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Question

List I has four entries and List II has five entries. Each entry of List I is to be matched with one or more than one entries of List II.

List IList II (A)The possible value(s) of a for which the largest(P)9value of sin2x−2asinx+a+3 is 7 is/are(B)The possible value(s) of a for which the smallest(Q)16value of x4−ax2+2a−1 for x∈[−1,2] is−7, is/are(C)If a relation R is defined on set of integers as(R)−3 R={(x,y):4x2+9y2≤36}, then possibleelement(s) in the domain is/are(D)If sinx+cosx=15, then |12tanx| is equal to(S)1 (T)11

Which of the following is the only CORRECT combination?

List IList II (A)The possible value(s) of a for which the largest(P)9value of sin2x−2asinx+a+3 is 7 is/are(B)The possible value(s) of a for which the smallest(Q)16value of x4−ax2+2a−1 for x∈[−1,2] is−7, is/are(C)If a relation R is defined on set of integers as(R)−3 R={(x,y):4x2+9y2≤36}, then possibleelement(s) in the domain is/are(D)If sinx+cosx=15, then |12tanx| is equal to(S)1 (T)11

Which of the following is the only CORRECT combination?

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Solution

The correct option is **D** (B)→(R),(T)

(A)

Let f(x)=sin2x−2asinx+a+3

Let sinx=t, where t∈[−1,1]

Given equation is t2−2at+(a+3)

In t∈[−1,1]

Largest value can occur at t=1 or t=−1

At t=1,−a+4=7

⇒a=−3

At t=−1, 1+3a+3=7

⇒a=1

(A)→(R),(S)

(B)

f(x)=x4−ax2+2a−1,x∈[−1,2]

Let t=x2,t∈[0,4]

g(t)=t2−at+(2a−1) in t∈[0,4]

Smallest value can occur at

g(0)=2a−1=−7

⇒a=−3g(4)=15−2a=−7

⇒a=11g(a2)=a24−a22+2a−1=−7⇒a2−8a−24=0⇒a=4±2√10

(B)→(R),(T)

(A)

Let f(x)=sin2x−2asinx+a+3

Let sinx=t, where t∈[−1,1]

Given equation is t2−2at+(a+3)

In t∈[−1,1]

Largest value can occur at t=1 or t=−1

At t=1,−a+4=7

⇒a=−3

At t=−1, 1+3a+3=7

⇒a=1

(A)→(R),(S)

(B)

f(x)=x4−ax2+2a−1,x∈[−1,2]

Let t=x2,t∈[0,4]

g(t)=t2−at+(2a−1) in t∈[0,4]

Smallest value can occur at

g(0)=2a−1=−7

⇒a=−3g(4)=15−2a=−7

⇒a=11g(a2)=a24−a22+2a−1=−7⇒a2−8a−24=0⇒a=4±2√10

(B)→(R),(T)

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