Page No 11.21:
Question 1:
Find the general solutions of the following equations:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
We have:
(i)
The value of x satisfying is .
∴
⇒
⇒ ,
(ii)
The value of x satisfying is .
∴
⇒
⇒ ,
(iii) (or)
The value of x satisfying is .
∴
⇒
⇒ ,
⇒
(iv) (or)
The value of x satisfying is .
∴
⇒
⇒ ,
(v)
The value of x satisfying is .
∴
⇒
⇒ ,
(vi)
⇒ (or)
The value of x satisfying is .
∴
⇒
⇒ ,
Page No 11.21:
Question 2:
Find the general solutions of the following equations:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
Answer:
We have:
(i)
⇒
⇒
⇒ ,
(ii)
⇒
⇒
⇒ ,
(iii)
⇒
⇒
⇒ or
⇒ or
⇒ , or ,
⇒ , or ,
(iv)
⇒
⇒
On taking positive sign, we have:
⇒
⇒
⇒,
Now, on taking negative sign, we have:
⇒
⇒
(v)
(vi)
(vii)
(viii)
(ix)
(x)
On taking positive sign, we have:
On taking negative sign, we have:
(xi)
or
Now,
(xii)
On taking positive sign, we have:
On taking negative sign, we have:
Page No 11.22:
Question 3:
Solve the following equations:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Answer:
(i)
or
or
is not possible.
(ii)
or,
is not possible.
(iii)
or
is not possible.
(iv)
or
Now,
(It is not possible.)
(v)
or
Now,
And,
(vi)
Now,
(vii)
On taking positive sign, we have:
On taking negative sign, we have:
Page No 11.22:
Question 4:
Solve the following equations:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
Answer:
(i)
Now,
or,
or
, or
or
(ii)
or
or
or
or
(iii)
or
or
or
or
(v)
or
or
, or
or
or (Taking negative sign will give absurd result.)
or
(vi)
or
or
or ,
(vii)
or
or
or,
or or
(viii)
or
or
, or
or
(ix)
or
, or
or
Page No 11.22:
Question 5:
Solve the following equations:
(i)
(ii)
(iii)
Answer:
(i) We have:
Now,
or
Now,
And,
∴ or
Here,
(ii) Given:
Now,
or
Now,
And,
∴ or
(iii) Given:
Now,
or, or,
And,
Or,
And,
∴ or
Page No 11.22:
Question 6:
Solve the following equations:
(i)
(ii)
(iii)
(iv)
Answer:
(i) Given:
...(i)
The equation is of the form , where and .
Let: and
Now, and
On putting and in equation (i), we get:
(ii) Given: ...(ii)
The equation is of the form of , where and .
Let: and
Now, and
On putting and in equation (ii), we get:
On taking positive sign, we get:
Now, on taking negative sign of the equation, we get:
(iii) Given: ...(iii)
The equation is of the form , where and .
Let: and
Now, and
On putting and in equation (iii), we get:
On taking positive sign, we get:
On taking negative sign, we get:
(iv) Given:
The equation is of the form , where and .
Let: and
Now, and
On putting and in equation (iv), we get:
On taking positive sign, we get:
On taking negative sign, we get:
Page No 11.22:
Question 7:
Solve the following equations:
(i) [NCERT EXEMPLAR]
(ii) [NCERT EXEMPLAR]
(iii) [NCERT EXEMPLAR]
(iv) [NCERT EXEMPLAR]
(v) [NCERT EXEMPLAR]
(vi) 4sinx cosx + 2 sin x + 2 cosx + 1 = 0
(vii) cosx + sin x = cos 2x + sin 2x
(viii) sin x tan x – 1 = tan x – sin x
(ix) 3tanx + cot x = 5 cosec x
Answer:
(i)
(ii)
But, is not possible.
Putting n = 0 and n = 1, we get
(iii)
Putting n = 0 and m = 0, we get
(iv)
(v)
But, is not possible.
