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 Bihar Board 12th Maths Important Questions Short Answer Type Part 1

Bihar Board 12th Maths Important Questions Short Answer Type Part 1

Relation & Functions

Question 1.
Prove that the function f :R→R given by f(x) = 2x is one-one & on to.
Solution:
For one-one – Take f(x1) = f(x2) .
⇒ 2x1 = 2x2
⇒ x1 = x2
∴ f is one-one.
For on to co-domain = R
Rage = y = 2x = 2 x R = R
∴ Co-domain = Rage
∴ f is on to
Hence f is one-one onto.

Question 2.
Show that the function f: N → N given by f(1) = f(2) = 1 & f(x) = x-1 for every x > 2 is on to but not one-one,
Solution:
∴ f(1) = f(2) = 1
∴ f is not one-one
∴ f(x) = x – 1
∵ Co -domain = N
∀ x > 2, Rage = N
∴ Co-domain = Rage
∴ f(x) is on to.

Question 3.
Let L be the set of all lines in a plane and# he the relation inL defined as R = { (L1 L2) : L1 is perpendicular to L2). Show that R is symmetric but neither reflexive nor transitive.
Answer:
a line L1 can not be ⊥r to itself.
R is not reflexive.
For symmetric — If L1 RL2
⇒ L2RL1

⇒ L1 is ⊥r to L2
⇒ L2 is ⊥r to L1
⇒ (L2,L1) ∈ R.
∴ R is symmetric
for transitive — If L1 RL2, L2RL3 = L1RL3
Given is ⇒ L1 is ⊥r L2
& . L2 is lr L3
⇒ L1||L3
⇒ L1 is ⊥r L2
⇒ L2 is ⊥r L3
⇒ L1||L3
∴ R is not transitive.

Question 4.
Show that the function f:R→R: f(x) =x3 is one-one and onto.
Solution:
Given, f(x1) = f(x2) ⇒ x13 = x23

⇒ x1 = x2
⇒ f is one-one.
Let y ∈ R and 3; = x3; Then x = y1/3 e R.
Thus, for each y in the codonain R there exists y1/3 in R such that
f(y1/3) = (y1/3)3 = y
⇒ f is onto.
Hence, f is one-one to.

Matrix

Question 1.
If A = [2342], B = [1−235] , C = [−2354] then find the value of (i) A- B (ii) 3A – C
Answer:

Question 2.
If A = [3−112] Prove that A2 – 5. A + 7I= 0.
Solution:
 

Question 3.
If Aα = [cosα−sinαsinαcosα], Prove that AαAβ = AβAα = Aα+β
Solution:

AαAβ = AβAα = Aα+β

Question 4.
If A = [3121]
Find the number a and b such that A2 + aA + bI = 0 where I is an identity matrix of order 2×2 and 0 is a 2 x 2 Zero matrix.
Solution:

Equation the respective elements of two matrices we get.
8 + 2a = 0 ⇒ a =—4
and 11 + 3a + b = 0 ⇒ 11 – 12 + b = 0 ⇒ b= 1
Hence .
a = -4, b = 1.

Determinants

Question 1.
Prove that ∣∣∣∣a+b+2cccab+c+2aabbc+a+2b∣∣∣∣ = 2(a + b + c)3
Solution:
Operating C1 + C2 + C3 is the determinant of L.H.S.

Question 2.
Find the area of the triangle with vertices (-2,-3), (3,2) and (-1,-8).
Answer:
Area of the triangle = 12∣∣∣∣x,x2x3y,y2y3111∣∣∣∣
= 12∣∣∣∣−23−1−32−8111∣∣∣∣=12∣∣∣∣−54−1−510−8001∣∣∣∣ (R1 – R2, R2 – R3
= – 15 = 15 5q.units.

Question 3.
Find Values of k if area of triangle is 4 Sq. units and vertices are : (k, 0), (4,0)’, (0,2).
Solution:
Δ = ∣∣∣∣1k0140102∣∣∣∣ or, 4 = ∣∣∣∣1k0140102∣∣∣∣
or, 4 = 2∣∣∣1k14∣∣∣ or, 4 = 2(4 – k)
or, 4 – k = 2 or, 4 = 2(4 – k)
or, 4 – k = 2 ∴ k = 2
4 – 2 = k

Inverse circular Function

Question 1.
Find the principal value of following :

Solution:
(i) Let sin-1([latex-\frac{1}{\sqrt{2}}][/latex]) = y
Then sin y = 12√ ≠ +ve (1st or 2nd quadram)
Last value of Angle π4
∴ P. V of sin−1(12√)=π4

(ii) Let y = cot-1([latex-\frac{1}{\sqrt{3}}][/latex]) or cot y = [latex-\frac{1}{\sqrt{3}}][/latex]
We know that the range of principal value of Cot-1 is (0, π)

Hence P.V 2π3

(iii) Ley y = sin-1 3√2
then sin y = 3√2
∴ Range of Principal value of sin-1 is [−π2,π2]
∴ sin y = 3√2sinπ3
∴ P.V 2π3

