Hence, there are 37 integers between 200 and 500 which are divisible by 8.

Given: PA and PB are the tangents to the circle.
PA = 12 cm
QC = QD = 3 cm

To find: PC + PD

PA = PB = 12 cm
(The lengths of tangents drawn from an external point to a circle are equal)
Similarly, QC = AC = 3 cm
and QD = BD = 3 cm.

Now, PC = PA − AC = 12 − 3 = 9 cm
Similarly, PD = PB − BD = 12 − 3 = 9 cm

Hence, PC + PD = 9 + 9 = 18 cm.

The given equation is x²+k(2x+k−1)+2=0.
⇒x²+2kx+k(k−1)+2=0
So, a = 1, b = 2k, c = k(k − 1) + 2
We know D=b²−4ac
⇒D=(2k)²−4×1×[k(k−1)+2]
⇒D=4k²−4[k2−k+2]
⇒D=4k²−4k2+4k−8
⇒D=4k−8=4(k−2)
For equal roots, D = 0
Thus, 4(k − 2) = 0
So, k = 2.