Ncert Solutions Maths class 11th

A triangle ABC lying in the first quadrant has two vertices as A(1,2) and B(3,1). If ∠BAC = 90°, and ar(?ABC) = 5√5 sq. units, then the abscissa of the vertex C is:

A triangle ABC lying in the first quadrant has two vertices as A(1,2) and B(3,1). If ∠BAC = 90°, and ar(?ABC) = 5√5 sq. units, then the abscissa of the vertex C is:

Let u = (2z+i)/(z-ki), z = x + iy and k > 0. If the curve represented by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is:

Given the following two statements:
(S₁): (q∨p) → (p ↔~ q) is a tautology.
(S₂): ~ q∧ (~ p ↔ q) is a fallacy. Then:

If 1 + (1 − 2² . 1) + (1 − 4² • 3) + (1 − 6² · 5) + … . + (1 − 20² · 19) = α − 220β, then an ordered pair (α, β) is equal to:

Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m ) above the line AC is:

Let x²/a² + y²/b² = 1(a > b) be given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, φ(t) = 5/12 + t − t², then a² + b² is equal to:

Let α and β be roots of x² – 3x + p = 0 and y and δ be the roots of x² – 6x + q = 0. If α, β, γ, δ, form a geometric progression. Then ratio (2q + p): (2q – p) is:

The shortest distance between the curves y² = 8x and x² + y² + 12y + 35 = 0 is :

Let C be the circle of minimum area enclosing the ellipse E: x²/a² + y²/b² = 1 with eccentricity 1/2 and foci (±2,0). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis of E and contains the point of intersection of E with the negative y -axis. Then the maximum area of the triangle PQR is :