SSA Chandigarh TGT Math Mock Test -5 - SSA Chandigarh MCQ

ਸਹੀ ਕਰਮਣੀ ਵਾਚ: ਫਿਲਮ ਲਈ ਅਜਿਹੀਆਂ ਕਤਾਰਾਂ ਘੱਟ ਹੀ ਦੇਖੀਆਂ ਜਾ ਸਕਦੀਆਂ ਹਨ।

Key Points

• ਦਿੱਤੇ ਗਏ ਵਾਕ "ਕਿਸੇ ਫਿਲਮ ਲਈ ਅਜਿਹੀਆਂ ਕਤਾਰਾਂ ਸ਼ਾਇਦ ਹੀ ਕਿਸੇ ਨੇ ਦੇਖੀਆਂ ਹੋਣ" ਨੂੰ ਕਰਮਣੀ ਵਾਚ (Passive Voice) ਵਿੱਚ ਪਾਉਣ ਲਈ, ਅਸੀਂ ਨੂੰ ਉਸ ਵਿਕਲਪ ਨੂੰ ਚੁਣਣ ਦੀ ਲੋੜ ਹੈ ਜੋ ਇਸ ਗੱਲ ਨੂੰ ਦਰਸਾਉਂਦਾ ਹੈ ਕਿ ਕਤਾਰਾਂ ਨੂੰ ਦੇਖਣ ਦਾ ਅਨੁਭਵ ਬਹੁਤ ਘੱਟ ਹੀ ਲੋਕਾਂ ਨੇ ਕੀਤਾ ਹੈ।
• ਵਿਕਲਪ 4) "ਫਿਲਮ ਲਈ ਅਜਿਹੀਆਂ ਕਤਾਰਾਂ ਘੱਟ ਹੀ ਦੇਖੀਆਂ ਜਾ ਸਕਦੀਆਂ ਹਨ।" ਇਹ ਦਰਸਾਉਂਦਾ ਹੈ ਕਿ ਅਜਿਹੀਆਂ ਕਤਾਰਾਂ ਨੂੰ ਦੇਖਣਾ ਇੱਕ ਦੁਰਲੱਭ ਘਟਨਾ ਹੈ ਅਤੇ ਇਸੇ ਵਿਚਾਰ ਨੂੰ ਵਾਪਰਨ ਵਾਲੇ ਨੇ ਵੀ ਲੈਂਦਾ ਹੈ।

1st term = 12
2nd term = 12 + 20 = 32
3rd term = 32 + 40 = 72
4th term = 72 + 80 = 152

Thus, the missing term is 5th term = 152 + 160 = 312

The reason behind Kritika's this behaviour shows that her thoughts get acknowledged in schools. Sometimes it happens with children, if they feel that they are not being given importance or no one will praise them then they adopt such kind of behaviour that they talk only to those people who praise them gives due importance. Hence, the correct answer is, 'Her thoughts get acknowledged in schools.'

Let I = ∫5x dx/[(x+1)(x2+9)]
Let 5x/[(x+1)(x2+9)] = A(x+1) + (Bx + C)/(x²+9)
⇒5x = (A+B)x² + (B+C)x + 9A + C
Comparing the coefficients of x2 on both sides, we get
A + B = 0    ...(1)
Comparing the coefficients of x on both sides, we get
B + C = 5    .....(2)
Comparing constants on both sides, we get
9A +C = 0    ....(3)
From (1), we get B = −A
From (3), we get C = −9A
now, from (2), we get 
−A − 9A = 5 
⇒A = −1/2
B = 1/2
C = 9/2
So, 5x/[(x+1)(x²+9)] = −1/2 × [1(x+1)] + 1/2 × (x + 9)/(x² + 9)
So, ∫5x dx/[(x+1)(x² + 9)] = −1/2∫dx/(x+1) + 1/2∫(x+9)/(x² +9) dx
=−1/2 log|x+1| + 1/2∫x dx/(x² +9) + 9/2∫dx (x² +9)
=−1/2 log|x+1| + 1/4∫2x dx/(x² +9) + 9/2∫dx/(x² +(3)²)
⇒−1/2 log |x+1| + 1/4log|x ² +9| + 9/2×1/3 tan⁻¹(x/3) + C
=−1/2 log |x+1| + 1/4log∣|x2+9| + 3/2 tan⁻¹(x/3) + C

