Arts (Hindi Medium) कक्षा १२ - CBSE Question Bank Solutions for Mathematics (गणित)

(x + 1)²(x + 2)³(x + 3)⁴

`cos^-1 ((sinx + cosx)/sqrt(2)), (-pi)/4 < x < pi/4`

`tan^-1 (sqrt((1 - cosx)/(1 + cosx))), - pi/4 < x < pi/4`

`tan^-1 (secx + tanx), - pi/2 < x < pi/2`

`tan^-1 (("a"cosx - "b"sinx)/("b"cosx - "a"sinx)), - pi/2 < x < pi/2` तथा `"a"/"b" tan x > -1`

`sec^-1 (1/(4x^3 - 3x)), 0 < x < 1/sqrt(2)`

`tan^-1 ((3"a"^2x - x^3)/("a"^3 - 3"a"x^2)), (-1)/sqrt(3) < x/"a" < 1/sqrt(3)`

`tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2) - sqrt(1 - x^2))), -1 < x < 1, x ≠ 0`

x = `"t" + 1/"t"`, y = `"t" - 1/"t"`

x = `"e"^theta (theta + 1/theta)`, y= `"e"^-theta (theta - 1/theta)`

x = 3cosθ – 2cos³θ, y = 3sinθ – 2sin³θ

sin x = `(2"t")/(1 + "t"^2)`, tan y = `(2"t")/(1 - "t"^2)`

x = `(1 + log "t")/"t"^2`, y = `(3 + 2 log "t")/"t"`

यदि x = e^cos2t और y = e^sin2t तो सिद्ध कीजिए कि `"dy"/"dx" = (-y log x)/(xlogy)` है।

 यदि x = asin2t (1 + cos2t)  और y = b cos2t (1–cos2t) तो दर्शाइए कि, x = `pi/4` पर;`("dy"/"dx") = "b"/"a"`

यदि x = 3sint – sin 3t और y = 3cost – cos 3t तो t = `pi/3` पर `"dy"/"dx"` ज्ञात कीजिए।

sinx के सापेक्ष `x/sinx`को अवकलित कीजिए।

tan^–1x के सापेक्ष `tan^-1 ((sqrt(1 + x^2) - 1)/x)` को अवकलित कीजिए, जब x ≠ 0.

`sin xy + x/y` = x² – y

sec(x + y) = xy