JEE Advanced · Mathematics · 21. ITF

Match List I with List II and select the correct answer using the code given below the lists :

List - I List - II

A.
${(\frac{1}{y^{2}} (\frac{cos ({tan}^{- 1} y) + y sin ({tan}^{- 1} y)}{cot ({sin}^{- 1} y) + tan ({sin}^{- 1} y)}) + y^{4})}^{1 / 2} takes value$ P. $\frac{1}{2} \sqrt{\frac{5}{3}}$

B.
If cosx + cosy + cosz = 0 = sinx + siny + sinz then possible value of cos $\frac{x - y}{2} is$ Q. $\sqrt{2}$

C.
If cos $(\frac{π}{4} - x)$ cos2x + sinx sin 2x secx = cosx sin2x secx + cos $(\frac{π}{4} + x)$ cos2x then possible value of secx is R. $\frac{1}{2}$

D.
If cot $({sin}^{- 1} \sqrt{1 - x^{2}}) = sin ({tan}^{- 1} (x \sqrt{6})), x \neq 0,$ then possible value of x is S. 1

	List - I		List - II
A.	${(\frac{1}{y^{2}} (\frac{cos ({tan}^{- 1} y) + y sin ({tan}^{- 1} y)}{cot ({sin}^{- 1} y) + tan ({sin}^{- 1} y)}) + y^{4})}^{1 / 2} takes value$	P.	$\frac{1}{2} \sqrt{\frac{5}{3}}$
B.	If cosx + cosy + cosz = 0 = sinx + siny + sinz then possible value of cos $\frac{x - y}{2} is$	Q.	$\sqrt{2}$
C.	If cos $(\frac{π}{4} - x)$ cos2x + sinx sin 2x secx = cosx sin2x secx + cos $(\frac{π}{4} + x)$ cos2x then possible value of secx is	R.	$\frac{1}{2}$
D.	If cot $({sin}^{- 1} \sqrt{1 - x^{2}}) = sin ({tan}^{- 1} (x \sqrt{6})), x \neq 0,$ then possible value of x is	S.	1

A a-r;b-p;c-s;d-q;
B a-p;b-q;c-r;d-s;
C a-p;b-r;c-s;d-q;
D a-s;b-r;c-q;d-p;

Verified Solution

Answer & Solution

Correct Answer

(D) a-s;b-r;c-q;d-p;

Step-by-step Solution

Detailed explanation

(P) - 4; (Q) - 3; (R) - 2 or 4; (S) - 1

(P) {(\frac{1}{y^{2}} {(\frac{cos ({tan}^{-1} y) + y sin ({tan}^{-1} y)}{cot ({sin}^{-1} y) + tan ({sin}^{-1} y)})}^{2} + y_{4})}^{1 / 2}

= {[\frac{1}{y^{2}} {[\frac{(\frac{1}{\sqrt{1 + y^{2}}} + \frac{y . y}{\sqrt{1 + y^{2}}})}{(\frac{\sqrt{1 - y^{2}}}{y} + \frac{y}{\sqrt{1 - y^{2}}})}]}^{2} + y^{4}]}^{1 / 2}

= \frac{y^{2}}{y^{2}} (1 - y^{2}) + y^{4}

= 1

(Q) cos x + cos y = - cos z

sin x + sin y = - sin z Square and add

2 + 2 cos (x - y) = 1

\Rightarrow cos (x - y) = - 1 / 2

\Rightarrow 2 {cos}^{2} (\frac{x - y}{2}) - 1 = - 1 / 2, \Rightarrow cos (\frac{x - y}{2}) = 1 / 2

(R) cos 2 x (cos (\frac{π}{4} - x) - cos (\frac{π}{4} + x)) + 2 {sin}^{2} x = 2 sin x cos x

cos 2 x (\sqrt{2} sin x) + 2 {sin}^{2} x = 2 sin x cos x

\sqrt{2} sin x [cos 2 x + \sqrt{2} sin x - \sqrt{2} cos x] = 0

Either sin x = 0 OR

{cos}^{2} x - {sin}^{2} x = \sqrt{2} (cos x - sin x)

sec x = 1 OR cos x = sin x

\Rightarrow sec x = \sqrt{2}

2 OR 4

(S) cot ({sin}^{-1} \sqrt{1 - x^{2}}) = sin ({tan}^{-1} (x \sqrt{6}))

\frac{x}{\sqrt{1 - x^{2}}} = \frac{x \sqrt{6}}{\sqrt{1 + 6 x^{2}}}

\Rightarrow 1 + 6 x^{2} = 6 - 6 x^{2}

\Rightarrow 12 x^{2} = 5

\Rightarrow x = \sqrt{\frac{5}{12}} = = \frac{1}{2} \sqrt{\frac{5}{3}}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From JEE Advanced

Consider the given data with frequency distribution

\begin{array}{l} x_{i} & 3 & 8 & 11 & 10 & 5 & 4 \\ f_{i} & 5 & 2 & 3 & 2 & 4 & 4 \end{array}

Match each entry in List-I to the correct entries in List-II.

	List-I		List-II
$(P)$	The mean of the above data is	$(1)$	$2.5$
$(Q)$	The median of the above data is	$(2)$	$5$
$(R)$	The mean deviation about the mean of the above data is	$(3)$	$6$
$(S)$	The mean deviation about the median of the above data is	$(4)$	$2.7$
		$(5)$	$2.4$

The correct option is

JEE Advanced 2023 Easy

Explore more questions on app