 limx→1x⁹−1/x⁵−1 has indeterminate initial form (it has form 0/0). Therefore 1is a zero and x−1 a factor of both the numerator and the denominator.
limx→1 x¹⁰−1/x⁵−1
= limx→1[(x−1)(x⁸+x⁷+x⁴….⋅+x+1)]/(x−1)(x⁵+x⁶+x⁴+⋅⋅⋅+x+1)
= (1+1+1+1+.....upto 10 terms)/(1+1+1+⋅⋅⋅upto 5 terms)
= 10/5 = 2 

f(x) = sinx.cosx
f’(x) = -sin²x + cos²x
-sin²x + cos²x = 0
sin²x = cos²x
tan²x = 1
tanx = +-1
for tan x = 1, x = π/4
for tan x = -1, x = 3π/4
At x = π/4, f(π/4) = ½
At x = 3π/4, f(3π/4) = -½
At π/4 is the local maxima.

Since, y = e^x and y = x do not meet for any x ∈ R ∴ A ∩ B = φ.

Option d is correct, because it is the property of definite integral
 ∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx

On comparing the given equations with: 
, we get: 

As n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
But in case of disjoint sets, n(A ∩ B) = 0
∴ n(A ∪ B) = n(A) + n(B)

y = V^3/3
V = x² - 2x + 3
y = 1/3*v³
dy/dx = d/dx{1/3*(x²-2x+3)³} 
= 1/3*3*(x²-2x+3)²*(2x-2) 
{d/dx(xⁿ)=nx^n-1*dx/dx} 
= 2(x-1)*(x²-2x+3)²

x³ - 18x² + 96x = x(x² - 18x + 96) = x[(x-9)² + 15]
 =x(x-9)² +15 ≥ 0

lim(x→0) [log10 + log1/10]/x
= [log10 + log10]/0
= 0/0 form
lim(x→0) [(1/(x+1/10) * 1]/1
lim(x→0) [(1/(0+1/10) * 1]/1
= 1/(1/10) => 10

Mutually exclusive means no common event.
if 2 comes on either dive then sum can't be greater than 8.It will always less than or equal to 8.
So mutually exclusive.

 d(x^y)=d(e^(y⋅log(x)))
=e^(y⋅log(x))d(y⋅log(x))
=(x^y)(dy⋅log(x) + y⋅d(log(x))
=(x^y) 
d(y^x) = (y^x)(log(y)dx + x/ydy). 
Since  d(x^y) = d(y^x) , and simplifying by  x^y = y^x , we get
log(y)dx + x/ydy = log(x)dy + y/xdx. 
Removing the denominators leads to:
xylog(y)dx + x²dy = xylog(x)dy + y²dx 
(xylog(y) − y²)dx = (xylog(x) − x²)dy 
dy/dx = (xylog(y)−y²)/(xy⋅log(x)−x²)

As y ∈ N, y can be 1, 2, 3, 4...
∴ x will be 1, 1/2, 1/3, 1/4...
⇒ 1 ∈ Q

Answer:  C
Solution: Variance= [summation (y^2×f) /N] -[ summation (yf) /N]^2
=(296/25) -(0/25) ^2
=11.84
standard deviation=√11.84=3.12

Probability that Amit will qualify the exam(E) = 0.06
Probability that Priyanka will qualify the exam(F) = 0.12
Probability that both will qualify the exam P(E⋂F) = 0.02
P(EUF) = P(E) + P(F) - P(EUF)
= 0.06 + 0.12 - 0.02   = 0.16
P(E’⋂F’) = 1 - P(EUF)
⇒ 1 - 0.16
= 0.84

As per the definition of LPP:
Linear Programming Problem is one that is concerned with finding the optimal value of a linear function of several variables, subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities.
Since the variables (x & y) are non-negative, the feasible solution of a LPP belongs to the 1st quadrant only.

 f(x) = a^x
Taking log bth the sides, log f(x) = xloga
f’(x)/a^x = loga
f’(x) = a^x loga   {a^x > 0 for all x implies R, 
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.