(vi)
(vii)
(viii)
(ix)
Page No 11.22:
Question 8:
Solve the following equations:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0
Answer:
Page No 11.22:
Question 9:
Solve the following equations:
3sin2x – 5 sin x cos x + 8 cos2 x = 2
Answer:
Page No 11.22:
Question 10:
Solve the following equations:
Answer:
Page No 11.26:
Question 1:
The smallest value of x satisfying the equation is
(a)
(b)
(c)
(d)
Answer:
(c)
Given:
To obtain the smallest value of x, we will put in the above equation.
Thus, we have:
Hence, the smallest value of x is .
Page No 11.26:
Question 2:
If
(a)
(b)
(c)
(d)
Answer:
(a)
Given: ...(i)
This equation is of the form , where and .
Let:
Now,
And,
On putting in equation (i), we get:
For , .
∴
Page No 11.26:
Question 3:
If , then the values of θ form a series in
(a) AP
(b) GP
(c) HP
(d) none of these
Answer:
(a) AP
Given:
Now,
Now, on putting the value of , we get:
= a1
= a2
= a3
= a4
And so on.
Also,
And so on.
Thus, forms a series in AP.
Page No 11.26:
Question 4:
If a is any real number, the number of roots of in the first quadrant is (are).
(a) 2
(b) 0
(c) 1
(d) none of these
Answer:
(c) 1
Given:
It is a quadratic equation.
If , then the equation becomes
There are two roots of the given equation, but we need to find the number of roots in the first quadrant.
There is exactly one root of the equation, that is, .
Page No 11.26:
Question 5:
The general solution of the equation is
(a)
(b)
(c)
(d) none of these
Answer:
(c)
Given:
Page No 11.26:
Question 6:
A solution of the equation , lies in the interval
(a)
(b)
(c)
(d)
Answer:
(d)
Given:
or
or
is not possible.
∴
The values of lies in the third and fourth quadrants.
Hence, lies in .
Page No 11.26:
Question 7:
The number of solution in [0, π/2] of the equation is
(a) 5
(b) 7
(c) 6
(d) none of these
Answer:
(c) 6
Given:
or
or
Now,
Or,
And,
Hence, there are six solutions.
Page No 11.26:
Question 8:
The general value of x satisfying the equation is given by
(a)
(b)
(c)
(d)
Answer:
(b)
Given:
...(i)
This equation is of the form , where and .
Let:
and
Now,
and
On putting and in equation (i), we get:
Page No 11.26:
Question 9:
The smallest positive angle which satisfies the equation is
(a)
(b)
(c)
(d)
Answer:
(a)
Given:
or,
∴ or, is not possible.
For n = 0, the value of .
Hence, the smallest positive angle is .
Page No 11.26:
Question 10:
If , then the values of x are
(a)
(b)
(c)
(d)
Answer:
(c)
Given:
Page No 11.27:
Question 11:
If , then, x is equal to
(a)
(b)
(c)
(d) none of these.
Answer:
(b)
Given equation:
or
or
Now,
And,
∴
Page No 11.27:
Question 12:
A value of x satisfying is
(a)
(b)
(c)
(d)
Answer:
(d)
Given equation:
...(i)
Thus, the equation is of the form , where and .
Let:
and
and
and
On putting and in equation (i), we get:
Page No 11.27:
Question 13:
In (0, π), the number of solutions of the equation is
(a) 7
(b) 5
(c) 4
(d) 2.
Answer:
(d) 2
Given equation:
Now,
, which is not possible, as it is not in the interval .
Hence, the number of solutions of the given equation is 2.
Page No 11.27:
Question 14:
The number of values of x in [0, 2π] that satisfy the equation
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2
or,
or
Here, is not possible.
∴
Now for n = 0 and 1, the values of .
Hence, there are two solutions in .
Page No 11.27:
Question 15:
If , then x =
(a) 0
(b)
(c) 1
(d) none of these
Answer:
(d) none of these
Given equation:
Let :
Now,
∴
And,
Taking log on both sides, we get:
Page No 11.27:
Question 16:
The equation has .... solution.