(iv) Let y = sin-1 ( −12√ )
or sin y = −12√
∵ Range of P.V of sin-1 is [−π2,π2]
∴ sin y = sin −π4
∴ P.V = −π4

(v) sin y = sin-1 (−12)
∵ Range of P. V. of sin -1 is [−π2,π2]
∴ sin y = sin (-π/4)
∴ P.V = −π6

Question 2.
Find the value of sin-1(sin 3π5
Solution:
we knoe that,

Question 3.
Prove that 2tan−113+tan−117=π4
Answer:

Question 4.
Prove that tan−1α−β1+αβ+tan−1β−γ1+βγ+tan−1γ−α1+γα

Question 5.
Prove that sin−145+sin−1513+sin−11625=π2
Answer:

Question 6.
Solve 2tan-1(cosx) = tan-1(2 cosec x)
Answer:
2tan-1(cosx)
= tan−1(2cosx1−cos2x)=tan−1(2cosxsin2x)
Putting this valu in (1)
∴ tan−1(2cosxsin2x) = s cosec x = sinx
cos x = sin x or tan x = 1
⇒ x = π4

Question 7.
tan−11−x1+x=12tan−1x, x > 0
Answer:
12tan−1x=tan−11−x1+x

Question 8.
sin-1(1 – x) – 2sin-1 x – π2 then x is equal to
(A) 0, 12
(B) 1, 12
(C) 0
(D) 12
Solution:
sin-1(1 – x) – 2sin-1 x – π2
Putting π2 = sin-1(1 – x) + cos-1 (1-x)
-2sin-1 = cos-1(x-1)
Let sin-1 = α
-2sin-1 = -2α = cos-1(x-1)
or cos 2α = 1 – x
1 – 2cos2α = (1 – x)
Putting sin α = x
1 – 2x2 = 1 – x
or 2x2-x = 0
x(2x – 1) = 0 x = 0, 12
But x = 12 does not satisfy the equation ∴ x = 0
∴ Part (C) is the correct answer.

Continuity and Differentiabiilty

Question 1.
f(x) = x sin 1x , x ≠ 0 and f(0) = 0.
Solution:

= Lh→0(hsin14) = 0
Also given that f(0) = O

So, function is continous at x = 0

Question 2.
Prove that every differentiable function is continuous also.
Solution:
Let the function f(x) is differentiable at x = a.

Lth→0 f(a -h) – f(a) = f1(a) x 0 = 0
Lth→0 f(a – h) = f(a)
Lth→0f(a + h) Lth→0f(a
So, Lth→0 f(a+h) = Lth→0 f(a-h) = f(a)
∴ f(x) is continuous at x = a.

Differentiation

Question 1.
Differentiate the following w.r.t to x.
(i) Cos(sin(log a))
(ii) log (x + a2+x2−−−−−−√
(iii) log 1−cosx
(iv) loge (logen
(v) ta-1(1−x1+x)
(vi) sin(x2 + 1)
(vii) tan-11+x2√−1x
(viii) (logx)logx
(ix) (x)x
(x) (logx)x + xlogx
(xi) cos (a cos x + b sin x)
(xii) xcos-1x
(xiii) esinx + (tanx)x
Solution:(i) Let Sin (logx) = t, log x = v
∴ y = cos t , t = sin v = log x

(ii) Let y = log(x + a2+x2−−−−−−√)
Diff w.r.to x

(iii) Let y = 1−cosx = log 2−sin2x/2
or y = log tan2x/2−−−−−−−√=log(tanx2)
Diff w.r.to x

(iv) Let y = log3 (log)xe
Diff w.r.to x
dydx=d(loge(logxe))dx=1logxe⋅1x

(v) Let y = tan−1(1−x1+x)
Diff w.r.to x

(vi) Let y = sin(x2 + 1)
Diff w.r.to x
dydx=d(sin(x2+1))dx=cos(x2+1)⋅d(x2+1)dx
cos (x2 + 1).(2x + 0) = 2xcos (x2 + 1)

(vii) Let y = tan-11+x2−1√x put x = tan θ 1 + x2 = sec2

Differentiability with respect for, we get
dydx=12(1+x2)

(viii) Let y = (logx)x
⇒ log y = log x log (log x)
differentiablity w. r. for , we get

(ix) Let y = xx = exlogex
With restpect to x

(x) Let y = (log)x + xlogx
y = elogex + elogex. logex
Diff With respect to x

(xi) Let y = cos (a cos x + bsin x)
⇒ dydx = -sin (a cos x + b sin x_) ddx (acos x +b sin x)
= -sin(a cos x + b sin x ). (-a sin x + b cos x)

(xii) Let y = ecos-1x
cos-1x.loex
or y = e
Diff With respect to x

(xiii) Let y = esinx + (tanx)x + (tan x)x = esinx + exloge tan x
Diff w.r.to x

Question 2.
If y = sin(tan-12x) then prove that dydx=2(1+4x2)3/2
Solution:
Put tan-1 2x = t
y = sin(tan-12x) = sin t
dydt = cos t
∴ tan -12x = t