(a) finite
(b) infinite
(c) one
(d) no
Answer:
(d) no
Given equation:
...(i)
Thus, the equation is of the form , where and .
Let:
and
Now,
Aso,
On putting and in equation (i), we get:
From here, we cannot find the value of .
Page No 11.27:
Question 17:
If , then general value of x is
(a)
(b)
(c)
(d)
Answer:
(d)
Given equation:
...(i)
This is of the form , where and .
Let:
and .
Now,
And,
Putting and in equation (i), we get:
Page No 11.27:
Question 18:
General solution of is
(a)
(b)
(c)
(d)
Answer:
(c)
Given:
Page No 11.27:
Question 19:
The solution of the equation lies in the interval
(a)
(b)
(c)
(d)
Answer:
Given equation:
or
or
Now, is not possible.
And,
For n = 0, , for n = 1, and so on.
Hence, lies in the interval .
Page No 11.27:
Question 20:
If and then the solutions are
(a)
(b)
(c)
(d)
Answer:
(b)
Given equation:
Or,
So, both lie in .
Page No 11.27:
Question 21:
The number of values of x in the interval [0, 5 π] satisfying the equation is
(a) 0
(b) 5
(c) 6
(d) 10
Answer:
(c) 6
Given:
or
Now, is not possible, as the value of lies between 1 and 1.
Also, sin x is positive only in first two quadrants. Therefore, sin x is positive twice in the interval .
Hence, it is positive six times in the interval , viz
Page No 11.27:
Question 22:
Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2π] is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
tan x + sec x = 2 cos x in [0, 2π]
i.e where x ∉(2x + 1) (∵ cos x is not defined)
i.e sin x + 1 = 2 cos2x
⇒ sin x + 1 = 2(1 − sin2x) (∵ sin2x + cos2x = 1)
i.e 2 sin2x + sin x − 1 = 0
i.e 2 sin2x + 2 sin x – sin x – 1 = 0
i.e 2 sin x (sin x + 1) −1 (sin x + 1) = 0
i.e (2 sin x − 1) (sin x + 1) = 0
i.e sin x = or sin x = −1
i.e for sin x =
we have x is I or II Quadrant
i.e x = and for sin x = −1
x = which is not possible
Hence, tan x + sec x = 2 cos x has 2 columns in [0, 2]
Hence, the correct answer is option C.
Page No 11.28:
Question 1:
The values of satisfying tan 5θ = cot 2θ are ______________.
Answer:
for
tan 5θ = cot 2θ
i.e tan 5θ =
⇒ tan 5θ tan2θ = 1 ...(1)
Since
tan (x + y) =
i.e tan(5θ + 2θ ) =
i.e tan 7θ = (from 1)
i.e tan 7θ is not defined
i.e tan 7θ =
⇒ 7θ = n
i.e θ =
If
Page No 11.28:
Question 2:
The number of values of satisfying is ______________.
Answer:
i.e (using identity :- 1 − cos2 = 2 sin2, 1 + cos 2 = 2cos2)
i.e tan2 = 3
i.e tan =
i.e
out of which lie in
Hence, number of value is 2.
Page No 11.28:
Question 3:
The most general value of θ satisfying 2sin2θ – 1 = 0 is ______________.
Answer:
Page No 11.28:
Question 4:
The number of values of x ∈ (–π, π) satisfying 2 tan2x = sec2x is ______________.
Answer:
For x∊ (−π, π)
2tan2x = sec2x
∴ Number of value of x in (−π, π)
Satisfying 2tan2x = sec2x is 4.
Page No 11.28:
Question 5:
The general value of x satisfying tan x tan 2x = 1 is ______________.
Answer:
tan x tan 2x = 1
i.e 1 − tan x tan 2x = 0 ...(1)
i.e tan 3x = ∞ ...(from 1)
i.e 3x =
Page No 11.28:
Question 6:
If cos mx = cos nx, m ≠ n, then x =______________.
Answer:
If cos mx = cos nx ; m ≠ n
i.e mx = 2rπ ± nx ; r∈Z
i.e x(m ± n) = 2rπ
i.e x =
Page No 11.28:
Question 7:
The number of values of x ∈ [0, 2π] satisfying the equation 2 sin2x = 4 + 3 cos x is______________.
Answer:
For x ∈ [0, 2]
2sin2x = 4 + 3cos x
i.e 2(1 − cos2x) = 4 + 3cos x
i.e 2 − 2cos2x = 4 + 3cos x
⇒ 2cos2x + 3cos x + 2 = 0
⇒ 2cos2x + 4cos x − cos x + 2 = 0
i.e No solution is possible
∴ No value of x satisfies
2sin2x = 4 + 3cos x
Page No 11.28:
Question 8:
The set of values of x satisfying the equation is______________.
Answer:
i.e tan (3x − 2x) = 1 ( using identity :- tan (x − y) = )
i.e tan x = 1
i.e x = tan−1
i.e x = ; n ∈ Z
Page No 11.28:
Question 9:
If then x = _____________.
Answer:
Page No 11.28:
Question 1:
Write the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π].
Answer:
Given:
tanx + secx = 2 cosx
Now,
And,
Hence, the given equation has two solutions in .
Page No 11.28:
Question 2:
Write the number of solutions of the equation .
Answer:
We have:
...(i)
The equation is of the form , where and .
Now,
Let:
and
Thus, we have:
and
By putting and in equation (i), we get:
The solution is not possible.
Hence, the given equation has no solution.
Page No 11.28:
Question 3:
Write the general solutions of tan2 2x = 1.
Answer:
Given:
Hence, the general solution of the equation is
Page No 11.28:
Question 4:
Write the set of values of a for which the equation has no solution.
Answer:
Given:
If or , then the equation will possess a solution.
For no solution, .
Page No 11.28:
Question 5:
If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.
Answer:
Given:
If , then
Now, for
If then
Now, for
If then
Now,
when
And,
when
Clearly, we can see that for , has exactly one solution.
∴
Page No 11.28:
Question 6:
Write the number of points of intersection of the curves and .
Answer:
Given curves: and
Now,
Also,
For the other value of n, the value of x will not satisfy the given condition.
Hence, the number of points of intersection of the curves is two, i.e., .
Page No 11.28:
Question 7:
Write the values of x in [0, π] for which and cos 2x are in A.P.
Answer:
(i)
This equation is of the form , where and .
Now,
Let:
and
Thus, we have:
and
On putting and in equation (1), we get:
For n = 0, the values of x are and for n = 1, the values of x are .
For the other value of n, the given condition is not true, i.e., [0, π].
Page No 11.28:
Question 8:
Write the number of points of intersection of the curves .
Answer:
Given:
and
Now,
Also,
The value of function lies between 1 and 1. Therefore, the two curves will not intersect at any point.
Hence, the number of points of intersection of the curves is 0.
Page No 11.28:
Question 9:
Write the solution set of the equation in the interval [0, 2π].
Answer:
Given:
Now, or
or
is not possible.
Thus, we have:
By putting n = 0 and n = 1 in the above equation, we get:
or in the interval
For the other value of n, x will not satisfy the given condition.
∴ and
Page No 11.28:
Question 10:
Write the number of values of x in [0, 2π] that satisfy the equation .
Answer:
Given equation:
Now,
Here, is not possible.
Or,
Taking positive sign,
Taking negative sign,
and will satisfy the given condition, i.e., x in [0, 2π].
Hence, two values will satisfy the given equation.
Page No 11.29:
Question 11:
If , , find .
Answer:
Given:
Applying componendo and dividendo, we have
Page No 11.29:
Question 12:
If , where , then find the value of x.
Answer:
The given equation is .
Now,
But, is not possible.
Putting n = 0 and n = 1, we get
Page No 11.29:
Question 13:
If secx cos5x + 1 = 0, where , find the value of x.
Answer:
The given equation is secx cos5x + 1 = 0.
Now,
Putting n = 0 and n = 1, we get
Also, putting m = 0, we get
Hence, the values of x are and .
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